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New Titles  |  FAQ  |  Keep Informed  |  Review Cart  |  Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education

Proceedings of the 1984 Vancouver Conference in Algebraic Geometry
Edited by: J. Carrell, A. V. Geramita, and P. Russell
A co-publication of the AMS and Canadian Mathematical Society.
 SEARCH THIS BOOK:
1986; 503 pp; softcover
Volume: 6
Reprint/Revision History:
reprinted 1990
ISBN-10: 0-8218-6010-0
ISBN-13: 978-0-8218-6010-6
List Price: US$88 Member Price: US$70.40
Order Code: CMSAMS/6

The research and expository papers in this book cover a cross-section of the latest developments in modern algebraic geometry. The volume will be especially beneficial to researchers and to those who want to keep abreast of the latest directions in the field. Aimed at those with an expert background in the field, the topics covered include algebraic groups and representation theory, enumerative geometry, Schubert varieties, rationality, compactifications, and surfaces.

Titles in this series are copublished with the Canadian Mathematical Society. Members of the Canadian Mathematical Society may order at the AMS member price.

• F. A. Bogomolov -- Rationality of the moduli of hyperelliptic curves of arbitrary genus
• S. Kumar -- Cohomology algebra of Schubert varieties associated to Kac-Moody groups
• E. Looijenga -- New compactifications of locally symmetric varieties
• L. E. Renner -- Binary quintics
• S. E. Kleiman and R. Speiser -- Enumerative geometry of cuspidal plane cubics
• F. Bien -- Affine Weyl groups with involutions
• P. Slodowy -- On the geometry of Schubert varieties attached to Kac-Moody Lie algebras
• A. Papantonopoulou -- Embeddings of geometrically ruled surfaces in the four-quartic
• B. Harbourne -- The geometry of rational surfaces and Hilbert functions of points in the plane
• M. Koras and P. Russell -- $$G_m$$-actions on $$A^3$$
• T. Urabe -- Singularities in a certain class of quartic surfaces and sextic curves and Dynkin graphs
• T. A. Springer -- Algebraic groups with involutions
• W. E. Lang -- An analogue of the logarithmic transform in characteristic $$P$$
• R. Smith and R. Varley -- The tangent cone to the discriminant
• S. Zucker -- On the monodromy of algebraic varieties
• J. B. Carrell -- Cohomology of Springer fibres, orbits of the Weyl group, and a theorem of DeConcini and Procesi
• S. L. Kleiman -- Tangency and duality
• M. Szurek -- Binary quintics and the icosahedron
• F. Ronga -- On the number of planes triply tangent to a surface in projective space
• B. H. Dayton and L. G. Roberts -- The seminormalization of an abstract configuration of lines
• L. Kaup and K.-H. Fieseler -- Singular Poincaré duality and intersection homology
• J. Scherk -- $$CR$$ structures on the link of an isolated singular point
• E. Ballico and A. V. Geramita -- The minimal free resolution of the ideal of $$s$$ general points in $$P^3$$
• S. J. Colley -- Lines having specified contact with projective varieties
• G.-M. Greuel -- Torsion free modules and simple curve singularities
• H. Kurke -- A vanishing theorem for vector bundles on the flag variety and moduli spaces of vector bundles with "small" Chern classes


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# Pyramid Problem

### Pyramid Problem

Suppose we have a pyramid with square base(side = x) and height h. The volume of the pyramid is 1000 cm3. Find the value of x if the surface area is minimum.
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### Re: Pyramid Problem

The volume of a pyramid with square base of side length x and height h is $$\frac{x^2h}{3}$$. The surface consists of four triangles and the square base. Obviously the square base has area $$x^2$$. The area of each triangular face is the base length, x, times the altitude of each triangle measured along the perpendicular from the vertex of the pyramid perpendicular to the base. Drop a line from the vertex of the pyramid to the center of the base of the pyramid, then the line from there to the center of a side of the base, then a the line back to the vertex. That describes a right triangle with legs of length h and x/2. The altitude of a triangular side is the hypotenuse of that right triangle so has length $$\sqrt{h^2+ \frac{x^2}{4}}= 2\sqrt{4h^2+ x^2}$$. The area of each such triangle is "1/2 base times height" so is $$\frac{1}{2}x(2\sqrt{4h^2+ x^2}= x\sqrt{4h^2+ x^2}$$. There are 4 such triangles so the total surface area of the pyramid, including all such triangles and the base is $$x^2+ 4x\sqrt{4h^2+ x^2}$$.

So the problem is to find the value of x that, for fixed h, minimizes $$V= x^2+ 4x\sqrt{4h^2+ x^2}$$ subject to the constraint $$A= \frac{x^2h}{3}= 1000$$.
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### Re: Pyramid Problem

A really good solution!

Math Tutor

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# Introduction to Equipotential Surface

Top

When you find a constant potential upon any part of the surface is known as an equipotential surface. We can also define it as the result of the potential difference between any two points is zero. A surface stands for the locus of all points. Each point delivers the same potential; then, it is called equipotential.

Have you ever wondered what is equipotential surface is? In this type of condition, you should put no effort to move a charge from one place to another. An equipotential surface has the benefit that each point has the same potential level.

### Define Equipotential Surface

In other terms, an equipotential surface is a surface that exists with the same electrical potential at each point. If any point lies at the same distance from the other, then the sum of all points will create a distributed space or a volume. Scientists have termed it as equipotential volume.

Do you know the work done on an equipotential surface?

Well, now it’s time to know the truth. Let’s consider that VA and VB both are two point charges in an equipotential surface. We need to calculate the work done in the moving charge. The relation is given below:

W = q0(VA –VB)

However, on an equipotential surface, you will find the difference between two significant points is zero.

That means VA –VB = 0

So, work done = 0

### Properties of Equipotential Surface

Here is the list that can help you to know the equipotential surface properties:

1. The electric field and an equipotential surface both are always perpendicular to each other.

2. The chance of the intersection of two equipotential surfaces is impossible.

3. Equipotential surfaces look like concentric spherical shells for a point charge.

4. When the Equipotential surfaces are under a uniform electric field, they act as normal towards the x-axis.

5. The potential of a hollow charged spherical conductor is constant. This is also named equipotential volume. There will be no work done to move a charge from its centre.

6. The equipotential surface acts as a sphere when the point charge is isolated. This says that concentric spheres that revolve around the point charge possess different equipotential surfaces.

7. The equipotential surface follows the direction from higher potential to lower potential.

8. When any plane is available normal to the uniform electric field acts as an equipotential surface.

9. Gaps that are available between equipotential surfaces help us to determine whether the regions are stronger or weaker.

Mathematically, E= −dV / dr ⇒ E ∝ 1/dr

### Some Problems on Equipotential Surface

Q1. A Charged Particle Possessing a 1.4 mC Charge Travels a Distance of 0.4m on an Equipotential Surface. The Voltage that the Equipotential Surface Carries is 10 V. Find Out the Work Done by the Field During the Motion of the Charged Particle.

Ans: Work done over the electric field in an equipotential surface has a mathematical expression:

W = -q ΔV

As we know, work done will be zero here because, in this case, ΔV = 0.

Therefore, W = 0

Q2. If an Electric Charge Having a Certain Mass and Charge is Free from Equilibrium to Motion. It Interacts with the Uniform Electric Field of 160 N/C. Calculate the Amount of Time Taken by the Electron in Achieving the Speed of 0.1 c. Given: c = the Velocity of Light (c = 3 ×108 meter/sec); Mass of the Electron m = 9.1 × 10–31 kg, e = 1.6 × 10–19 Coulomb.

Ans: The amount of force applied over the electron,

F = E * e

= 106 * (1.6 × 10–19) N

= 1.6 * 10-13 N

We can find the acceleration of the electron by the following method:

a = $\frac{F}{m}$ = $\frac{1.6 \times 10^{-13}}{9.6 \times 10^{-31}}$ = 1.8 * 1017 m/sec2

As mentioned in the question, the initial velocity is zero

Consider the time taken by the electron to achieve a speed of 0.1 c.

So, we can insert the formula

v = u + at

or v = at

or, t = $\frac{v}{a}$ = $\frac{0.1 \times c}{a}$ = $\frac{0.1 \times(3 \times 10^{8})}{1.8 \times 10^{17}}$

### Some Basic Ideas about Equipotential Surface

We can represent the equipotential process with ease. Electric potentials or Voltage is the form of electric fields. Both of the terms, such as Voltage and equipotential surface, are interrelated. They are very significant terms that can help you to learn.

You don’t have to put effort or supply any power to move a charge upon equipotential surfaces, as here you will find the change in Voltage is zero.

