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My research interest is in three main areas: ‘Predictive Psychometrics’, ‘Regression models for categorical response variables’ and ‘Longitudinal Data Analysis’.
Predictive PsychometricsMost psychological research focusses on explanation. That is the ‘why’ question: why do things happen as they happen? In practical situations it is not only important to
understand why things happen but it is also important to predict, i.e. to be able to tell what will happen in the near future for a specific person. This line of research focusses on statistical
methods for prediction.
Regression models for categorical response variablesMany measurements in Psychology are categorical. Main examples are
Although these measurements are categorical often the categorical nature is not taken into account in the statistical analysis. This might lead to fallible results and wrong conclusions.
Longitudinal Data AnalysisTo really understand the manner in which people are developing we need to obtain measurements over time. Having such data, however, invalidates an important assumption of
most statistical techniques: the independence assumption. This line of research applies and develops methodology for the statistical analysis of repeated measurements. | {"url":"https://markderooij.info/research/","timestamp":"2024-11-03T22:10:51Z","content_type":"text/html","content_length":"32781","record_id":"<urn:uuid:d6d4b810-09d0-4812-9f04-6438351d634e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00452.warc.gz"} |
Einstein’s 1905 paper on the Photoelectric Effect
Einstein’s Paper on the Photoelectric Effect (1905)
The paper that earned Einstein the Nobel Prize
About six weeks before he submitted his doctoral thesis at the University of Zürich in 1905, Einstein on March 18th submitted the paper ‘Über einen die Erzeugung und Verwandlung des Lichtes
betreffenden heuristischen Gesichttspunkt’ (“On a Heuristic Viewpoint Concerning the Production and Transformation of Light”) to Annalen der Physik, the world’s premier physics journal at the
time. In the paper, Einstein proposes the existence of energy quanta, light particles now called photons, motivated by Max Planck’s earlier derivation of Planck’s law of black body radiation.
He was 26 years old at the time.
This week’s newsletter discusses the contents of Einstein’s Nobel Prize-winning paper, including derivations of his arguments based on the wonderful paper:
This post is for paid subscribers | {"url":"https://www.privatdozent.co/p/einsteins-1905-paper-on-the-photoelectric-8e3?s=w","timestamp":"2024-11-13T04:14:59Z","content_type":"text/html","content_length":"161636","record_id":"<urn:uuid:3aadbb54-a64a-4302-bc68-ae0b040c576a>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00555.warc.gz"} |
Central limit theorem for non-stationary random products of SL(2, R) matrices
Tuesday, October 15, 2024 - 1:00pm to 2:00pm
Consider a sequence of independent and identically distributed SL(2, R) matrices. There are several classical results by Le Page, Tutubalin, Benoist, Quint, and others that establish various forms of
the central limit theorem for the products of such matrices. In our work, we generalize these results to the non-stationary case. Specifically, we prove that the properly shifted and normalized
logarithm of the norm of a product of independent (but not necessarily identically distributed) SL(2, R) matrices converges to the standard normal distribution under natural assumptions. A key
component of our proof is the regularity of the distribution of the unstable vector associated with these products. | {"url":"https://www.math.uci.edu/node/38191","timestamp":"2024-11-11T08:26:45Z","content_type":"text/html","content_length":"37635","record_id":"<urn:uuid:cd77ab5c-a724-4c4f-835f-193d63ee28de>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00658.warc.gz"} |
There has to be a better way to compute integrals than looking at paper with vacant eyes and waiting for inspiration or...
Matthew Rogers
There has to be a better way to compute integrals than looking at paper with vacant eyes and waiting for inspiration or burning a hole through paper with trial-and-error. Share your secrets with a
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Nope, integration is harder than differentiating. IIRC, this was actually shown mathematically (meta stuff, the likes of Gödel).
Way back in Freshman Calc, I had a prof who said that while any old fool can differentiate, 'Integration is an art'
Like with everything, I guess, just practice. In physics, especially, you'll find the same integrals coming up again and again, so eventually you'll just be able to knock them out without even
thinking about it
the problem is, that there are loads of stuff that cannot be integrated properly and instead you get non-trivial functions like erg(x) or Si(x) or maybe the occasional arctan(x).
it's called simpson's rule
I have HEARD that there's an algorithm which can perform general integration. I mean, analytically, not numerically.
They tested it on a big book of integrals and found a huge number of them were wrong. Mostly typos. A plus where a minus should be. That sort of thing.
Never came across it again. Most people (except physicists and mathematicians) just brute-force it with a computer nowadays.
Try Wolfram.
You mean the Risch algorithm.
It's implemented in Wolfram alpha
The specification is about 100 pages long
Are there any integrals that aren't tabulated or given methods/algorithms to solve them?
example, integral table off google.
>integration by parts
>arctan, arcsin, etc.
>some other shit I fucked up so bad I forgot what it even is already
absolute bane of my existence in high school calc and probably will continue into college. I assume practice is really the best way to fix this, right?
There is a homotopy type of method that introduces an extra parameter and then utilizes differentiation under an integral.
Feynman used it to calculate definite integrals that other people struggled with.
yep, just lots of practice.
But calc practice is easy cuz the problems are usually solvable in half a page.
Thanks. I never would have located that on my own. I'll probably never use it, but it's nice to know it exists.
's prof plagiarized that bon mot. :)
We were taught at calculus classes that integration is an art, not a science (in contrast to differentiation—even a monkey can be trained to take derivatives). And we were taught wrong. The Risch
algorithm (which is known for decades) allows one to find, in a finite number of steps, if a given indefinite integral can be taken in elementary functions, and if so, to calculate it.
>There has to be a better way to compute integrals than looking at paper with vacant eyes and waiting for inspiration or burning a hole through paper with trial-and-error.
Yes, it is called a computer, specifically the Risch algorithm.
arctan(x) is a trivial function though
>homootopy type
Put don the algebraic topology user.
If you don't mind me asking, where are you from that you do integration in high school calc? I took AP calculus in high school and the farthest we got was differentiation. The only time we did
integration was when the instructors were showing us what we would do when we get to college. The integration we did in high school was entirely extra credit.
california bay area, doing calc BC. AB had only a little integration, and the majority of kids don't even take calc at all, let alone BC. in BC we've done a lot of integration, but this is a class
which students chose specifically to take.
that's what I figured, although I'm a brainlet so I like to use an entire page for some problems just so I can be very careful with it
He does calculus without reals
I also took AB calc in high school, shit wasn't good.
But integration by parts as only a little bit? I didn't do that until Calculus II in college. Hell we didn't even do vector calculus until Calculus IV.
Actually when I talked to my Canadian friends about Calculus IV, they had no idea what I was talking about. Is this an American thing? Are Canadians just retarded? Am I retarded? Who is retarded?
I didn't mean it in that sense.
Would you prefer I say perturbation?
>calc 4
You go to a brainlet uni
I took IB Math HL in Canada. We covered this.
>shit wasn't good
what exactly do you mean?
well we did quite a bit of integration by parts this year, but I can't really say how in-depth we actually went. In fact we just finished up with vector calculus (pretty basic stuff, dot products,
cross products and the like, practically not even calculus really) and that's it for this year. we're done with the curriculum and looking towards reviewing everything before college and for finals
and AP testing.
I'm pretty sure I'm retarded since I struggle to do work properly in all of my classes but who knows, you could be retarded too.
>In fact we just finished up with vector calculus (pretty basic stuff, dot products, cross products and the like, practically not even calculus really)
What you are describing is algebra.
Holy shit America.
Definition: the set of classical analytic functions is defined as [math]\sin[/math], [math]\cos[/math], [math]\tan[/math], [math]\csc[/math], [math]\sec[/math], [math]\cot[/math], [math]\arcsin[/
math], [math]\arccos[/math], [math]\arctan[/math], [math]\exp[/math], [math]\ln[/math], [math]\left|\cdot\right|[/math], [math]\sqrt[n]\cdot[/math] for [math]n\,\in\,\mathbf Z[/math] constant,
polynomials, and their linear combinations, products, reciprocals and compositions.
Theorem 1: all classical analytic functions are computably differentiable and their derivatives are classical analytic functions.
Proof: Cf. your calc 1 & 2 lessons.
Theorem 2: there exists a classical analytic function whose counter-derivative isn't a classical analytic function.
Proof: [math]\begin{array}[t]{cccl} f: & \mathbf R & \longrightarrow & \mathbf R \\ & x & \longmapsto & \exp\left(-x^2\right) \end{array}[/math].
Corollary: integrating classical analytic functions is more difficult than differentiating them.
Dude, take the obvious derivatives as tangent lines and then just do the opposite. In fact, if you do that to an iterative depth on a function, you'll get a shape with sharp edges. Compare x^2 with x
^6. Then you can find Area with geometry
Integration by parts is taught at highschool in Europe. Then again, more rigerously in a first year course, which goes all the way up to vector calculus.
i.e Your country is retarded.
Method of smallest squares, user
>like with everything I guess, just practice
Or use a calculator, that's much more efficient. Honestly, the mathematician's obsession with solving problems by hand is something I doubt I'll ever understand.