W = −ΔPE = −q ΔV = 0

So, all the efforts (work) will be just a waste of time, isn’t it! When You notice that the field in motion is perpendicular with force, the work is also zero.

When the force F acts in the same direction as E (electric field), the motion is always lying perpendicular to E.

The relation is, W = Fd cos θ = qEd cos θ = 0.


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# Finding the sum of squares of roots of a quartic polynomial.

What is the sum of the squares of the roots of $x^4 - 8x^3 + 16x^2 - 11x + 5$ ?

This question is from the 2nd qualifying round of last year's Who Wants to be a Mathematician high school competition which can be seen here:

I know the answer (32) because that is also given in the link, and I have checked by brute force that the given answer is correct.

However, I have made no progress at all in figuring out how to calculate the sum of squares of the roots - either before or after knowing the answer! I was expecting there to be a nice "trick" analagous to the situation if they had given a quadratic and asked the same question -- in that case I know how to get the sum and product of the roots directly from the coefficients, and then a simple bit of algebraic manipulation to arrive at the sum of squares of the roots.

In this case (the quartic) I have no idea how to approach it, and I have not spotted any way to simplify the problem (e.g. I cannot see an obvious factorisation, which might have helped me).

I've looked on the web at various articles which dicuss the relationships between the coefficients of polynomials and their roots and - simply put - I found nothing which gave me inspiration for this puzzle.

Given the audience for this test, it should be susceptible to elementary methods ... I would appreciate any hints and/or solutions!

Thank you.

• $(\sum a_i)^2 = \sum a_i^2 + \sum_\limits{i\neq j} a_ia_j$ – AnotherJohnDoe Sep 17 '18 at 13:20
• This is a wonderful post, a rare essay for a new contributor. Since finding the solution is in this case more important than having it, here is a hint: Use the relations of Vieta for the four roots $a,b,c,d$ - say - of the given equation, which show how to extract the elementary symmetric polynomials, first two being $\sum a =a+b+c+d$ and $\sum ab=ab+ac+ad+bc+bd+cd$. Can you now find $(a+b+c+d)^2$? en.wikipedia.org/wiki/Vieta%27s_formulas – dan_fulea Sep 17 '18 at 13:21
• Thank you! This, and the answers posted soon afterwards were very helpful. – BBO555 Sep 17 '18 at 13:55

We have that

$$(x-a)(x-b)(x-c)(x-d)=$$ $$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$

then by

• $S_1=a+b+c+d$
• $S_2=ab+ac+ad+bc+bd+cd$
• $S_3=abc+abd+acd+bcd$
• $S_4=abcd$

$$a^2+b^2+c^2+d^2=S_1^2-2S_2$$

and more in general by Newton's sums we have that

• $P_1=a+b+c+d=S_1$
• $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2$
• $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3$
• $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4$
• Thanks for this very quick and helpful answer ! – BBO555 Sep 17 '18 at 13:56
• You are welcome! Bye – gimusi Sep 17 '18 at 14:21

Hint:

\begin{align} \sum_{i=1}^4a_i^2 &= \left(\sum_{i=1}^4a_i \right)^2-2\sum_{i< j}a_ia_j \end{align}

Also Vieta's formula might help.

• Yes indeed - thank you! – BBO555 Sep 17 '18 at 13:57

Hint:

$$r_0^2+r_1^2+r_2^2+r_3^2=(r_0+r_1+r_2+r_3)^2-2(r_0r_1+r_0r_2+r_0r_3+r_1r_2+r_1r_3+r_2r_3)$$

and Vieta.

• Thank you! It appears the solution was much closer to hand than I realised at first ... – BBO555 Sep 17 '18 at 13:57
• @BBO555: maths are full of tricks... – Yves Daoust Sep 17 '18 at 15:39


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# Current Rates for Condition Inspections on NON-Builder EAB Aircraft (in the US)

### Help Support Homebuilt Aircraft & Kit Plane Forum:

#### TFF

##### Well-Known Member
Looking in the logbooks is not giving them away. I wouldn’t do one without reading through the logbooks. I have been burned with doctored logs before, that took a lot to fix. The owner had no clue.

If you send your plane to a big shop, the only thing they care about is business. They will hold books for a check, because that’s the agreement of business. That they sign is irrelevant; both sides have made an agreement. Work done for money. Not signature. No money, what is the shop going to do?You still have to pay even if the outcome is not what you want. Your not a fool sending a Cub to someone doing King Airs? Do people have to pretend to be in a militia during their annuals?

Truth from what most A&Ps have learned the hard way is airplane owners pull some of the dirtiest tricks not to pay. Not their fault your hobby requires things to be official. That is why most A&Ps like airlines, only one person screwing you. I would also say 50% of people who have an A&P, quit because owners and airlines are asses, they can make more working on road vehicles without being personally insulted. The A&Ps that stay are the ones who love aviation and put up with it.

#### Kyle Boatright

##### Well-Known Member
Supporting Member
They will hold books for a check, because that’s the agreement of business.

The FAA is very clear that you cannot do that. Yeah, I know, it is done, but it is a great way for a shop to get the wrong kind of attention from the FAA.

If a customer doesn't pay a shop, a mechanic's lien and/or other legal remedies are available.

#### BBerson

##### Light Plane Philosopher
Supporting Member
There is no requirement for logbooks at all for TC'd aircraft. The requirement is records. 91.417 requires the inspection records be kept just 12 months. So a loose leaf binder could be used to discard the inspection records after a year.
Now the EA-B operating limitation 23 requires a condition inspection entry in both the logbook and record. What does that mean?
“23. Condition inspections must be recorded in the aircraft logbook and maintenance records showing the following, or a similarly worded, statement…”

It should say “or“ not and. Or just record to be precise. Who writes this stuff at the FAA?

#### Marc Zeitlin

##### Exalted Grand Poobah
I know they don’t require it. But I also don’t see any evidence that it would be prohibited.
So if you mean that it's not prohibited to provide a list of issues to address, of course it's not prohibited. But as others have pointed out, _IF_ any of the issues prevent the aircraft from being "in a condition for safe operation", then the CI doesn't get signed off. Either the owner, someone else, or I have to fix the issue before I'll sign off the CI. So what _I_ provide to my customers is a list of all issues that don't prevent me from signing the CI, and I provide it in a separate email, NOT in the logbook.

It should be possible for an owner to pay for a single inspection sign off with defects listed, and then do the repairs himself and not return for another inspection. A second inspection puts the repair liability back on the inspector.
See above - if the defects prevent a CI signoff, then no signoff occurs prior to the defect mitigation.

Sometimes, if I know and trust the customer, I'll let them complete the fix, send me some pics or description of what they did, and then send them the signoff for the logbook (which is almost always, but not always, the "maintenance record" as well).

As others have pointed out, there is no such thing as a signoff that says "in a condition for safe operation except for <x>, which you can fix later".

With respect to logbooks/maintenance records, I will not sign off a CI for a customer that doesn't let me review the logbooks - there's no way for me to know WTF is going on with the plane with respect to calendar items or time items without that info. If I don't see logbooks, I make the assumption that EVERYTHING is out of date or past TBO (which is not regulatory, but still ends up with me recommending an awful lot of part replacements).

#### BBerson

##### Light Plane Philosopher
Supporting Member
See above - if the defects prevent a CI signoff, then no signoff occurs prior to the defect mitigation.
No sign off is the direct opposite of 43.11(b) that a mechanic who normally inspects TC'd aircraft only would use to comply with FAR43.11(a)(5).
Where is this contrary guidance for condition inspections found in documents?

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#### BBerson

##### Light Plane Philosopher
Supporting Member
Guidance for maintenance records is in AC 43-9C change 2. https://www.faa.gov/documentLibrary/media/Advisory_Circular/AC_43-9C_CHG_2.pdf

Yes, this is for Standard aircraft. But the operating limitations include parts of 43. So, lacking EA-B maintenance guidance, I think this AC applies unless other guidance is offered.

start with this, page 2 par 5.
5. MAINTENANCE RECORD REQUIREMENTS.
a. Responsibilities. Part 91, § 91.417 states that an aircraft owner/operator shall keep and maintain aircraft maintenance records. However, part 43, § 43.9 states that each person who maintains, performs preventive maintenance, rebuilds, or alters an aircraft, airframe, aircraft engine, propeller, appliance, or component part shall make an entry in the maintenance record of that equipment. Section 43.11 states that the person approving or disapproving for return to service an aircraft, airframe, aircraft engine, propeller, appliance, or component part after any inspection performed in accordance with part 91, 125, 135, § 135.411(a)(1), or § 135.419 shall make an entry in the maintenance record of that equipment. The persons that are authorized to perform maintenance can be found in §§ 43.3 and 43.7.

#### Marc Zeitlin

##### Exalted Grand Poobah
No sign off is the direct opposite of 43.11(b) that a mechanic who normally inspects TC'd aircraft only would use to comply with FAR43.11(a)(5).
Where is this contrary guidance for condition inspections found in documents?
There is none. The OL's are the only guidance for CI's on E-AB aircraft. After that, it's up to the person with the RC or the A&P to decide how to deal with it - there's no such thing as an official "disapproval". There is no requirement to do anything if the person decides not to sign off the CI because of a discrepancy - there's just no signoff in the logbook, until someone decides that the plane is "in a condition for safe operation", and that's a fairly subjective decision.

While you list an AC and references to Part 43, AC's are not regulatory and any reference to Part 43 doesn't apply, because (as we discussed) Part 43 doesn't apply to Experimental aircraft. Now, the OL's will usually state, as you saw in paragraph 23 of the website you referenced earlier, that the wording for the CI signoff should be something like:

“I certify that this aircraft has been inspected on [insert date] in accordance with the scope and detail of FAR 43 Appendix D, and was found to be in a condition for safe operation.”

This implies that Part 43 Appendix D should be referenced for determining whether the aircraft is "in a condition for safe operation", but if you read Appendix D, it's pretty **** vague - about the only thing that's required is a <relatively useless, if you read Mike Busch> compression check - everything else is just "look at this thing and make sure it's OK", for the most part.

So if it doesn't say it in the OL's and if it isn't in 14 CFR Part 91.417 with respect to maintenance, it doesn't apply to E-AB aircraft. The RC holder or the A&P has a LOT of discretion here. And most of the time, that's a good thing.

#### crusty old aviator

##### Well-Known Member
Supporting Member
Like Dana’s A&P/IA, I travel as needed, and I do only owner-assisted inspections. If the owner isn’t interested in his/her aircraft, why should I? I charge $50/hr for the inspection and$25/hr for travel. The FAA doesn’t like traveling IA’s because the “industry” protects those with big shops and high overheads; but for EABs, they don’t get involved unless there’s an issue.
I‘ve been involved with EAA and homebuilts constructed of all materials for four and a half decades and haven’t found an “airworthy” one yet! (airworthy is a legal term, but you’d be surprised how many IAs sign off EAB aircraft as “airworthy”)
I insist on seeing the logs and the documents required to be carried in the aircraft before performing the inspection. I don’t take them with me: they belong with the aircraft. If the owner ”forgets“ to bring them, I don’t perform the inspection, but still charge for my travel time. I’ve had a few owners not pay me: it’s rare, and they were ostracized off their airports.
My overhead is low, so I don’t have to charge much. Most of my customers wouldn’t be able to afford to keep their aircraft if they had to rely on a big box maintenance facility, and that’s what keeps me going. The Book says we’re placed here to serve others...
Finding a mechanic like me isn’t easy, so I’m told, which is a shame, but most A&Ps don‘t want anything to do with homebuilts: EABs are outside their comfort zones, mostly due to their own ignorance.
Good luck figuring out what you’ll be flying!

#### BBerson

##### Light Plane Philosopher
Supporting Member
So if it doesn't say it in the OL's and if it isn't in 14 CFR Part 91.417 with respect to maintenance, it doesn't apply to E-AB aircraft. The RC holder or the A&P has a LOT of discretion here. And most of the time, that's a good thing
It is in 14CFR 91.417 where it says records of inspections shall be kept for the period specified. It doesn’t say only inspections found airworthy, it just says inspections. The AC fills in the details for clarification.
The incomplete operating limitations could be clarified by the FAA, I suppose.

Reading 91.417 carefully again, I see the owner shall keep records of inspections.
If I contact a mechanic for inspections, I will require a record of that inspection to comply. Or seek another.
I don’t expect many mechanics will understand my request. It may take some instruction.

#### BBerson

##### Light Plane Philosopher
Supporting Member
Finding a mechanic like me isn’t easy, so I’m told, which is a shame, but most A&Ps don‘t want anything to do with homebuilts: EABs are outside their comfort zones, mostly due to their own ignorance.
Good luck figuring out what you’ll be flying!
Yes, like you, for a short time I was a traveling mechanic that only did owner assisted annuals (never any condition inspection requested) after selling my higher overhead shop. It was difficult to find owners with so many other inspectors in the state, so I gave it up and made far more building houses with less liability.

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#### Marc Zeitlin

##### Exalted Grand Poobah
It is in 14CFR 91.417 where it says records of inspections shall be kept for the period specified. It doesn’t say only inspections found airworthy, it just says inspections. The AC fills in the details for clarification.
Agreed - 91.417 states that all maintenance and inspection shall be kept for either 1 year, when they're superseded, or forever, as described.

A couple of points:

First, as @crusty old aviator pointed out, an E-AB aircraft CANNOT be "airworthy", so lets not use that term - there is no TC, and to be "airworthy", an aircraft must be in compliance with its TC. Since there is none, "airworthy" cannot be achieved - just the 2nd part of the TC'd aircraft's annual inspection which is to be found "in a condition for safe operations", which is what the OL's for E-AB aircraft require.

Second, the only required (or available, for that matter) "inspection" of an E-AB aircraft is the annual "Condition Inspection". There are no 100 hour inspections, 50 hour inspections, Maintenance Program inspections, or anything else. So the only "inspection" that needs to be logged is the CI, and the OL's point that out and state what it needs to say.

And again, the AC43.9C is not mandatory nor regulatory, as it says in its first paragraph.

Reading 91.417 carefully again, I see the owner shall keep records of inspections.
If I contact a mechanic for inspections, I will require a record of that inspection to comply. Or seek another.
I don’t expect many mechanics will understand my request. It may take some instruction.
So the "record" of the inspection is just the signoff in the logbook. The "record" does not, and does not have to, list all and every thing that I saw during the CI - just the signoff, if I did so. If I didn't sign it off because I found something to prevent it, then as stated before, it'll get fixed and then I'll sign it off.

I don't mind having this discussion, but I have to say that I'm having a hard time figuring out what point you're trying to make or to what understanding you're trying to come. What's bothering you about all this? The relative undefined nature of what a CI is? The leeway that the RC holder/A&P has? I might be able to help more if I had a better understanding about where you're trying to go...

#### BBerson

##### Light Plane Philosopher
Supporting Member
I don't mind having this discussion, but I have to say that I'm having a hard time figuring out what point you're trying to make or to what understanding you're trying to come. What's bothering you about all this? The relative undefined nature of what a CI is? The leeway that the RC holder/A&P has? I might be able to help more if I had a better understanding about where you're trying to go...
Glad to know you don’t mind having the discussion. I think you have extensive experience with condition inspections. I don’t know if you do annuals.
I have not yet done a condition inspection, only annual inspections. So I may do a condition inspection in the future and want to learn more about condition inspection procedure from both an owner and a mechanics perspective. Which is the topic asked in post one.

Here is a sample discrepancy list from an annual inspection I would use as template, but I would change “unairworthy“ to something else, perhaps, “found the following discrepancies“. Since I can’t think of a single word like “uncondition”.

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#### Dana

##### Super Moderator
Staff member
The difference is that when an IA signs off an annual with a list of discrepancies, the annual is "good" once the discrepancies have been addressed, and the IA doesn't have to see it again. The required condition inspection for experimentals doesn't allow that; at the conclusion of the inspection the aircraft is either "in a condition for safe operation" and it gets signed off as such or it isn't. If there are discrepancies the person doing the condition inspection can only sign off the CI after the discrepancies are fixed.

#### Toobuilder

##### Well-Known Member
Supporting Member
And to further Dana's point, in the TC world, a "discrepancy" is defined by engineering documentation. The A&P/IA has very little latitude to decide if the condition is "safe" or not... And that's not their job anyway. Their job is to inspect the airplane and determine if the airplane CONFORMS to the TC and other engineering data. That's what an "Annual Inspection" is - verifying conformance to engineering data.

The E-AB world has little to zero "engineering data" to conform to, so the determination of "safe" is up to the inspector. One A&P may decide one condition is a deathtrap, and the next guy might be just fine with it. Who is right? Could be both or none... The only thing that matters is the ink in the book and the character of the inspector who penned it.

#### BBerson

##### Light Plane Philosopher
Supporting Member
Several commenters have said that Dana, please show the FAA document that says that, other than the operating limitation which doesn’t say that and repeating opinions.
The operating limitation is worded suspiciously similar to the airworthy sign off. Even to the point of the phrase “the following or similar worded statement”.
When I was doing owner assisted annuals I changed the wording of the sign off to suit owner/maintainers.
Where it says ”list of discrepancies and unairworthy items” I deleted unairworthy because owners don’t like that word in the log. The FAA approved it, because discrepancies and unairworthy items is redundant.