My mother never let me use a calculator as a kid. On standardized tests I was given one and the only operators I used were the elementary ones (only + and • before I learned how to manipulate - and
÷--around 6th or 7th grade iirc). I hated it but I was able to pick up on patterns easily compared to my peers who had used a calculator throughout their academic career.
For me, it was mainly a matter of
>I'm tired of doing so many steps. Is there a faster way?
reminder that CAS and automated provers are becoming better mathematicians than mathematicians were ever capable of being
yeah well that was our final mini- "unit" for this year, I know it's just algebra. although we tied it back in to calc with some derivatives, but that was about it for vectors this year.
Bentley Green
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American Mathematical Society
Book Review
The AMS does not provide abstracts of book reviews. You may download the entire review from the links below.
Full text of review: PDF
This review is available free of charge.
Book Information: Editors: H. Bass
A. Buium
P.J. Cassidy Title:
Selected works of Ellis Kolchin
Additional book information:
edited by
H. Bass
A. Buium
P.J. Cassidy
, American Mathematical Society, Providence, RI, 1999, xiii + 639 pp., ISBN 0-8218-0542-8, $120.00$
Bertrand, D., Review of Lectures on Differential Galois Theory by A. Magid, Bull. Amer. Math. Soc., vol. 33, 1996, pp. 289-294.
Borel, A., Algebraic groups and Galois theory in the work of Ellis R. Kolchin, in Selected Works of Ellis Kolchin with Commentary, eds. Bass, H., et al., American Mathematical Society,
Providence, 1999, pp. 505-526. CMP 99:09
Buium, A. and Cassidy, P., Differential algebraic geometry and differential algebraic groups: From algebraic differential equations to Diophantine geometry, in Selected Works of Ellis Kolchin
with Commentary, eds. Bass, H., et al., American Mathematical Society, Providence, 1999, pp. 567-636. CMP 99:09
P. Deligne, Catégories tannakiennes, The Grothendieck Festschrift, Vol. II, Progr. Math., vol. 87, Birkhäuser Boston, Boston, MA, 1990, pp. 111–195 (French). MR 1106898
Janelidzhe, G., Galois theory in categories: the new example of differential fields, in Categorical Topology and Its Relation to Analysis Algebra and Combinatorics: Prague, Czechoslovakia, 22-27
August 1988, eds. Adamek, J. and Mac Lane, S., World Scientific (1989), pp. 369-380.
Poizat, B., Les corps différentiellment clos, compagnons de route de la théorie des modéles, in Selected Works of Ellis Kolchin with Commentary, eds. Bass, H., et al., American Mathematical
Society, Providence, 1999, pp. 555-566. CMP 99:09
Singer, M., Direct and inverse problems in differential Galois theory, in Selected Works of Ellis Kolchin with Commentary, eds. Bass, H., et al., American Mathematical Society, Providence, 1999,
pp. 527-554. CMP 99:09
Carol Tretkoff and Marvin Tretkoff, Solution of the inverse problem of differential Galois theory in the classical case, Amer. J. Math. 101 (1979), no. 6, 1327–1332. MR 548884, DOI 10.2307/
Garrett Birkhoff and Morgan Ward, A characterization of Boolean algebras, Ann. of Math. (2) 40 (1939), 609–610. MR 9, DOI 10.2307/1968945
Review Information: Reviewer:
Andy R. Magid
University of Oklahoma
Bull. Amer. Math. Soc.
(2000), 337-342
Published electronically:
February 1, 2000
Review copyright:
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NCERT Solutions Class 2 Maths Chapter 11 - Lines and Lines | Free PDF
Lines and Lines deals with the concept of lines. This chapter have exercises on: Drawing and identifying straight lines,curved lines, Zigzag lines. The solutions for Math-Magic Chapter-11 have been
created and verified by our experienced subject matter experts, according to the CBSE syllabus and guidelines of NCERT. For practice, our subject matter experts have created very interactive,
activity-based, and Image-based worksheets on these topics to enhance learning. | {"url":"https://www.orchidsinternationalschool.com/ncert-solutions-class-2-maths/chapter-11-lines-and-lines","timestamp":"2024-11-06T17:51:07Z","content_type":"text/html","content_length":"54410","record_id":"<urn:uuid:d196f48c-317c-4149-ac75-d81382fbae12>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00619.warc.gz"} |
TGD diary{{post.title}}String world sheets, partonic 2-surfaces and vanishing of induced (classical) weak fields
Well-definedness of the em charge is the fundamental on spinor modes. Physical intuition suggests that also classical Z
field should vanish - at least in scales longer than weak scale. Above the condition guaranteeing vanishing of em charge has been discussed at very general level. It has however turned out that one
can understand situation by simply posing the simplest condition that one can imagine: the vanishing of classical W and possibly also Z
fields inducing mixing of different charge states.
1. Induced W fields mean that the modes of Kähler-Dirac equation do not in general have well-defined em charge. The problem disappears if the induced W gauge fields vanish. This does not yet
guarantee that couplings to classical gauge fields are physical in long scales. Also classical Z^0 field should vanish so that the couplings would be purely vectorial. Vectoriality might be true
in long enough scales only. If W and Z^0 fields vanish in all scales then electroweak forces are due to the exchanges of corresponding gauge bosons described as string like objects in TGD and
represent non-trivial space-time geometry and topology at microscopic scale.
2. The conditions solve also another long-standing interpretational problem. Color rotations induce rotations in electroweak-holonomy group so that the vanishing of all induced weak fields also
guarantees that color rotations do not spoil the property of spinor modes to be eigenstates of em charge.
One can study the conditions quite concretely by using the formulas for the components of
spinor curvature
1. The representation of the covariantly constant curvature tensor is given by
R[01]= e^0 ∧ e^1-e^2∧ e^3 , R[23]= e^0∧ e^1- e^2∧ e^3 ,
R[02]=e^0∧ e^2-e^3 ∧ e^1 , R[31] = -e^0∧
e^2+e^3∧ e^1 ,
R[03] = 4e^0∧ e^3+2e^1∧ e^2
, R[12 = 2e^0∧ e^3+4e^1∧ e^2 .
R[01]=R[23] and R[03]= R[31] combine to form purely left handed classical W boson fields and Z^0 field corresponds to Z^0=2R[03].
Kähler form is given by
J= 2(e^0∧e^3+e^1∧ e^2) .
2. The vanishing of classical weak fields is guaranteed by the conditions
e^0∧ e^1-e^2∧e^3 =0 ,
e^0∧ e^2-e^3 ∧e^1 ,
4e^0∧ e^3+2e^1∧e^2 .
3. There are many manners to satisfy these conditions. For instance, the condition e^1= a× e^0 and e^2=-a×e^3 with arbitrary a which can depend on position guarantees the vanishing of classical W
fields. The CP[2] projection of the tangent space of the region carrying the spinor mode must be 2-D.
Also classical Z^0 vanishes if a^2= 2 holds true. This guarantees that the couplings of induced gauge potential are purely vectorial. One can consider other alternaties. For instance, one could
require that only classical Z^0 field or induced Kähler form is non-vanishing and deduce similar condition.
4. The vanishing of the weak part of induced gauge field implies that the CP[2] projection of the region carrying spinor mode is 2-D. Therefore the condition that the modes of induced spinor field
are restricted to 2-surfaces carrying no weak fields sheets guarantees well-definedness of em charge and vanishing of classical weak couplings. This condition does not imply string world sheets
in the general case since the CP[2] projection of the space-time sheet can be 2-D.
How string world sheets could emerge?
1. Additional consistency condition to neutrality of string world sheets is that Kähler-Dirac gamma matrices have no components orthogonal to the 2-surface in question. Hence various fermionic would
flow along string world sheet.
2. If the Kähler-Dirac gamma matrices at string world sheet are expressible in terms of two non-vanishing gamma matrices parallel to string world sheet and sheet and thus define an integrable
distribution of tangent vectors, this is achieved. What is important that modified gamma matrices can indeed span lower than 4-D space and often do so (massless extremals and vacuum extremals
representative examples). Induced gamma matrices defined always 4-D space so that the restriction of the modes to string world sheets is not possible.
3. String models suggest that string world sheets are minimal surfaces of space-time surface or of imbedding space but it might not be necessary to pose this condition separately.
In the proposed scenario string world sheets emerge rather than being postulated from beginning.
1. The vanishing conditions for induced weak fields allow also 4-D spinor modes if they are true for entire spatime surface. This is true if the space-time surface has 2-D projection. One can expect
that the space-time surface has foliation by string world sheets and the general solution of K-D equation is continuous superposition of the 2-D modes in this case and discrete one in the generic
2. If the CP[2] projection of space-time surface is homologically non-trivial geodesic sphere S^2, the field equations reduce to those in M^4× S^2 since the second fundamental form for S^2 is
vanishing. It is possible to have geodesic sphere for which induced gauge field has only em component?
3. If the CP[2] projection is complex manifold as it is for string like objects, the vanishing of weak fields might be also achieved.
4. Does the phase of cosmic strings assumed to dominate primordial cosmology correspond to this phase with no classical weak fields? During radiation dominated phase 4-D string like objects would
transform to string world sheets.Kind of dimensional transmutation would occur.
Right-handed neutrino has exceptional role in K-D action.