An IA is required to follow specific procedures and entries if an aircraft is not approved for return to service.
I won’t accept that a “condition inspection“ procedure would allow inspectors to neglect discrepancy documention.

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#### Turd Ferguson

##### Well-Known Member
Supporting Member
The term "airworthy" for FAA purposes is two-part. It's when "an aircraft or one of its component parts meets its type design (engineering standards) and is in a condition for safe operation.” (mostly non specific).

The "condition for safe operation" part is quite often a judgement call on the part of the inspector. For example, you won't find any criteria for how much slack in control hinges is acceptable on most light planes. How much brake lining must remain, or how much corrosion is acceptable. Those are all judgement calls. No measurements where once can show on a calibrated scale why the amount of wear and tear present is unacceptable. It's just tacit knowledge of the situation and the mechanic declaring "I'm not okay with that"
That is kinda one of the reasons why the FAA wants a person with at least an A&P doing condition inspections on homebuilts. They are trained on what condition for safe operation means even where there is no specific criteria.

#### BBerson

##### Light Plane Philosopher
Supporting Member
I did a search of Code of Federal Regulations for ”condition inspection”. Nothing listed.

#### Marc Zeitlin

##### Exalted Grand Poobah
I think you have extensive experience with condition inspections. I don’t know if you do annuals.
Well, as of today, I've done about 171 (about) CI's, and zero "annuals". I don't work on TC'd aircraft - only E-AB and E-R&D.

I have not yet done a condition inspection, only annual inspections. So I may do a condition inspection in the future and want to learn more about condition inspection procedure from both an owner and a mechanics perspective. Which is the topic asked in post one.
Gotcha.

So @Dana has stated that:

The difference is that when an IA signs off an annual with a list of discrepancies, the annual is "good" once the discrepancies have been addressed, and the IA doesn't have to see it again. The required condition inspection for experimentals doesn't allow that; at the conclusion of the inspection the aircraft is either "in a condition for safe operation" and it gets signed off as such or it isn't. If there are discrepancies the person doing the condition inspection can only sign off the CI after the discrepancies are fixed.
And you've asked for FAA references for the inability of the A&P to do that. I assume, but don't know, that if discrepancies are listed as being required to be fixed prior to the Annual being in effect, that someone legally allowed to perform the fixing must sign the logbooks prior to the next flight, saying that the discrepancies have been fixed. That person might be an IA, an A&P, or the owner, if the discrepancy is of a low enough level that it's legal for the owner to do the work.

My position is that whether there is FAA documentation (which there isn't) that addresses this specific question or not, it's a logical impossibility to state, in the logbook, that the aircraft "is in a condition for safe operation", if there's any discrepancy/issue that prevent that statement from being true. I suppose that in theory, an A&P could provide a signoff in the logs as well as a list of things to fix prior to the next flight, and that the plane wouldn't be legal unless the discrepancies were signed off. But since it's legal for your grandmother's dog to work on E-AB aircraft (although an A&P is required for E-R&D), it's unlikely that any A&P (me included) would be willing to leave themselves open to that sort of situation - not requiring proof of remediation prior to providing the signoff.

As I said earlier, if there are safety discrepancies, I just won't provide a signoff until their fixed, and I don't know why anyone would. The fix can be demonstrated in a number of ways or performed by any number of folks (whom I'd need to trust). @Toobuilder 's comments about latitude and subjectivity is spot on. I've seen signoffs of things that I assumed would have killed the pilot almost immediately, and I've seen A&P's reject stuff that I know to be perfectly safe and acceptable over a million hours of canard composite flight time.

I did a search of Code of Federal Regulations for ”condition inspection”. Nothing listed.
That's because the only place it's referenced is in the aircraft's Operating Limitations. Read Paragraph 23 again, or look through Order 8130.2J for the relevant OL paragraphs for E-AB aircraft.

If you're asked to perform a CI and agree to do so (and if you've been an A&P for a long time and know the build technology of the aircraft you're being asked to inspect, there's exactly zero reason for you not to do it), you'll have all the rope you want to hang yourself and your customer . YOU will get to decide whether you like something or not; whether you want to make the owner fix a thing or not; and whether you're willing to let someone else fix a thing either before or after you put your signature in their logbook or not. Now, hopefully, you'll review as much information and documentation about the plane in question as is available so that you're going in with SOME amount of education and what to look for. Rattling around in my head is pretty much every modification and fix that's been done to canard composite aircraft; why, and whether or not they're "mandatory". I DON'T have that info for RV's, but all the RV plans changes are on-line, so I review them all and include them all in my inspection checklist, so I know what I'm looking for. I'd hope that all A&P's do the same...

I don't think there's as much definition of what to do and how, and what not to do and how, as you're expecting or looking for . Maybe that's a good thing, and maybe that's a bad thing, but I think that's what we're all saying here.

Hope this helps.

#### BBerson

##### Light Plane Philosopher
Supporting Member
Thanks Marc, it comes down to I don’t want to ever end up in court where the question put to me is: “did you in fact sign that the aircraft was in a condition for safe operation?” That's all the jury will hear.
I likely won’t participate in assuming such risk.

#### Toobuilder

##### Well-Known Member
Supporting Member
Well, the fact is that the "condition for safe operation" only lasts as long as it takes the ink to dry. The reason is that unlike TC'd aircraft, ANYBODY can perform work on the E-AB during the next 12 months, trained or not. And there is no requirement to document that work. The owner can get your CI signature on monday, and on Tuesday drill a bunch of holes in the spar web and then pull the wings off in flight on Wednesday... It WAS safe when you saw it and yours is the last entry in the book, but that is not evidence that the airplane was not altered after the fact.


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### el_heffeh's blog

By el_heffeh, history, 5 weeks ago, ,

Hello,

Recently I had just participated in Codeforces Round #654 (Div. 2). I submitted a solution for D at the last second (1:59:59) but it said "Contest is over" because the submit button has a 1-2 second delay. @MikeMirzayanov can you please add some functionality to have a small 1-2 second grace period after the contest is over for people who submit at the last second? I know I still have to get better but for some I think this would be very beneficial.

Thanks.

• +19

 » 5 weeks ago, # | ← Rev. 3 →   +21 I don't think mike will add this feature.Contests are like exams.The duration is $2$ hours, $120$ minutes, $7200$ seconds, exactly, for everyone.We cannot give you one more second to submit your solution, or it'll be unfair to others.Every second during the contest is the same valuable, even for the first second, as well for the last second.
•  » » 5 weeks ago, # ^ |   -36 During an exam, when teacher starts collecting the answer sheets, students sitting in the last row get a minute or two extra... So, analogically, why not here !? :)
•  » » 5 weeks ago, # ^ |   +4 Sorry, I didn't just mean this for me. I also meant for everyone, because if submissions are locked pretty much after three seconds left, then the contest (technically) is 1 hour, 59 minutes, and 57 seconds instead of 2 hours. So I think that it would be a good idea to — Have the submit button be instant without delay or — Allow submissions that pressed submit at the last second. Let me know what you think.
•  » » » 5 weeks ago, # ^ |   0 I don't think it's a bad idea, but it could be technically hard to implement.Once ago I was frustrated by the delay too(although that solution wasn't correct), but the contests are fair, so the only option is to get experience from it.In my opinion people should be aware of the time left, submit the solutions earlier but not at the very last second.Actually you can get your under-pressure performance trained, then that pity-mistake would never be made again.
•  » » » 5 weeks ago, # ^ |   0 I was also saying for everyone and not only you but people misunderstood my point.
•  » » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 I'm sorry if it looks confusing but I didn't reply to your comment. I was replying to dapingguo8's first comment.
 » 5 weeks ago, # |   +20 No but then there will be some users who gonna say can't we get 2-3 more seconds ... because no matter there will be some end and at that there will be someone standing there and asking Oh what if I got those extra seconds ...
 » 5 weeks ago, # |   +13 In reality though, most probably the page loads for 2-3 seconds in your case, which causes the contest timer in the user interface display the wrong time. So, the contest lasts for 2 hours exactly, but when the contest timer shows 00:00:10 left, it might actually be 00:00:07 or so because the page didn't load instantly.
 » 5 weeks ago, # |   +10 The problem isn't a delay in codeforces. Is your internet. It takes some time to your code reach the server, wich I assume it is in Russia. So the delay may be bigger to you and smaller to others. By the time your code reaches the server, the contest may already be finished. There is nothing Mike can do.
•  » » 5 weeks ago, # ^ |   0 Hmm... I guess that's true. I probably just have to be faster and better in the future...


text:
# QUESTION 5 All of the following reactions concerning alkali metals are correct except 2Na (s) +...