1. Electroweak gauge potentials do not couple to ν[R] at all. Therefore em neutrality condition is un-necessary if the induced gamma matrices do not mix right handed neutrino with left-handed one.
This is guaranteed if M^4 and CP[2] parts of Kähler-Dirac operator annihilate separately right-handed neutrino spinor mode. Also ν[R] modes can be interpreted as continuous superpositions of 2-D
modes and this allows to define overlap integrals for them and induced spinor fields needed to define WCW gamma matrices and super-generators.
2. For covariantly constant right-handed neutrino mode defining a generator of super-symmetries is certainly a solution of K-D. Whether more general solutions of K-D exist remains to be checked out.
See the chapter
The recent vision about preferred extremals and solutions of the modified Dirac equation
of "TGD: "Physics as infinite-dimensional geometry" or the
12 comments:
Dear Matti,
If classical em field is associated with long ranged color field as TGD say it, i try to clear my imagination by ask my questions:
em wave has tensor product of spin group and color group?
Why every em wave with special wavelength has special color or in other words there is 1-1 correspondence between them?
Dear Hamed,
one can say that color field is associated with induced Kähler form, which does not always coincide with em field which contains also other term proportional to classical Z^0 field. If classical
weak fields vanish as they should at string world sheets carrying fermions so that they have well-defined em charge and couple purely vectorially then Kähler form is proportional to em field. Em
field is accompanied by classical long ranged color field.
What matters is whether classical color gauge potentials couple to induced spinor fields. The classical color field is not visible in the couplings of spinors since they couple directly only to
the ew gauge potentials.
Color is visible only at quantum level. Conformal algebra generators carrying color and color partial waves assignable to imbedding space spinors and appearing as ground states of super-conformal
representations. These imbedding space spinor modes correspond to the spinors of QFT and QCD appears as QFT approximation to the dynamics.
I would not assign to em waves color. Color is quale which in TGD inspired theory of consciousness corresponds to increment of color
quantum numbers (this is not a joke!) in quantum jump. Quite concretely the flow of color quantum numbers from subsystem in question or to it give rise to sensation of color. There are many
manners to induce this flow. Photons have specialised to induce this flow but photons as such do not have any color. Qualia are not properties of physical objects but characterise their change in
quantum jump. This saves from logical paradoxes.
Dear Matti,
this is correct? for example when a photon with the wavelength of 700nm come to my eyes, although it hasn't any color but it leads to increment of red color quantum number in the retina.
But if photons haven't any color, how they can induce the flow of color quantum numbers in retina? conservation of color quantum number is violated?
Maybe i can improve my last question in this form: Why does every photon with special wave length can increase special color quantum number in the retina?
This comment has been removed by the author.
This comment has been removed by the author.
Dear Hamed.
Color sensation can quite well induced the flow of color quantum numbers. The total color quantum numbers of system plus complement should vanish. It is also possible that color polarisation
occurs so that the sensory receptor and its complement have opposite but increasing color quantum numbers. I have modelled sensory perception as analog of di-electric breakdown for the analog of
capacitor: now color capacitor.
It is interesting that for instance an object with color is surrounded by narrow stripe with complementary color: does this follow from vanishing of total color quantum numbers.
One can of course ask whether gluons are generated in color perception by incoming photons and create the flow of color quantum numbers.
If one believes that dark QCD is present in scales of living matter (four Gaussian Mersennes (1+i)^n-1 in length scale range 10 nm, 2.5 micrometer which is number theoretical miracle, suggest
that both scaled/dark versions of QCD and and ew interactions are there and correspond to these Mersennes).
An interesting thing is also that we can observe color only if there are at least two colours. Completely homogenous light to screen does not create sensation of color. I think here one must
build a model for for color perception is. For years ago I considered this problem: do not remember anymore the explanation.
Dear Hamed,
as you say, an interesting question is why color
of certain wavelength (actually distribution of wavelengths) can induce color sensation.
Could one speak of Bose-Einstein condensation color receptors have Bose-Einstein condensate with definite color quantum numbers and this is amplified by polarisation induced by incoming photons
around resonance energies. Photons give only the energy. Total color remains zero but color quantum numbers at plates of color capacitor increase. The color quantum numbers of color receptors
would correlate with visual colours.
Could one consider the situation in terms
dark QCD in which string like objects having opposite colors and realised as color magnetic
flux tubes are generated between the plates
of color capacitor. Kind of resonant burst of mesons generated by incoming photon energy.
Could this relate to dark photons decaying to biophotons in turn providing energy for creating color polarisation? Biophotons are just ordinary photons in eneryy range including visible and part
of UV (molecular excitation energies).
Dear Matti,
Does state function reduction, occurs for massless extremals too? as I understand, In standard QM, only wave function of electron reduce after measurement and it can’t describe what occurs for
photons after the measurement. Although if we do the double-slit experiment for a single photon, we see the wave function of photons reduce at a point of screen. Hence one can say that it is a
problem for standard QM that can’t describe wave function evolution and reduction of wave function of a photon.
After reduction of wave function of an electron in some boundary of CD(suppose in lower boundary),
if at this moment of consciousness we go from lower to upper boundary and observe the wave function of the electron(In other words we are observing geometric time evolution of electron in this
process), what does we see? Is this correct? We see the electron wave function just over a small 3-surface in lower boundary and when we are going from lower to upper at the moment of
consciousness we see electron wave function propagates over more regions or over more and more small 3-surfaces.
State function reduction is completely general process. All particles, not only electrons. For photons it occurs for instance in double slit experiments. When both slits are open the photons are
in waves going through both slits. Measurement redoes to localised state in screen.In TGD this initial state would mean two brached 3-surface going through slits. When only single slit is open
they are initially in superposition involving only single slit.
An important correction to the original picture about state function reduction in ZEO. I have mentioned about this few times but maybe I have not emphasized enough the change.
Originally I thought that reduction occurs alternately to the boundaries of CD. This might be the case in very short scales where time arrow is not yet present.In longer scales - larger CDs- this
is not the case: microscopy and arrow of time enter into the picture.
Then I realised that state function reduction at given boundary of CD can occur arbitrary many times repeatedly. In ordinary QM this means repeated state function reduction giving again the
reduced states ( quantum mechanical Zeno paradox relates to this and is empirical fact).
In TGD each repeated state function reduction leaves the state at already reduced boundary invariant but changes the state at other boundary. This gives rise to experience flow and arrow of time.
The changes at second boundary however occur and give to conscious experience something: the experience about flow of time at least. Interesting problem is to try to understand exactly what is
this contribution! IS this just our experience about classical everyday world as opposed to experience of state function reduction which brings to my mind moment of birth about which I do not
remember much;-).
Returning to your question. The information about electron's state is coded by the wave function at reduced boundary. Effective 2-dimensionality reduces this information to partonic 2-surfaces
and tangent space data at then.
When we observe electron localisation it is a state function reduction to opposite boundary after sequence of reduction to same boundary. Boundary of CD is changed. What the scale of CD
is in this case. Scale of electrons CD of .1 seconds if one assumes that it corresponds to secondary p-adic time scale? Amusingly, this is also the time scale of our sensory perception. Maximal
time resolution of sensory perception. In smaller scales time ordering can fluctuate.
The best manner to speak about state function is to speak about WCW spinor fields. Quantum superpositions of space-time surfaces. These change in reduction reducing them to eigenstates of
measured observables at either boundary of CD.
*Superposition* of massless externals (MEs) changes to a new one for instance. It is easily misleading to talk about state function reduction for ME. If one wants to do this, one must assume that
reduction occurs in scales shorter than that characterising ME so that ME is unchanged in the process. ME serves as an arena of quantum physics in this case.
Sorry I must stop. I am going to Helsinki! I continue later!
State function reduction is completely general process. All particles, not only electrons. For photons it occurs for instance in double slit experiments. When both slits are open the photons are
in waves going through both slits. Measurement redoes to localised state in screen.In TGD this initial state would mean two brached 3-surface going through slits. When only single slit is open
they are initially in superposition involving only single slit.
An important correction to the original picture about state function reduction in ZEO. I have mentioned about this few times but maybe I have not emphasized enough the change.
Originally I thought that reduction occurs alternately to the boundaries of CD. This might be the case in very short scales where time arrow is not yet present.In longer scales - larger CDs- this
is not the case: microscopy and arrow of time enter into the picture.
Then I realised that state function reduction at given boundary of CD can occur arbitrary many times repeatedly. In ordinary QM this means repeated state function reduction giving again the
reduced states ( quantum mechanical Zeno paradox relates to this and is empirical fact).
In TGD each repeated state function reduction leaves the state at already reduced boundary invariant but changes the state at other boundary. This gives rise to experience flow and arrow of time.
The changes at second boundary however occur and give to conscious experience something: the experience about flow of time at least. Interesting problem is to try to understand exactly what is
this contribution! IS this just our experience about classical everyday world as opposed to experience of state function reduction which brings to my mind moment of birth about which I do not
remember much;-).
Returning to your question. The information about electron's state is coded by the wave function at reduced boundary. Effective 2-dimensionality reduces this information to partonic 2-surfaces
and tangent space data at then.