###### Question:

QUESTION 5 All of the following reactions concerning alkali metals are correct except 2Na (s) + 2H2O (I) 2NaOH (aq) + H2 (g) 2NaH (s) 2Na (s)+ H2 (g) 2Na (s) + Cl2 (g) 2NaCl (s) Na (s) + O2 (g)NaO2 (s) 2Na (s)+ S (s)Na2S (s)

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# Does rotational weight matter on a bike

My last post sparked some interesting discussion among some of my cycling friends: Not all weight is equal. The consensus was that weight on the wheels was more important (2x-7x more important) than weight of the frame.  I'm having a hard time buying this... but wanted to clearly write out the arguments here.

### Acceleration and Rotational Inertia

Work is the change in kinetic energy ($W=\Delta K$). For a mass on your frame or wheels, there is work done when you accelerate from rest to your riding velocity (we'll say here 15 mph). We'll stick with the 10 grams from the previous post, and assume our riding velocity is around 15 mph.  That means for both wheels and frame, the work done to accelerate 10 grams to 15 mph is given by:

$W=\frac{1}{2}mv^2$

This ends up being around $5.4\times 10^{-5}$ Calories. The wheels have an additional component of work being done... they also have rotational energy:

$k_{rot}=\frac{1}{2}I*\omega^2$.

here, we'll take the radius of the wheel to be 350 mm*. This gives us a rotational velocity $\omega=\frac{v}{2\pi r}\approx 3 \frac{rev}{sec}$.  The rotational inertia is given by $I=mr^2$.  If we plug in both $I$ and $\omega$, the radius ends up canceling out and we're left with:

$k_{rot}=\frac{1}{2}m*(\frac{v}{2\pi})^2$.

so the work done to bring that 10 gram wheel spinning up to 15 mph is an additional $1.4\times 10^{-6}$ Calories! This means it takes about 2.5% more energy to accelerate a mass on your wheel to 15 mph than it does on your frame.

1. Wheel Radius doesn't matter. 29ers get a bad wrap for being slow accelerators. It's clear from a physics argument this acceleration penalty isn't due to the larger diameter of the wheels.  While the 29" wheels do have higher rotational inertia, they also spin slower and the effect cancels out. (I'd buy an argument that maybe 29" wheels are less stiff than 26" wheels; or, by default, 29" wheels are heavier than 26" wheels...)
2. How often do you accelerate? While acceleration might be important if you're racing matched doubles on a track, trying to break away from a peloton, or win a sprint finish... it really doesn't seem that important for the casual rider. This is especially true on a road bike ride, where you're riding a long distance at a relatively constant velocity...

### Turning and Stability

The other time I can see wheel mass to matter more than frame mass is in cornering. Bicycle dynamics and stability are a pretty complicated matter, but we can think about a few things:

1. Conservation of Angular Momentum When cornering, you want to lean your bike into the turn. You do this for two reasons. First, if you lean a spinning wheel over, it will cause rotation to conserve angular momentum. Second, turning the
2. Work done by steering the work done when turning a wheel is given by
$W=\tau \Delta \theta$

where turning our wheel is given by the change in angular momentum (L):

$\tau=\frac{dL}{dt}$

and $L=Iw$

*700c rims are normally 622mm in diameter. the actual diameter of the wheel depends on the tire size. 350 mm seems like a good round guess.


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# Proper Nouns

Please note : In Pali, the Proper nouns (the names of persons, animals, places and things) also decline. Thus not only all the ‘a’-ending masculine common nouns but also all the ‘a’-ending masculine proper nouns decline like the noun ‘buddha’. Some examples of ‘a’ ending masculine proper nouns are :

• Sāriputta,
• Ānanda,
• Rāhula,
• Anāthapiṇḍika,
• Kanthaka (the horse of bodhisatta Prince Siddhattha)
• Kammāsadhamma (the place where the Buddha gave Mahāsatipaṭṭhāna discourse),
• Gijjhakūṭa (name of a mountain, the vulture’s peak)
Audio for pronunciation of Masculine ‘a’ ending Proper Nouns

In the next Unit, we will learn the eight cases and their role in Pali sentences.

For an overview of usage of cases and alternate way of studying declension tables (with the help of terminations) please refer to [Declension of nouns] in the menu at this link : https://learning.pariyatti.org/mod/page/view.php?id=1874.

Memorisation helps a great deal in learning Pali and translating the suttas. Hence, the students are encouraged to memorise at least a few declension tables. The declension of ‘a’-ending masculine noun can be taken as a basic format. If we memorise this table, declension of other masculine and neuter nouns can be understood quite easily by comparison with this basic format.
Last modified: Tuesday, 10 January 2023, 11:02 AM


text:
## Duke Mathematical Journal

### On the proper moduli spaces of smoothable Kähler–Einstein Fano varieties

#### Abstract

In this paper we investigate the geometry of the orbit space of the closure of the subscheme parameterizing smooth Kähler–Einstein Fano manifolds inside an appropriate Hilbert scheme. In particular, we prove that being K-semistable is a Zariski-open condition, and we establish the uniqueness of the Gromov–Hausdorff limit for a punctured flat family of Kähler–Einstein Fano manifolds. Based on these, we construct a proper scheme parameterizing the S-equivalent classes of $\mathbb{Q}$-Gorenstein smoothable, K-semistable $\mathbb{Q}$-Fano varieties, and we verify various necessary properties to guarantee that it is a good moduli space.

#### Article information

Source
Duke Math. J., Volume 168, Number 8 (2019), 1387-1459.

Dates
Revised: 26 November 2018
First available in Project Euclid: 3 May 2019

https://projecteuclid.org/euclid.dmj/1556848995

Digital Object Identifier
doi:10.1215/00127094-2018-0069

Mathematical Reviews number (MathSciNet)
MR3959862

Zentralblatt MATH identifier
07080115

#### Citation

Li, Chi; Wang, Xiaowei; Xu, Chenyang. On the proper moduli spaces of smoothable Kähler–Einstein Fano varieties. Duke Math. J. 168 (2019), no. 8, 1387--1459. doi:10.1215/00127094-2018-0069. https://projecteuclid.org/euclid.dmj/1556848995

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• [53] G. Tian, K-stability and Kähler-Einstein metrics, Comm. Pure Appl. Math. 68 (2015), no. 7, 1085–1156.
• [54] E. Viehweg, Quasi-Projective Moduli for Polarized Manifolds, Ergeb. Math. Grenzgeb. (3) 30, Springer, Berlin, 1995.
• [55] X. Wang and C. Xu, Nonexistence of asymptotic GIT compactification, Duke Math. J. 163 (2014), no. 12, 2217–2241.
• [56] S.-T. Yau, On the Ricci curvature of a compact Kähler manifold and the complex Monge-Ampère equation, I, Comm. Pure Appl. Math. 31 (1978), no. 3, 339–441.


text:
# Right inverses of L\'evy processes and stationary stopped local times

August, 1999
Report Number:
565
Authors:
Steven N. Evans
Citation:
Electronic Journal of Probability</em>, Vol. 5 (2000) Paper no. 7, pages 1-17
Abstract:

If \$X\$ is a L\'evy process on the line, then there exists a non--decreasing, c\`adl\`ag process \$H\$ such that \$X(H(x)) = x\$ for all \$x \ge 0\$ if and only if \$X\$ is recurrent and has a non--trivial Gaussian component. The minimal such \$H\$ is a subordinator \$K\$. The law of \$K\$ is identified and shown to be the same as that of a multiple of the inverse local time at \$0\$ of \$X\$. When \$X\$ is Brownian motion, \$K\$ is just the usual ladder times process and this result extends the classical result of L\'evy that the maximum process has the same law as the local time at \$0\$. Write \$G_t\$ for last point in the range of \$K\$ prior to \$t\$. In a parallel with classical fluctuation theory, the process \$Z := (X_t - X_{G_t})_{t \ge 0}\$ is Markov with local time at \$0\$ given by \$(X_{G_t})_{t \ge 0}\$. The transition kernel and excursion measure of \$Z\$ are identified. A similar programme is carried out for L\'evy processes on the circle. This leads to the construction of a stopping time such that the stopped local times constitute a stationary process indexed by the circle.

PDF File:
Postscript File:


text:
# Is this wave spatially coherent?

Is the following wave (the solid lines representing the intensity maxima) spatially coherent?

I've been seeing different, seemingly contradictory definitions of spatial coherence.