When we observe electron localisation it is a state function reduction to opposite boundary after sequence of reduction to same boundary. Boundary of CD is changed. What the scale of CD
is in this case. Scale of electrons CD of .1 seconds if one assumes that it corresponds to secondary p-adic time scale? Amusingly, this is also the time scale of our sensory perception. Maximal
time resolution of sensory perception. In smaller scales time ordering can fluctuate.
The best manner to speak about state function is to speak about WCW spinor fields. Quantum superpositions of space-time surfaces. These change in reduction reducing them to eigenstates of
measured observables at either boundary of CD.
*Superposition* of massless externals (MEs) changes to a new one for instance. It is easily misleading to talk about state function reduction for ME. If one wants to do this, one must assume that
reduction occurs in scales shorter than that characterising ME so that ME is unchanged in the process. ME serves as an arena of quantum physics in this case.
Sorry I must stop. I am going to Helsinki! I continue later!
Well, Have a good time there. | {"url":"https://matpitka.blogspot.com/2014/05/string-world-sheets-partonic-2-surfaces.html?showComment=1401889701274","timestamp":"2024-11-11T13:00:52Z","content_type":"application/xhtml+xml","content_length":"158294","record_id":"<urn:uuid:3491348f-2724-4f9e-be86-6bd260725392>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00750.warc.gz"} |
Precalculus: An Investigation of Functions
by David Lippman, Melonie Rasmussen
Publisher: Lulu.com 2011
ISBN/ASIN: B005J2KM16
Number of pages: 568
A textbook covering a two-quarter precalculus sequence including trigonometry. An emphasis is placed on modeling and interpretation, as well as the important characteristics needed in calculus.
Exercises are included in the book.
Download or read it online for free here:
Download link
(6.1MB, PDF)
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Vesuvius Challenge
Segmentation - a different approach
This high-level, code-free tutorial explains the main steps of the Thaumato Anakalyptor ^1 pipeline for automated segmentation in full 3D. For a more technical walkthrough of the pipeline, see this
walkthrough video. For an introduction to one of the core problems in this pipeline where we need your help, check out the Sheet Stitching Problem Playground.
Thaumato is one approach but there may be others - for example, volumetric instance segmentation might be useful as an input to later mesh stitching algorithms. If interested, check out this notebook
providing easy access to volumetric segmentation labels.
We will tackle the task of segmentation from another perspective: that of a Neapolitan fisherman.
Neapolitan fisherman.
As we saw in the "Segmentation and Flattening" tutorial, the ultimate purpose of segmentation is that of identifying a triangular mesh on the wrapped papyrus sheet.
Imagine being Neapolitan fishermen, casting our nets not into schools of fish... but onto sheets! We want the net to adhere perfectly on the inner part of the sheet, that is next to where ink is
supposed to be located. Indeed, the ink lies on the face of the papyrus that looks towards the pole of rotation of the scroll.
Here, adherence means every node of the net should be placed on the inner surface of the sheet, with adjacent nodes also being adjacent on the sheet. Therefore the problem is twofold: node placement
and connectivity establishment.
But how can we cast the net in? After all, the data we have is a 3D volumetric image composed of colored blocks. You can think of it as a big volume made of single units of grayscale LEGO blocks
(Figure 1, left). The color of a block represents the density of the material in that region of space: lighter means more dense, hence could be papyrus, while darker means less dense, and could be
air. The ink is somewhere in there.
Figure 1. Scroll data as LEGO blocks and sagittal view of a subvolume slice.
The right image in Figure 1 displays the view of a section of the volumetric image. Imagine cutting the volume in subvolumes. Take one and look at it from above. Every block composing the subvolume
(called voxels) will look like a colored square. That is the color of the pixel in the image.
Node placement
To perform node placement we have to detect all the blocks composing the inner surface of the sheet, and forget about the rest. This accounts to creating a mask where non-edge blocks are flagged with
0s and edge blocks are flagged with 1s. In layman's terms, color in black all the non-edge voxels, and in white the edge voxels.
Surface detection
What Thaumato Anakalyptor does is exactly this: it performs surface detection in 3D by convolving the volumetric image with a 3D Sobel Kernel. In simpler terms, it computes a discrete approximation
of the 3D color gradient. Then, driven by the intuition that the greatest variation in color will be on an edge, it filters out all the voxels for which the magnitude of a gradient is small (below a
given threshold).
At the same time, considering for every voxel a vector that points towards the umbilicus of the scroll (the pole of rotation), and taking the dot product between this vector and the gradient computed
just before, one can filter out as well all the voxels that are on the face of the sheet that don't look towards the center. In Figure 2 we display a sagittal view of the mask for the subvolume
showed in Figure 1. The image was obtained with a modified version of Thaumato Anakalyptor.
Figure 2. Sheet inner surface mask for a subvolume (sagittal view).
Point Cloud Extraction
Unfortunately, our net still cannot get in. Instead of having grayscale LEGO blocks, we now have black/white LEGO blocks. The colors of the blocks changed, but we are still talking about a dense
volume. In order for the net to get in we need to create some empty space. This means performing a change of representation. Instead of dealing with LEGO blocks located at positions (i,j,k) on a
tridimensional grid, we will transform the blocks into infinitesimal points located at positions (x,y,z) in Euclidean space.
At the same time, we throw away all the black blocks (that now are points), and we keep just the white ones (those that lie on the sheet surface). We extracted what is called a point cloud. In Figure
3 we show a point cloud obtained with a modified version of Thaumato Anakalyptor, colors are added just for the purpose of display, since the cloud is very dense. As you may notice, the subvolume
looks like a jungle of points that kind of align in parallel sheets. Remember to keep a copy of the computed gradients since they will be eventually needed in the next steps.
Exploiting the created empty space, our fisherman net can now enter the scroll. Every node of the fisherman's net will be placed on a point in the point cloud. But how exactly? We want the net to
overlay a contiguous patch of surface, without connecting parallel sheets. This problem will be addressed in the next section.
Figure 3. Point cloud, colors added for visualization's sake.
Connectivity establishment
We now have a cloud of points, and it's a mess! In order to fish a single, continguous patch of a sheet, we first have to identify adjacent parts of the sheets, dispatching every point in its proper
We don't want to select points that lie on parallel surfaces, since this would mean to identify as contiguous points that are not really contiguous in the scroll! We diplay the nature of this problem
in the left part of Figure 4, where correctly grouped points are circled in red, and incorrectly grouped points are circled in purple.
Figure 4. Identifying patches. (Left) Correct and incorrect grouping; (right) identified patches colored in the point cloud.
Thaumato Anakalyptor performs this operation with Mask3D ^2, a Deep Neural Network specifically trained for the purpose. The results should look somewhat like the right part of Figure 4 (that
represents a different subvolume, less densely packed). The algorithm segments point clouds in contiguous patches of sheet.
Unfortunately, we have to face two unexpected problems:
1. the algorithm creates patches with holes and in the holes there are other smaller patches;
2. the area of these patches is too small.
Thaumato Anakalyptor tackles this issue with stitching.
Stitching is an extension of the point cloud segmentation problem addressed in the previous step, but on patches rather than points.
Figure 5. Stitching adjacent patches (with the same winding number) together.
In Figure 5 different patches have different colors and patches that are stitched together are linked with red arcs. In the end, all the patches that are stitched together will form a segment,
hopefully big enough to read on it some columns of text with ink detection models. It is worth saying that the segmentation performed by Thaumato Anakalyptor generates overlapping patches. The patch
overlap is exploited during the stitching process. Thaumato Anakalyptor performs stitching in a sort of Monte Carlo fashion. It first builds an uncertainty graph where nodes of this graph are patches
and edges are weighted by the amount of overlap between patches. Many random walks are launched of the graph to build a subgraph (to select a cover of nodes/patches). The covers that selected the
edges with the maximal overlap are chosen as final segments.
Mesh reconstruction
Now that we have a few big patches, represented as groups of point clouds made of contiguous points on the surface of the sheet, we can select one, and perform mesh reconstruction.
This means finally to establish a connectivity relationship between points: to decide which nodes that are connected in the fisherman net are connected in the point cloud as well. In layman's term,
we find a way to connect points with edges!
Surface reconstruction can be performed with several algorithms, such as Poisson Surface Reconstruction ^3 (that leverages both the positions and the normals to the points, i.e. the gradients that we
computed in the previous steps) and Delaunay's triangulation ^4. Thaumato Anakalyptor uses a mixture of both.
If instead of working on subvolumes we work on the full volume of the scroll, we can envisage ending up with a large contiguous mesh that is wrapped several times around the umbilicus. This is what
Thaumato Anakalyptor can do, visualized in Figure 6.
Figure 6. Large mesh wrapped several times around the center. (Left) Side view; (center) from above; (right) zoomed-in view.
In Figure 6, the vertices of the mesh are shown in yellow and the edges connecting them in black. It is worth noting from the zoomed-in view that nearby points are connected by edges and form a
triangular mesh.
Congratulations! You automatically fished the scroll, or a subpart of it!
Further steps
In this extra section we describe how from a triangular mesh we can obtain the 2D image of a sheet. This process involves two main steps: flattening and rendering.
Now that we have a mesh, we must compute a 3D->2D map to obtain a UV (two-dimensional) parametrization for its vertices. This is a fancy way to say that you have to flatten the mesh. After all, our
purpose is to eventually read the ink on a flat image.
A community fork ^5, recently merged to Thaumato Anakalyptor, implements an algorithm called SLIM ^6 that aims to find a map that minimizes an isometric distortion energy. This means that all the
points that are equidistant in 3D will be mapped as equidistant as possible in 2D. The vertices of the mesh shown in Figure 6 are displayed using the obtained UV parametrization in Figure 7.
Figure 7. UV parametrization of the vertices of the mesh displayed in Figure 6.
The mesh, along with its UV parametrization, are finally saved in a .obj file.
It is extremely important that the obtained UV coordinates are still real numbers. This means that between points there is a lot of empty space! In order to go back to a "pixel-style" representation,
we have to convert the plot in Figure 7 to a pixelized image. Notice that in Figure 7 we displayed axes with ticks, this was not a mistake.
Per-pixel-map and layers
For every integer couple (x,y) in the figure we insert a new point. These coordinates (x,y) will be the integers' positions (i,j) of pixels in the rendered image. Eventually, we will end up with a
number of points way higher than that of the vertices of the mesh. We will identify in which triangle of the triangular mesh (in 2D) these points fall and compute their barycentric coordinates: three
coefficients that allow to define the position of every point in a triangle as a linear combination of the positions of the vertices. In Figure 8 we show a red point whose position will be 0.2 *
vertex1 + 0.3 vertex2 + 0.5 vertex3.
Figure 8. A red point in a triangle represented with its barycentric coordinates.
Exploiting the barycentric coordinates, we can then map back every additional point that we inserted in the UV map to its alleged position in 3D space. This step is called obtaining a per-pixel-map
and it's performed in Thaumato Anakalyptor using a recently merged community fork ^7.
Why do we need the 3D positions of the new points? Remember that during the first step, Node Placement, we totally forgot about the density of the material obtained via the CT scan (its original
color) and started working on a black and white mask, that later became a point cloud, and so on. All the colors displayed in Figures 3-7 were only for representation purposes.
Now we need to recover that information, and to do so we need to know the 3D position of every new point that is going to become a pixel of our rendered image. Unfortunately, the map from integer 2D
positions to 3D points will result in points with non-integer positions. What does it mean? After all, the original voxels (Figure 1) only have integer positions. What "color" will these new points/
pixels then have?
We are going to compute the color of the new points by trilinear interpolation with their closest ones in the volume. This will allow us to obtain images with a smooth varying color. The segment from
Figure 7 rendered as a 2D image is displayed in Figure 9.
Figure 9. The segment from Figure 7 rendered as a 2D image.
We forget to mention that since ink prediction is performed not on a single layer, but in what is called a surface volume, that is a stack of surfaces parallel to the surface we obtained along this
pipeline, we need to store and transform as well the values of the normals to the surface at each point. We already computed them for the vertices of the mesh, so we only need to compute the ones for
the new points using barycentric coordinates.
In Figure 10 we show the ink prediction from the Phase 1 Grand Prize winning model ^8 on the stack of rendered layers obtained from the segment displayed in Figure 7 - the composite image is shown in
Figure 9.
Figure 10. Ink prediction on the layers rendered from the UV in Figure 7.
Additional notes
Please notice that every single one of the previously mentioned steps can be improved, both in terms of mesh generation quality and computational efficiency. For further information on Thaumato
Anakalyptor and its roadmap, please read its official Technical Report. | {"url":"https://scrollprize.org/tutorial4","timestamp":"2024-11-10T17:07:56Z","content_type":"text/html","content_length":"40773","record_id":"<urn:uuid:33e1e0cb-1161-4746-8ef1-dbe4d8c55105>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00018.warc.gz"} |
Euclid Book VI
These proofs can all be found in Book $\text {VI}$ of Euclid's The Elements.
The specific form of these has been adapted from Sir Thomas L. Heath's Euclid: The Thirteen Books of The Elements: Volume 2, 2nd ed.
The following 36 pages are in this category, out of 36 total. | {"url":"https://proofwiki.org/wiki/Category:Euclid_Book_VI","timestamp":"2024-11-02T17:36:56Z","content_type":"text/html","content_length":"46248","record_id":"<urn:uuid:ea00a2a2-bb0f-429e-a03c-f54312b224c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00616.warc.gz"} |
Differential equation - Wikipedia Republished // WIKI 2
In mathematics, a differential equation is an equation that relates one or more unknown functions and their derivatives.^[1] In applications, the functions generally represent physical quantities,
the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Such relations are common; therefore, differential equations play a prominent
role in many disciplines including engineering, physics, economics, and biology.
The study of differential equations consists mainly of the study of their solutions (the set of functions that satisfy each equation), and of the properties of their solutions. Only the simplest
differential equations are solvable by explicit formulas; however, many properties of solutions of a given differential equation may be determined without computing them exactly.
Often when a closed-formexpression for the solutions is not available, solutions may be approximated numerically using computers. The theoryofdynamicalsystems puts emphasis on qualitative
analysis of systems described by differential equations, while many numericalmethods have been developed to determine solutions with a given degree of accuracy.
YouTube Encyclopedic
• 1/5
• Differential equations, a tourist's guide | DE1
• Physical Models and their Differential Equations
• Differential Equation - Introduction (5 of 15) Real Example of a Differential Equation - 1
• What are differential equations?
• Differential equation introduction | First order differential equations | Khan Academy
Differential equations came into existence with the inventionofcalculus by IsaacNewton and GottfriedLeibniz. In Chapter 2 of his 1671 work MethodusfluxionumetSerierumInfinitarum,^[2] Newton
listed three kinds of differential equations:
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=f(x)\\[4pt]{\frac {dy}{dx}}&=f(x,y)\\[4pt]x_{1}{\frac {\partial y}{\partial x_{1}}}&+x_{2}{\frac {\partial y}{\partial x_{2}}}=y\end{aligned}}}
In all these cases, y is an unknown function of x (or of x[1] and x[2]), and f is a given function.
He solves these examples and others using infinite series and discusses the non-uniqueness of solutions.
JacobBernoulli proposed the Bernoullidifferentialequation in 1695.^[3] This is an ordinarydifferentialequation of the form
${\displaystyle y'+P(x)y=Q(x)y^{n}\,}$
for which the following year Leibniz obtained solutions by simplifying it.^[4]
Historically, the problem of a vibrating string such as that of a musicalinstrument was studied by JeanleRondd'Alembert, LeonhardEuler, DanielBernoulli, and Joseph-LouisLagrange.^[5]^[6]^[7]^[
8] In 1746, d’Alembert discovered the one-dimensional waveequation, and within ten years Euler discovered the three-dimensional wave equation.^[9]
The Euler–Lagrangeequation was developed in the 1750s by Euler and Lagrange in connection with their studies of the tautochrone problem. This is the problem of determining a curve on which a
weighted particle will fall to a fixed point in a fixed amount of time, independent of the starting point. Lagrange solved this problem in 1755 and sent the solution to Euler. Both further developed
Lagrange's method and applied it to mechanics, which led to the formulation of Lagrangianmechanics.
In 1822, Fourier published his work on heatflow in Théorie analytique de la chaleur (The Analytic Theory of Heat),^[10] in which he based his reasoning on Newton'slawofcooling, namely, that the
flow of heat between two adjacent molecules is proportional to the extremely small difference of their temperatures. Contained in this book was Fourier's proposal of his heatequation for conductive
diffusion of heat. This partial differential equation is now a common part of mathematical physics curriculum.
In classicalmechanics, the motion of a body is described by its position and velocity as the time value varies. Newton'slaws allow these variables to be expressed dynamically (given the position,
velocity, acceleration and various forces acting on the body) as a differential equation for the unknown position of the body as a function of time.
In some cases, this differential equation (called an equationofmotion) may be solved explicitly.
An example of modeling a real-world problem using differential equations is the determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The
ball's acceleration towards the ground is the acceleration due to gravity minus the deceleration due to air resistance. Gravity is considered constant, and air resistance may be modeled as
proportional to the ball's velocity. This means that the ball's acceleration, which is a derivative of its velocity, depends on the velocity (and the velocity depends on time). Finding the velocity
as a function of time involves solving a differential equation and verifying its validity.
Differential equations can be divided into several types. Apart from describing the properties of the equation itself, these classes of differential equations can help inform the choice of approach
to a solution. Commonly used distinctions include whether the equation is ordinary or partial, linear or non-linear, and homogeneous or heterogeneous. This list is far from exhaustive; there are many
other properties and subclasses of differential equations which can be very useful in specific contexts.
Ordinary differential equations
An ordinarydifferentialequation (ODE) is an equation containing an unknown functionofonerealorcomplexvariable x, its derivatives, and some given functions of x. The unknown function is
generally represented by a variable (often denoted y), which, therefore, depends on x. Thus x is often called the independentvariable of the equation. The term "ordinary" is used in contrast with
the term partialdifferentialequation, which may be with respect to more than one independent variable.
Lineardifferentialequations are the differential equations that are linear in the unknown function and its derivatives. Their theory is well developed, and in many cases one may express their
solutions in terms of integrals.