Some places define seem to be defining spatially incoherent light as light that has "spatially varying temporal coherence". I.e. the Wavefront (defined as the surface of constant phase) shape changes from Wavefront to Wavefront. From that definition, this would be spatially coherent because while the Wavefront is distorted, the Wavefront doesn't change over time.

Other places seem to define it as measuring phase correlation perpendicular to the direction of propagation of the wave. In this case I wouldn't consider this example wave to be spatially coherent because if you move horizontally along the wave, the phase changes rapidly. (I could have picked a more noisy variation in this direction, but I picked a sine wave for simplicity.)

• just to double check, in the image the lines are intensity (as you say) or field maxima? – scrx2 Dec 18 '15 at 15:18
• That should be essentially the same thing, no? – aquirdturtle Dec 19 '15 at 21:11
• No, the principle is different. Imagine for simplicity the red lines are straight. If you show intensity maxima you have basically some interference fringes. If you show the field, the intensity is constant. – scrx2 Dec 20 '15 at 13:29
• If you look at the instantaneous field (not time-averaged) of a plane wave, which is what I'm suggesting, the intensity is most certainly NOT constant, but varies just as the field varies with maxima at both the maxima and minima (negative maxima) of the field. So the only difference is the wavelength of the light, which is irrelevant for my question. – aquirdturtle Dec 21 '15 at 1:46
• Ok, it's a matter of definition, I'm used to think the intensity as the average over a period of the modulus squared of the field. This definition is based on experience: if you place a detector in your wave, it measures a constant signal, because no detector is able to see variations of "intensity" (in your definition) in the order of $10^{16}$ Hz. – scrx2 Dec 21 '15 at 11:00

I assume in your figure the red lines represent the phase profile of the wave (the points where the phase is constant), and not the intensity. And that the wave propagates from top to bottom.

Is the following wave spatially coherent? Yes.

If those lines were straight, the wave would be a plane wave, perfectly coherent in time and space. I consider then your plot as a propagating wave with a periodically undulated phase profile. Again, it's perfectly coherent in time and space.

The definition of coherence which I consider accepted says that a field is coherent where (when) there is a fixed phase relation between the electric field in different locations (times).

Mathematically this is described by the cross correlation of the field at two points in space or time.
Experimentally, we measure the intensity $|E|^2$, not the field $E$, so coherence is tested using interference as follows:

Spatial coherence: you measure the intensity fringe contrast (max-min) in the double slit experiment, as a function of the distance between the slits. The maximum distance between the slits at which there is still some interference is the spatial coherence length of the wavefront (the contrast depends on the phase correlation between two different points in space).

Temporal coherence: in an interferometer with 2 different arm lengths (e.g. Michelson) each point, or beam, of the wavefront interferes with a delayed copy of itself. The maximum delay (difference between the lengths of the 2 arms) at which there is still some interference is the temporal coherence length of the field.

Applying the above 2 criteria to your wave, you can see that it's both time and space coherent. This will hold true if you distort the red lines even more (you can make them random), IF they remain one identical to the others and equispaced (like the figure below, left). To decrease the temporal coherence, you can make the distance between the lines, i.e. the wavelength, more randomly changing (fig. below, middle). In this way, the phase difference $\Delta \phi$ between the field $E(x_o,t)=P1$ and the field $E(x_o,t+\tau)=P'1$ is not constant anymore, and the spectrum is not a delta but a broader peak. To decrease the spatial coherence, while keeping the temporal one, you can change the wavelength across the wavefront (fig. below right).

Probably a useful way to see if a wave is coherent is to consider the local wavelength $\lambda(x,t)$, and asking if it constant in $t$ (vertical direction in the figure) and / or in $x$ (horizontal direction in the figure), etc.

• I like this answer & definition, but could you more specifically address the other possible definition of spatial coherence? This definition also seems like a natural definition of spatial coherence, and I'd like to better understand why it's not useful / used. – aquirdturtle Mar 29 '16 at 17:18

What you have is a realization of a random process. Calling a realization coherent or not-coherent is a mistake, indeed I believe this mistake was made once on Wikipedia.

The traditional definition of spatial coherence asks how related the values at one detector are to the values at another detector. We then assume homogeneity which means that the relation doesn't change spatially and can write down the typical length of this relation which is often converted into an area taking the value as a radius.

To answer your question, spatial coherence can be measured by two probes along the wave-front but without seeing more realizations of the wave we really can't say anything about its coherence!

Ps. If you are using one probe and looking at different times that would be temporal coherence, which sounds like what you imagine for the perpendicular case.

• What exactly are you calling a random process? My wave? The Measurement? I'm pretty sure that it's fair to consider the coherence or non-coherence of the given wave. – aquirdturtle Dec 14 '15 at 7:19
• The 'field' is intrinsically random and you drew one possible realization. Notice that the squiggles you drew could be due to a 100% coherent source that scattered off some kind of grid. As you noted, its hard to understand the coherence of a bunch of squiggles (because thats not conceptually correct). – Mikhail Dec 14 '15 at 7:44


text:
# 4-vectors and tensor algebra

1. Jul 14, 2009

### smithg86

How would you write the following using index-notation?

$$s \cdot F \cdot u$$

Given that $$s = (s^0, \vec{s})$$ and $$u = \gamma(1, \vec{v})$$ are 4-vectors;
$$F$$ is a skew tensor of components $$F^{ik}$$ indicated as $$F = - ( \vec{F}^t, \vec{F}^s )$$, $$\vec{F}^t= \left\{F^{0 \alpha}\right\}$$, $$\vec{F}^s = \left\{F^{\beta \gamma}\right\}$$, with $$\alpha, \beta, \gamma = 1,2,3$$;
the dot between symbols indicates a contraction of neighboring indices with the metric tensor (+ ---).


text:
# 2d wave equation with finite differences blowing up [duplicate]

I am (naively) trying to solve the 2d wave equation with finite differences. But the system blows up instantly.

For simplicity I set the constant $$c=1$$, then I am left with $$\Delta u =u_{tt}.$$

I set up the FDM with

N = 9
u = np.zeros((N**2,1))
u.reshape(N,N)[N//2,N//2] = 1

Now, at the time t=0 the function looks like this

u_0 =
[[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]]

To which I apply the five point diffence stencil Lh

import scipy.sparse as sp
import numpy as np
h = 1/(N+1)
T = np.diag(N*[-2]) + np.diag((N-1)*[1],k=-1) +np.diag((N-1)*[1],k=1)
spt = sp.coo_matrix(T)
spi = sp.eye(N)
Lh = (sp.kron(spi,spt) + sp.kron(spt,spi) ) / h**2

Since Lh is my discretized Laplacian, for every timestep I add the acceleration that it gives me to some book-keeping vector v to change the velocity accordingly:

v = np.zeros_like(u)
print(u.reshape(N,N).astype(int))
for i in range(3):
v += Lh.dot(u)
u += v
print(u.reshape(N,N).astype(int))

Here is what happens:

[[   0    0    0    0    0    0    0    0    0]
[   0    0    0    0    0    0    0    0    0]
[   0    0    0    0    0    0    0    0    0]
[   0    0    0    0   99    0    0    0    0]
[   0    0    0   99 -398   99    0    0    0]
[   0    0    0    0   99    0    0    0    0]
[   0    0    0    0    0    0    0    0    0]
[   0    0    0    0    0    0    0    0    0]
[   0    0    0    0    0    0    0    0    0]]

In the next timestep:

[[     0      0      0      0      0      0      0      0      0]
[     0      0      0      0      0      0      0      0      0]
[     0      0      0      0   9999      0      0      0      0]
[     0      0      0  19999 -79699  19999      0      0      0]
[     0      0   9999 -79699 198800 -79699   9999      0      0]
[     0      0      0  19999 -79699  19999      0      0      0]
[     0      0      0      0   9999      0      0      0      0]
[     0      0      0      0      0      0      0      0      0]
[     0      0      0      0      0      0      0      0      0]]

What is the problem here? I have not set up a boundary condition to my solution. Maybe I should?

greetings

• Have you tried with smaller time step size dt? Then, the code reads Lh.dot(u)*dt and v*dt. – Hui Zhang Dec 2 '18 at 14:07
• The wave speed is 1. You should use 1, but better at least 2 time steps for any space step, thus $dt < dx/2$. – Lutz Lehmann Dec 2 '18 at 16:15
• thank you,Hui Zhang. @LutzL why is that? – dba Dec 3 '18 at 12:23
• To be at least qualitatively close the wave has to be able to travel at speed $c$. With the Euler method that you employ, changes propagate at speed $dx/dt$. Thus you need $dx>c\,dt$, and the larger the gap the better the quantitative accuracy. With implicit methods the picture is not as clear, as the inverse of a sparse matrix is usually not sparse, and you would have to look at the eigenvalues of the propagation matrix. – Lutz Lehmann Dec 3 '18 at 13:07


text:
# JEE-2015

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $$d_1$$ and $$d_2$$ are filled in the tube. Each liquid subtends $$90^\circ$$ angle at centre. Radius joining their interface makes an angle $$\alpha$$ with vertical. Ratio $$\frac {d_1}{d_2 }$$ is:

×


text:
CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   FLUENT (http://www.cfd-online.com/Forums/fluent/)
-   -   How to transform (mass flow rate) boundary condition from 3D to 2D? (http://www.cfd-online.com/Forums/fluent/108251-how-transform-mass-flow-rate-boundary-condition-3d-2d.html)

 nurul efha October 18, 2012 03:26

How to transform (mass flow rate) boundary condition from 3D to 2D?