Most ODEs that are encountered in physics are linear. Therefore, most specialfunctions may be defined as solutions of linear differential equations (see Holonomicfunction).
As, in general, the solutions of a differential equation cannot be expressed by a closed-formexpression, numericalmethods are commonly used for solving differential equations on a computer.
Partial differential equations
A partialdifferentialequation (PDE) is a differential equation that contains unknown multivariablefunctions and their partialderivatives. (This is in contrast to ordinarydifferentialequations,
which deal with functions of a single variable and their derivatives.) PDEs are used to formulate problems involving functions of several variables, and are either solved in closed form, or used to
create a relevant computermodel.
PDEs can be used to describe a wide variety of phenomena in nature such as sound, heat, electrostatics, electrodynamics, fluidflow, elasticity, or quantummechanics. These seemingly distinct
physical phenomena can be formalized similarly in terms of PDEs. Just as ordinary differential equations often model one-dimensional dynamicalsystems, partial differential equations often model
multidimensionalsystems. Stochasticpartialdifferentialequations generalize partial differential equations for modeling randomness.
Non-linear differential equations
A non-linear differential equation is a differential equation that is not a linearequation in the unknown function and its derivatives (the linearity or non-linearity in the arguments of the
function are not considered here). There are very few methods of solving nonlinear differential equations exactly; those that are known typically depend on the equation having particular symmetries.
Nonlinear differential equations can exhibit very complicated behaviour over extended time intervals, characteristic of chaos. Even the fundamental questions of existence, uniqueness, and
extendability of solutions for nonlinear differential equations, and well-posedness of initial and boundary value problems for nonlinear PDEs are hard problems and their resolution in special cases
is considered to be a significant advance in the mathematical theory (cf. Navier–Stokesexistenceandsmoothness). However, if the differential equation is a correctly formulated representation of a
meaningful physical process, then one expects it to have a solution.^[11]
Linear differential equations frequently appear as approximations to nonlinear equations. These approximations are only valid under restricted conditions. For example, the harmonicoscillator
equation is an approximation to the nonlinear pendulum equation that is valid for small amplitude oscillations.
Equation order and degree
The order of the differential equation is the highest orderofderivative of the unknown function that appears in the differential equation. For example, an equation containing only
first-orderderivatives is a first-orderdifferentialequation, an equation containing the second-orderderivative is a second-order differential equation, and so on.^[12]^[13]
When it is written as a polynomialequation in the unknown function and its derivatives, its degree of the differential equation is, depending on the context, the polynomialdegree in the highest
derivative of the unknown function,^[14] or its totaldegree in the unknown function and its derivatives. In particular, a lineardifferentialequation has degree one for both meanings, but the
non-linear differential equation ${\displaystyle y'+y^{2}=0}$ is of degree one for the first meaning but not for the second one.
Differential equations that describe natural phenomena almost always have only first and second order derivatives in them, but there are some exceptions, such as the thin-filmequation, which is a
fourth order partial differential equation.
In the first group of examples u is an unknown function of x, and c and ω are constants that are supposed to be known. Two broad classifications of both ordinary and partial differential equations
consist of distinguishing between linear and nonlinear differential equations, and between homogeneousdifferentialequations and heterogeneous ones.
In the next group of examples, the unknown function u depends on two variables x and t or x and y.
Existence of solutions
Solving differential equations is not like solving algebraicequations. Not only are their solutions often unclear, but whether solutions are unique or exist at all are also notable subjects of
For first order initial value problems, the Peanoexistencetheorem gives one set of circumstances in which a solution exists. Given any point ${\displaystyle (a,b)}$ in the xy-plane, define some
rectangular region ${\displaystyle Z}$, such that ${\displaystyle Z=[l,m]\times [n,p]}$ and ${\displaystyle (a,b)}$ is in the interior of ${\displaystyle Z}$. If we are given a differential equation
${\textstyle {\frac {dy}{dx}}=g(x,y)}$ and the condition that ${\displaystyle y=b}$ when ${\displaystyle x=a}$, then there is locally a solution to this problem if ${\displaystyle g(x,y)}$ and ${\
textstyle {\frac {\partial g}{\partial x}}}$ are both continuous on ${\displaystyle Z}$. This solution exists on some interval with its center at ${\displaystyle a}$. The solution may not be unique.
(See Ordinarydifferentialequation for other results.)
However, this only helps us with first order initialvalueproblems. Suppose we had a linear initial value problem of the nth order:
${\displaystyle f_{n}(x){\frac {d^{n}y}{dx^{n}}}+\cdots +f_{1}(x){\frac {dy}{dx}}+f_{0}(x)y=g(x)}$
such that
{\displaystyle {\begin{aligned}y(x_{0})&=y_{0},&y'(x_{0})&=y'_{0},&y''(x_{0})&=y''_{0},&\ldots \end{aligned}}}
For any nonzero ${\displaystyle f_{n}(x)}$, if ${\displaystyle \{f_{0},f_{1},\ldots \}}$ and ${\displaystyle g}$ are continuous on some interval containing ${\displaystyle x_{0}}$, ${\displaystyle y}
$ exists and is unique.^[15]
Related concepts
Connection to difference equations
The theory of differential equations is closely related to the theory of differenceequations, in which the coordinates assume only discrete values, and the relationship involves values of the
unknown function or functions and values at nearby coordinates. Many methods to compute numerical solutions of differential equations or study the properties of differential equations involve the
approximation of the solution of a differential equation by the solution of a corresponding difference equation.
The study of differential equations is a wide field in pure and appliedmathematics, physics, and engineering. All of these disciplines are concerned with the properties of differential equations of
various types. Pure mathematics focuses on the existence and uniqueness of solutions, while applied mathematics emphasizes the rigorous justification of the methods for approximating solutions.
Differential equations play an important role in modeling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons.
Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closedform solutions. Instead, solutions can be approximated using
Many fundamental laws of physics and chemistry can be formulated as differential equations. In biology and economics, differential equations are used to model the behavior of complex systems. The
mathematical theory of differential equations first developed together with the sciences where the equations had originated and where the results found application. However, diverse problems,
sometimes originating in quite distinct scientific fields, may give rise to identical differential equations. Whenever this happens, mathematical theory behind the equations can be viewed as a
unifying principle behind diverse phenomena. As an example, consider the propagation of light and sound in the atmosphere, and of waves on the surface of a pond. All of them may be described by the
same second-order partialdifferentialequation, the waveequation, which allows us to think of light and sound as forms of waves, much like familiar waves in the water. Conduction of heat, the
theory of which was developed by JosephFourier, is governed by another second-order partial differential equation, the heatequation. It turns out that many diffusion processes, while seemingly
different, are described by the same equation; the Black–Scholes equation in finance is, for instance, related to the heat equation.
The number of differential equations that have received a name, in various scientific areas is a witness of the importance of the topic. See Listofnameddifferentialequations.
Some CAS software can solve differential equations. These are the commands used in the leading programs:
See also
1. ^ Dennis G. Zill (15 March 2012). AFirstCourseinDifferentialEquationswithModelingApplications. Cengage Learning. ISBN 978-1-285-40110-2.
2. ^ Newton, Isaac. (c.1671). Methodus Fluxionum et Serierum Infinitarum (The Method of Fluxions and Infinite Series), published in 1736 [Opuscula, 1744, Vol. I. p. 66].
3. ^ Bernoulli,Jacob (1695), "Explicationes, Annotationes & Additiones ad ea, quae in Actis sup. de Curva Elastica, Isochrona Paracentrica, & Velaria, hinc inde memorata, & paratim controversa
legundur; ubi de Linea mediarum directionum, alliisque novis", ActaEruditorum
4. ^ Hairer, Ernst; Nørsett, Syvert Paul; Wanner, Gerhard (1993), Solving ordinary differential equations I: Nonstiff problems, Berlin, New York: Springer-Verlag, ISBN 978-3-540-56670-0
5. ^ Frasier, Craig (July 1983). "ReviewofTheevolutionofdynamics,vibrationtheoryfrom1687to1742,byJohnT.CannonandSigaliaDostrovsky" (PDF). Bulletin of the American Mathematical
Society. New Series. 9 (1).
6. ^ Wheeler, Gerard F.; Crummett, William P. (1987). "The Vibrating String Controversy". Am.J.Phys. 55 (1): 33–37. Bibcode:1987AmJPh..55...33W. doi:10.1119/1.15311.
7. ^ For a special collection of the 9 groundbreaking papers by the three authors, see
FirstAppearanceofthewaveequation:D'Alembert,LeonhardEuler,DanielBernoulli.-thecontroversyaboutvibratingstrings Archived 2020-02-09 at the WaybackMachine (retrieved 13 Nov 2012).
Herman HJ Lynge and Son.
8. ^ For de Lagrange's contributions to the acoustic wave equation, can consult Acoustics:AnIntroductiontoItsPhysicalPrinciplesandApplications Allan D. Pierce, Acoustical Soc of America,
1989; page 18.(retrieved 9 Dec 2012)
9. ^ Speiser, David. DiscoveringthePrinciplesofMechanics1600-1800, p. 191 (Basel: Birkhäuser, 2008).
10. ^ Fourier, Joseph (1822). Théorieanalytiquedelachaleur (in French). Paris: Firmin Didot Père et Fils. OCLC 2688081.
11. ^ Boyce, William E.; DiPrima, Richard C. (1967). Elementary Differential Equations and Boundary Value Problems (4th ed.). John Wiley & Sons. p. 3.
12. ^ Weisstein,EricW. "Ordinary Differential Equation Order." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/OrdinaryDifferentialEquationOrder.html
13. ^ Orderanddegreeofadifferentialequation Archived 2016-04-01 at the WaybackMachine, accessed Dec 2015.
14. ^ Elias Loomis (1887). ElementsoftheDifferentialandIntegralCalculus (revised ed.). Harper & Bros. p. 247.Extractofpage247
15. ^ Zill, Dennis G. (2001). A First Course in Differential Equations (5th ed.). Brooks/Cole. ISBN 0-534-37388-7.
16. ^ "dsolve-MapleProgrammingHelp". www.maplesoft.com. Retrieved 2020-05-09.
17. ^ "DSolve-WolframLanguageDocumentation". www.wolfram.com. Retrieved 2020-06-28.
18. ^ Schelter,WilliamF. Gaertner, Boris (ed.). "DifferentialEquations-SymbolicSolutions". The Computer Algebra Program Maxima - a Tutorial (in Maxima documentation on SourceForge). Archived
from the original on 2022-10-04.
19. ^ "BasicAlgebraandCalculus—SageTutorialv9.0". doc.sagemath.org. Retrieved 2020-05-09.
20. ^ "ODE". SymPy 1.11 documentation. 2022-08-22. Archived from the original on 2022-09-26.
Further reading
External links
This page was last edited on 20 August 2024, at 15:16 | {"url":"https://wiki2.org/en/Differential_equation","timestamp":"2024-11-05T03:22:04Z","content_type":"application/xhtml+xml","content_length":"226419","record_id":"<urn:uuid:a09deea9-53dc-48c2-9d11-aedd90d6af77>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00025.warc.gz"} |
<p><code>pow</code> Confusion</p>
In my first summer of graduate school my code suddenly stopped working because Fortran and Python (via pow() in C) do exponentiation differently. Once I debugged and understood the problem, I learned
about the highly optimized assembly code produced by Fortran for integer exponents.
To give a sample of the tool I wrote, here is an analysis of the assembly generated by the gfortran compiler for b = a**5:
$ python ./describe_assembly.py --exponent 5
| %xmm0[:64] | %xmm1[:64] |
| a | | f2 0f 10 07 movsd [(%rdi), %xmm0 ]
| a | a | 66 0f 28 c8 movapd [ %xmm0, %xmm1 ]
| a | a * a | f2 0f 59 c8 mulsd [ %xmm0, %xmm1 ]
| (a * a) * a | a * a | f2 0f 59 c1 mulsd [ %xmm1, %xmm0 ]
| (a * a) * ((a * a) * a) | a * a | f2 0f 59 c1 mulsd [ %xmm1, %xmm0 ]
| b | | f2 0f 11 06 movsd [ %xmm0, (%rsi)]
In my first summer of graduate school, I had a very productive June. I was working on a research project^1 on optimization of triangular meshes for compression and storage. A core part of this
involved taking fourth powers of floating point numbers^2. Once I reached a point where my slow Python code worked well enough, I started porting to Fortran 95. I foolishly expected bit-for-bit
identical behavior but instead the changes — which I later learned were exclusively from the fourth power operation — were so dramatic I lost a full week of productivity debugging the existing test
As with most optimization problems, this involved applying gradient descent^3 to a number of handpicked objective functions. Since this involved triangles and compression, we were trying to modify
the mesh^4 by "binning" triangles into similar shapes. For our early objective functions, this could lead to flattened out triangles that looked more like lines than triangles. Luckily there exist a
number of ways to measure the "quality" $q$ of a triangle as a number between $0$ and $1$; a very low quality corresponds exactly to this flatness. To penalize low-quality triangles we included
functions of $1 / q$ in our objective functions and several of the most successful used a loop over all triangles $T$ in the mesh $\mathcal{M}$
$\sum_{T \in \mathcal{M}} \frac{1}{q(T)^4}$
What's the Difference?
When the exponent n is an integer, Fortran x**n behaves much differently than Python. In Fortran, optimized assembly is generated to minimize the number of multiplications, whereas Python calls out
to pow() from math.h in C. Compare an explicit vs. implicit implementation in Python:
def fourth_explicit(value):
squared = value * value
return squared * squared
def fourth_pow(value):
return value ** 4
For well-behaved values, these two functions compute $1 / q^4$ without much difference
>>> 1 / fourth_explicit(0.25)
>>> 1 / fourth_pow(0.25)
>>> quality = 0.2689565746627065
>>> 1 / fourth_explicit(quality)
>>> 1 / fourth_pow(quality)
however when $q$ becomes small (the type of values we are penalizing) the magnitude of $1 / q^4$ means that small relative differences are still large absolute differences in our objective function:
>>> quality = 2.3824755061912883e-06
>>> 1 / fourth_explicit(quality)
>>> 1 / fourth_pow(quality)
>>> 1 / fourth_explicit(quality) - 1 / fourth_pow(quality)
The difference is caused by the fact that pow() treats everything like a floating point number and uses exp(), log() and multiplication via
Fortran's Nifty Algorithm
Breaking down the assembly for b = a**4 in Fortran we see the repeated squaring
movsd xmm0, QWORD PTR [rdi]
mulsd xmm0, xmm0
mulsd xmm0, xmm0
movsd QWORD PTR [rsi], xmm0
as compared to a jump (jmp) instruction in C
movsd xmm1, QWORD PTR .LC0[rip]
jmp pow
.long 0
.long 1074790400
Similarly for q**7 the assembly shows only multiplications (vs. a jmp)
movsd xmm0, QWORD PTR [rdi]
movapd xmm1, xmm0
mulsd xmm1, xmm0
mulsd xmm0, xmm1
mulsd xmm1, xmm1
mulsd xmm0, xmm1
movsd QWORD PTR [rsi], xmm0
I Don't Speak Assembly
Like most professional software engineers (and science PhDs who write code) I have no formal training, so reading assembly is something I've never done. So I wrote a script to generate a Fortran
subroutine that computes b = a**n for a fixed value of n, compiles the code to an object file and then disassembles it. For example:
$ gfortran -c -O3 ./pow7.f90 -o ./pow7.o -J ./
$ objdump --disassemble ./pow7.o
However, these instructions are still hard to visualize so I parsed the disassembled instructions from the object file and tracked all of the active registers (%xmm{N}) as they were updated. In
addition to the xmm registers, the %rsi (source) and %rdi (destination) registers are used for copying data from a and to b after the computation is done.
Being able to visualize this really helped! For n = 4 we can see the generated assembly is efficient enough that it only requires one register
$ python ./describe_assembly.py --exponent 4
| %xmm0[:64] |
| a | f2 0f 10 07 movsd [(%rdi), %xmm0 ]
| a * a | f2 0f 59 c0 mulsd [ %xmm0, %xmm0 ]
| (a * a) * (a * a) | f2 0f 59 c0 mulsd [ %xmm0, %xmm0 ]
| b | f2 0f 11 06 movsd [ %xmm0, (%rsi)]
and n = 7 can do all of its work in two registers
$ python ./describe_assembly.py --exponent 7
| %xmm0[:64] | %xmm1[:64] |
| a | | f2 0f 10 07 movsd [(%rdi), %xmm0 ]
| a | a | 66 0f 28 c8 movapd [ %xmm0, %xmm1 ]
| a | a * a | f2 0f 59 c8 mulsd [ %xmm0, %xmm1 ]
| (a * a) * a | a * a | f2 0f 59 c1 mulsd [ %xmm1, %xmm0 ]
| (a * a) * a | (a * a) * (a * a) | f2 0f 59 c9 mulsd [ %xmm1, %xmm1 ]
| ((a * a) * (a * a)) * ((a * a) * a) | (a * a) * (a * a) | f2 0f 59 c1 mulsd [ %xmm1, %xmm0 ]
| b | | f2 0f 11 06 movsd [ %xmm0, (%rsi)]
PS: It's worth noting here that black box analysis of the generated assembly is not the only way to understand what gfortran will do. I attempted to dive into the source for gfortran to
understand exactly how this is generated (in particular if it differs when the exponent n is not known until runtime). However, I was not able to find any conclusive section of code responsible
and ran out of time digging. | {"url":"https://blog.bossylobster.com/2020/11/pow-confusion.html","timestamp":"2024-11-10T01:25:01Z","content_type":"text/html","content_length":"35786","record_id":"<urn:uuid:60fa6561-2a39-4add-b436-21bed287f19f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00523.warc.gz"} |
seminars - State estimation for a high-dimensional nonlinear system by particle-based filtering methods
The performance of the particle-based schemes is compared with the convergent results from sequential Importance resampling method( SIR) based on measurement errors, observation locations, and
particle sizes in various sets of twin experiments. The sensitivity analysis shows strength and weakness of each filtering method when applied to multimodal non-linear systems. As the number of
particles is increased, SIR achieves the convergent results that are mathematically optimal solutions. Ensemble Kalman Filter (EnKF) shows suboptimal results regardless of sample sizes, and the
Maximum Entropy Filter (MEF) achieves the optimal solution even with a small sample size. Both EnKF, and MEF produces robust results with a relatively small sample size or increased measurement
locations. Small measurement errors or short intervals of observations (or, more frequent observations) significantly improve the performances of SIR and EnKF, and MEF still show robust results even
with a relatively small sample size or sparse measurement locations when the system experiences the transition between one region to the other region. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&page=45&l=en&document_srl=803449&sort_index=date&order_type=desc","timestamp":"2024-11-09T23:10:52Z","content_type":"text/html","content_length":"48700","record_id":"<urn:uuid:2b2ba219-f958-4786-acbc-8e76fe6649ba>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00398.warc.gz"} |
Guidelines for the Design of Small Bridges and Culvert Draft 2020
(Draft as on August, 2020)
SP 13
Special Publication No. 13 “ Guidelines for the Design of Small Bridges and
Culverts” was published in 1973, based on the paper entitled “ Guidelines for
Design of Small Bridges and Culverts” by Shri Goverdhan Lall, Additional Director
General (Roads) (Retd.) Ministry of Road Transport & Highways Government of
India. It is sincerely acknowledged that this Special Publication has been serving as
a text book and as one of main sources of knowledge to bridge engineers specially
with regard to discharge calculations.