Hello everybody,

I'm using CFD-Fluent to simulate gasification. In my case, mass flow rate that enter gasifier column is 0,0074 kg/s of coal. I want to simulate with 2D geometry. Only half of the domain with is modeled because of symmetry. Is it same the number of mass flow rate in 3D and 2D? Is the mass flow rate used in 2D unchanged (0,0074 kg/s)? Or the mass flow rate must be devided to transform in 2D? I hope you can help me. Thanks.

Hi.
Are you using 2D Ax-symmetric? If yes, mass flow should be divided by (2*PI). You are just modeling 1 rad of the whole circle.

 misagh October 21, 2012 17:59

Quote:
 Originally Posted by nurul efha (Post 387244) Hello everybody, I'm using CFD-Fluent to simulate gasification. In my case, mass flow rate that enter gasifier column is 0,0074 kg/s of coal. I want to simulate with 2D geometry. Only half of the domain with is modeled because of symmetry. Is it same the number of mass flow rate in 3D and 2D? Is the mass flow rate used in 2D unchanged (0,0074 kg/s)? Or the mass flow rate must be devided to transform in 2D? I hope you can help me. Thanks.
HI
You had better use intensive property for your simulation and further calculation.it is an easy way to get rid of changing units in 2.d or 3.d . i mean instead of using mass flow rate(which is an extensive property) just use velocity(intensive property)
good luck

 nurul efha October 21, 2012 20:47

Quote:
 Originally Posted by eng_s_sadeghi (Post 387678) Hi. Are you using 2D Ax-symmetric? If yes, mass flow should be divided by (2*PI). You are just modeling 1 rad of the whole circle.
I'm using 2D planar...

 nurul efha October 21, 2012 20:49

Quote:
 Originally Posted by misagh (Post 387797) HI You had better use intensive property for your simulation and further calculation.it is an easy way to get rid of changing units in 2.d or 3.d . i mean instead of using mass flow rate(which is an extensive property) just use velocity(intensive property) good luck

I think so. Thanks...

 Nickul October 27, 2012 14:03

Quote:
 Originally Posted by eng_s_sadeghi (Post 387678) Hi. Are you using 2D Ax-symmetric? If yes, mass flow should be divided by (2*PI). You are just modeling 1 rad of the whole circle.

Can you expand on this ? :) I have always entered the full mass flow for axisymetric problems, not a 1 rad slice...

Quick search to find a Fluent manual online, reveals the following, (v6.3)

http://hpce.iitm.ac.in/website/Manua...ug/node227.htm

Quote:
 http://hpce.iitm.ac.in/website/Manua.../ug/img741.gif Note that for axisymmetric problems, this mass flow rate is the flow rate through the entire ( http://hpce.iitm.ac.in/website/Manua.../ug/img742.gif-radian) domain, not through a 1-radian slice.

 villager July 31, 2016 19:11

Quote:
 Originally Posted by Nickul (Post 388853) Can you expand on this ? :) I have always entered the full mass flow for axisymetric problems, not a 1 rad slice... Quick search to find a Fluent manual online, reveals the following, (v6.3) http://hpce.iitm.ac.in/website/Manua...ug/node227.htm
Totally agree with Nickul about Mass flow rate (e.g. for v16.2):
"Note that for axisymmetric problems, this mass flow rate is the flow rate through the entire ( $2\pi$-radian) domain, not through a 1-radian slice."
https://www.sharcnet.ca/Software/Ans...low_Inlet.html

Only Mass Flux is through the 1-radian slice: "mass flux is the flux through a 1-radian slice of the domain."

 All times are GMT -4. The time now is 01:55.


text:
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 September 28th, 2019, 06:32 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Quadratic polynomial Find the quadratic equation which satisfies the condition $\displaystyle x_1 x_2 =x_1 +x_2$.
 September 28th, 2019, 07:34 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 I don't understand the problem statement. Quadratic equation in what? You could write the condition itself as $\displaystyle x_1 = \frac{x_2}{x_2-1}$ or $\displaystyle x_2 = \frac{x_1}{x_1-1}$ or $\displaystyle (x_1-1)(x_2-1)-1 = 0$ Are any of those what you were looking for? Thanks from topsquark and idontknow
 September 28th, 2019, 08:00 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 The equation is : $\displaystyle x^2 -2x+2=0$. It satisfies the condition $\displaystyle x_1 x_2 = x_1 + x_2$.
September 28th, 2019, 08:28 AM   #4
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,304
Thanks: 961

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by idontknow The equation is : $\displaystyle x^2 -2x+2=0$. It satisfies the condition $\displaystyle x_1 x_2 = x_1 + x_2$.
I presume that you have a quadratic equation of the form $\displaystyle y = ax^2 + bx + c$ that has zeros at $\displaystyle x_1$ and $\displaystyle x_2$ such that $\displaystyle x_1 x_2 = x_1 + x_2$, right?

By Vieta's formula we have that
$\displaystyle x_1 x_2 = \dfrac{c}{a}$
and
$\displaystyle x_1 + x_2 = - \dfrac{b}{a}$

Thus $\displaystyle \dfrac{c}{a} = -\dfrac{b}{a}$.

Clearly then, c = -b, so your quadratic will be $\displaystyle y = a x^2 + bx - b$. Your example is of the same form.

-Dan

September 28th, 2019, 04:05 PM   #5
Senior Member

Joined: Sep 2016
From: USA

Posts: 670
Thanks: 440

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by idontknow Find the quadratic equation which satisfies the condition $\displaystyle x_1 x_2 =x_1 +x_2$.
The equation $x_1x_2 = x_1 + x_2$ Is equivalent to $x_1x_2 - x_1 - x_2 = 0$ and the left side of this is already a quadratic polynomial so I have no clue what you are asking.

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text:
# Parity Theorem for Permutations - An Enabling Lemma

Author(s):
John O. Kiltinen

One way to prove the Parity Theorem is first to prove the

Even Identity Lemma. If e is the identity permutation in Sn that maps every element to itself, and if where the ti’s are transpositions, then k is an even number.

Before considering a proof of this lemma, let us show how it leads very directly to a proof of the Parity Theorem. Suppose that we have two expressions, and , for a permutation a in terms of transpositions. Then, since the inverse of a composition of a sequence of permutations is the composition of their inverses in the reverse order, and since every transposition is its own inverse, it follows that

This shows that e can be expressed using k + m transpositions. Once the lemma is established, we will know that k + m is an even number. This assures that k and m are either both even numbers or they are both odd numbers.

John O. Kiltinen, " Parity Theorem for Permutations - An Enabling Lemma," Convergence (December 2004)

## JOMA

Journal of Online Mathematics and its Applications


text:
# Ngô Quốc Anh

## April 18, 2010

### The Pohozaev identity: Elliptic problem with biharmonic operator

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 14:19

We now consider another kind of problem involving biharmonic operator. Let us assume $u>0$ a solution of the equation

$\displaystyle (\Delta^2u)(x)+Q(x)f(u(x))=0$

in $\mathbb R^n$. We shall prove the following result

Theorem. The following identity

$\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx}\hfill \\ \end{gathered}$

holds.

Proof. Multiplying the PDE by $x \cdot \nabla u$ on the both sides

$\displaystyle (x \cdot \nabla u){\Delta ^2}u + (x \cdot \nabla u)Q(x)f(u) = 0$

and integrating the resulting equation over the ball $B_r(0)$, one has

$\displaystyle\int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} + \int_{{B_r}(0)} {(x \cdot \nabla u)Q(x)f(u)dx} = 0$.

Clearly, by the divergence theorem, one gets

$\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u)Q(x)f(u)dx} = \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {{u_{{x_i}}}{x_i}Q(x)f(u)dx} } \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {{\nu ^i}u{x_i}Q(x)f(u)d\sigma } - \int_{{B_r}(0)} {u{{\left( {{x_i}Q(x)f(u)} \right)}_{{x_i}}}dx} } \right]} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {\frac{{{x_i}}}{r}u{x_i}Q(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + {x_i}{Q_{{x_i}}}f(u) + {x_i}Q(x)f'(u){u_{{x_i}}}} \right)dx} } \right]} \hfill \\ \qquad= \boxed{\int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x \cdot \nabla Q)f(u) + (x \cdot \nabla u)Q(x)f'(u)} \right)dx}}. \hfill \\ \end{gathered}$

Regarding to the first term we still use the divergence theorem, however, we are dealing with the biharmonic operator $\Delta^2 u$ which is defined to be $\Delta(\Delta u)$. We have

$\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} = - \int_{{B_r}(0)} {\nabla (x \cdot \nabla u) \cdot \nabla (\Delta u)dx} + \int_{\partial {B_r}(0)} {(x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} - \int_{\partial {B_r}(0)} {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {(x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }} - (x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } . \hfill \\ \end{gathered}$

Note that

$\displaystyle\begin{gathered} \Delta (x \cdot \nabla u) = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {{{\left( {{x_i}{u_{{x_i}}}} \right)}_{{x_j}{x_j}}}} } \hfill \\ \qquad= \sum\limits_{j = 1}^n {{{\left( {{u_{{x_j}}} + \sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}}}} } \right)}_{{x_j}}}} \hfill \\ \qquad= \sum\limits_{j = 1}^n {\left( {{u_{{x_j}{x_j}}} + {{\left( {\sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}}}} } \right)}_{{x_j}}}} \right)} \hfill \\ \qquad= \sum\limits_{j = 1}^n {\left( {{u_{{x_j}{x_j}}} + {u_{{x_j}{x_j}}} + \left( {\sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}{x_j}}}} } \right)} \right)} \hfill \\ \qquad= 2\Delta u + (x \cdot \nabla (\Delta u)). \hfill \\ \end{gathered}$

Therefore

$\displaystyle\int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} = 2\int_{{B_r}(0)} {{{(\Delta u)}^2}dx} + \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx}$.

We actually can estimate further as follows

$\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} = \int_{{B_r}(0)} {\sum\limits_{i = 1}^n {({x_i}{{(\Delta u)}_{{x_i}}}\Delta u)} dx} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {({x_i}\Delta u){{(\Delta u)}_{{x_i}}}dx} } \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ { - \int_{{B_r}(0)} {{{({x_i}\Delta u)}_{{x_i}}}(\Delta u)dx} + \int_{\partial {B_r}(0)} {{\nu ^i}({x_i}\Delta u)(\Delta u)d\sigma } } \right]} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ { - \int_{{B_r}(0)} {(\Delta u + {x_i}{{(\Delta u)}_{{x_i}}})(\Delta u)dx} + \int_{\partial {B_r}(0)} {\frac{{x_i^2}}{r}{{(\Delta u)}^2}d\sigma } } \right]} \hfill \\ \qquad= - \int_{{B_r}(0)} {{{(\Delta u)}^2}dx} - \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} + \int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } \hfill \\ \end{gathered}$

which follows

$\displaystyle\int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} = \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } - \frac{1}{2}\int_{{B_r}(0)} {{{(\Delta u)}^2}dx}$.

Hence

$\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} \hfill \\ \qquad\qquad= \boxed{\frac{3}{2}\int_{{B_r}(0)} {{{(\Delta u)}^2}dx} +\frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma }+ \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }} - (x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma}. } \hfill \\ \end{gathered}$

Multiply the PDE by $u$ on both sides and integrating it over the ball $B_r(0)$ together with the fact that

$\displaystyle - \int_{{B_r}(0)} {f\Delta gdx} = \int_{{B_r}(0)} {\nabla f \cdot \nabla gdx} - \int_{\partial {B_r}(0)} {f\frac{{\partial g}}{{\partial \nu }}d\sigma }$

we obtain

$\displaystyle\begin{gathered} \int_{{B_r}(0)} {{{(\Delta u)}^2}dx} = \int_{{B_r}(0)} {\Delta u\Delta udx} \hfill \\ \qquad= - \int_{{B_r}(0)} {\nabla u \cdot \nabla \Delta udx} + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {u{\Delta ^2}udx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } \hfill \\ \qquad= \boxed{\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma }} . \hfill \\ \end{gathered}$

Combining all gives

$\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx} . \hfill \\ \end{gathered}$

Cf. Exact solutions of nonlinear conformally invariant integral equations in $\mathbb R^3$Adv. Math. 194 (2005),  no. 2, pp. 485-503.

1. Hello!

I am interested in the Pohozaev identity but not good at it. Please say more about its applications! It seems that it is basically used to prove non-existence?

Thanks.

Comment by unknown — May 3, 2011 @ 21:54

• You are right, this is a great tool for proving non-existence result. The simplest example, originally due to Pohozaev, can be found here http://wp.me/p4uAr-Nu.

Comment by Ngô Quốc Anh — May 3, 2011 @ 22:28

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text:
Articles written in Journal of Earth System Science

• Suppressed biological production in the coastal waters off Visakhapatnam, India under the impact of the very severe cyclonic storm Hudhud

Tropical cyclones generally enhance biological production due to the increase in nutrients input due to vertical mixing. In contrast, the very severe cyclonic storm (VSCS)Hudhud decreased primary production due to the strong stratification associated with torrential rainfall and high suspended load from the major city where the cyclone made landfall. The study region received nutrients from the cold core eddy and coastal upwelling in the offshore and inshore regions, respectively, during pre-cyclone period and the same was suppressed under the influence of cyclonic winds led to convergence by shoreward Ekman transport. The land run-off brought nutrients to the coast during cyclone Hudhud; however, their concentrations were less than that during other cyclones (Orissa supercyclone, Sidr and Phailin). Such low nutrient levels resulted from the VSCS Hudhud crossing the urban region (Visakhapatnam city) whereas other cyclones crossed the coast over fertile agricultural lands which led to high nutrients input associated with phytoplankton blooms. Therefore, the biological response to a cyclone not only depends on the intensity of the cyclone but also on the region of the land it crosses.

• Influence of phytoplankton pigment composition and primary production on pCO$_{2}$ levels in the Indian Ocean

The tropical Indian Ocean is a net sink for the carbon dioxide (CO$_{2}$) in the atmosphere and phytoplankton production plays a crucial role in CO$_{2}$ fixation and determines the direction of CO$_{2}$ flux at the sea-to-air interface. In order to assess the influence of phytoplankton composition and primary production on pCO$_{2}$ levels in the Indian Ocean, sampling was conducted at 25 stations during the spring intermonsoon period under the auspices of the Indian GEOTRACES program. The pCO$_{2}$ was significantly correlated with salinity due to the discharge of low pCO$_{2}$ water by major rivers to the Bay of Bengal (BoB). The stronger negative correlation observed between pCO$_{2}$ and major phytoplankton marker pigments, net primary production and oxygen saturation levels suggesting significant influence of biological processes on pCO$_{2}$ levels in the Indian Ocean. This study indicates that pCO$_{2}$ levels are strongly modulated by biological processes than hitherto hypothesized as solubility pump in the Indian Ocean.

$\bf{Highlights}$

$\bullet$ Surface pCO$_{2}$ levels are undersaturated in the Indian Ocean compared to atmosphere, except Arabian Sea.

$\bullet$ Oligotrophic conditions prevailed in the entire tropical Indian Ocean.

$\bullet$ Picophytoplankton (cyanobacteria) is the dominant phytoplankton in the Indian Ocean.

$\bullet$ Significant relation between phytoplankton groups and pCO$_{2}$ indicates strong biological control on surface pCO$_{2}$.

• Journal of Earth System Science

Volume 132, 2023
All articles
Continuous Article Publishing mode

• Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
# How do you graph y=2x^2+6?

Sep 4, 2015

Refer the explanation section

#### Explanation:

Its vertex is at
$x = \frac{- b}{2 a}$

'b' term is not there in the function.
Then we shall write the function like this
$a x 2 + b x + c$ - This is the general form of the function.
The our function shall be writter as -
$2 {x}^{2} + 0 x + 6$

Now find the vertex -
$x = \frac{- b}{2 a} = \frac{- 0}{2 \times 2}$= 0

Since $2 {x}^{2}$ term is positive, It must be an upward facing curve.

Take a few points on either side of 0. Calculate the y value.

x :y
-3: 24
-2: 14
-1 :8
0: 6
1 :8
2 :14
3 :24

Plot the points. Join them with the help of a smooth curve. graph{2x^2+6 [-58.5, 58.55, -29.25, 29.2]}