The first revision of this Special Publication was taken up in 1994 under the
Chairmanship of Shri C. R. Alimchandani. The work was continued by the
reconstituted Committee of General design Features (B 2) in 2000. The revised
draft was prepared by Mr C V Kand who was helped by S/ Shri K S Jangde, S M
Sabnis, A S Khare and S P Bodhe of earlier Committee. The draft was discussed by
the reconstituted B 2 Committee and was given the final shape by a Sub
committee comprising Dr B P Bagish, S/ Shri A D Narain, G S Taunk, Ashok Basa,
and S K Nirmal. As per directions of the Council and BSS committee, the draft
document was further reviewed by Shri G Sharan, Secretary IRC and Convener B 2
Committee and was finally published in 2004.
The present revision has been taken up by IRC General Design Features
Committee (B 1) in the year 2018. The draft of this revised version has been
prepared by a Subcommittee of the B 1 Committee comprising Shri G Sharan,
(Chairman) , Dr B P Bagish, Shri Ashok Basa, Shri H C Arora, Shri Dinesh Kumar and
Shri P V Mayur. The draft was approved by B1 Committee in the meeting held on
15th Feb 2020 and authorized the Convener to incorporate any comments
received from Members. The document has been finalized by S/ Shri N K Sinha,
Convener, G Sharan and P V Mayur, and presented to IRC.
List of Plates
Personnel of the Bridges Specifications & Standards Committee
Article 1
General Aspects
Article 2
Site Selection & Inventory
Article 3
Collection of Design Data
Article 4
Empirical and Rational Formulae for Peak Run-off from Catchment 13
Article 5
Estimating Flood Discharge from the Conveyance
Factor and Slope of the Stream
Article 6
Design Discharge
Article 7
Alluvial Streams Lacey's Equations
Article 8
Linear Waterway
Article 9
Normal Scour Depth of Streams
Article 10
Maximum Scour Depth
Article 11
Depth of Foundations
Article 12
Span and Vertical Clearance
Article 13
Geometric Standards, Specifications and Quality Control
Article 14
Structural Details of Small Bridges and Culverts
Article 15
Elements of the Hydraulics of Flow through Bridges
Article 16
Article 17
Worked out Examples on Discharge Passed by Existing Bridges from
Flood Marks
Article 18
Overtopping of the Banks
Article 19
Pipes and Box Culverts
Article 20
Protection Work and Maintenance
Article 21
Raft Foundations
Article 22
CD. Works in Black Cotton Soils
Article 23
Box Cell Structures
Filling Behind Abutments, Wing and Return Walls
Plate No.
Chart for Time of Concentration
Run-off Chart for Small Catchments
Hydraulic Mean Depth R (METRES)
Typical Method of Determination of Weighted Mean Diameter of Particles (dm)
Abutment and Wing Wall Sections for Culverts
Details of segmental Masonry Arch Bridges without Footpaths- Effective Span
6m and 9m
7. Typical Details of Floor Protection Works for Box Cell Structures -General
8. RCC Pipe Culvert with Single Pipe of 1.2 Metre Dia and Concrete Cradle
Bedding for Heights of Fill Varying from 4.0 m to 8.0 m
9. RCC Pipe Culvert with Single Pipe of 1.2 Metre Dia and First class Bedding for
Heights of Fill Varying from 0.6 m-4.0 m
10. RCC Pipe Culvert with 2 Pipes of 1.2 Metre dia and Concrete Cradle Bedding for
Heights of Fill Varying from 4.0-8.0 m
11. RCC Pipe Culvert with 2 Pipes of 1.2 Metre dia and First class Bedding for
Heights of Fill Varying from 0.6-4.0 m
12. Circular and Rectangular Pipes Flowing Full
(As on 00.00.2020)
Adoption of Limit State Design instead of Working Stress method
Design of concrete small bridges and culverts as per IRC 112
Enhanced Design life of culverts and small bridges
Return periods of floods for design of culverts and small bridges
Adoption of Multi – lane bridges and culverts
Withdrawal of standard drawings included in the Revision 1, as these are
based on obsolete codes.
Precast construction for accelerated construction.
A large number of small bridges and culverts form part of most of our highways. For the
implementation of massive road development programmes, it is necessary to update the
changes with regard to various codes of practice of Indian Roads Congress, so as to
standardise the design methodology of such structures and to reduce the time spent on
project preparation and construction. With this objective in view, the Special Publication
No. 13 “Guidelines for Design of Small Bridges and Culverts” has been updated taking
into account a number of major changes in the provisions therein. Important features are
brought out below:
ARTICLE 1
1.1. General: Occurrence of culverts and small bridges on roads and highways depends
upon the type of region and terrain. The location, size and other details of such structures
should be decided judiciously to cater for the requirements of discharge and balancing of
water level on either side of road embankment. Number of culverts in 1 km length of road
in India varies from one (flat country) to three in undulating regions whereas one small
bridge (upto 30 m) is found within 1 to 4 km length of the road. Number of culverts in
hilly/undulating terrain Is generally more than in plain region.
1.2. Definitions
1.2 1. Bridge: A bridge is a structure having a total length above 6 m for carrying traffic or
other moving loads across a channel, depression, road or railway track or any other
1.2.2. Minor Bridge: A minor bridge is a bridge having a total length of upto 60 m.
1.2.3. Small Bridge: A small bridge is a bridge where the overall length of the bridge
between the inner faces of dirt walls is upto 30 m and individual span is not more than 10
1.2.4. Culvert: Culvert is a structure having a total length of upto 6 m between the outer
faces of walls measured at right angles. Cross drainage structures with pipes will be
termed as culvert, irrespective of length.
1.3 The Small Bridges and Culverts can be of following types:
a) RCC Hume Pipes
b) RCC slab on masonry/concrete abutment and piers
c) Stone slab on masonry/concrete abutment and piers
d) RCC box cell structure
e) RCC/masonry arches on masonry/concrete abutment and piers
Stone slabs can be used upto 2 m span when good quality stones having 200 mm
thickness are available.
1.4 Design philosophy
1.4.1 GENERAL
Small bridges and Culverts are as important in the infrastructure as a major
bridge. Each structural component of the small bridge / culvert shall therefore
satisfy the requirements of design as spelt out in various IRC codes and
standards as applicable.
Serviceability Limit State (SLS): The serviceability limit state shall be
considered as per IRC:112.
Ultimate Strength Limit State (ULS): Strength limit state shall be considered as
per IRC:112
Accidental Limit States: The accidental limit state shall consider as per IRC:112
Bridges shall be designed for specified limit states to achieve the objectives of
constructability, safety, serviceability, with due regard to issue of economy,
aesthetics and sustainability. Provision of IRC:112 shall be followed for design of
concrete structures. Sizes of opening in culverts shall be fixed in a manner which
provides sufficient space for inspection and maintenance and to avoid clogging.
The limit states specified herein are intended to provide for a constructible,
serviceable bridge, capable of safely carrying design loads for a specified lifetime
as given in IRC:5. Small bridges and culverts need not be checked for seismic
1.4.3 Riding Quality
In order to improve the riding quality, number of expansion joints should be
minimized. It is preferable to go for Integral structures / continuous structures.
1.5 Standard Designs
1.5.1 Standard designs drawings for slab bridges and RCC Boxes published by Ministry
of Road Transport & Highways are presently under revision because of change in design
philosophy from working stress method to limit state design besides incorporating
changes in the codal provisions. Hence the revised drawings of slab bridges and RCC
boxes, whenever published, shall be applicable for adoption.
1.3.11.5.2 Mean while parameters of loadings for design of small bridges and culverts are
furnished below:
1.4. H.P. Culverts: RCC pipe culverts having minimum 1200 mm diameter of type NP4
conforming to IS:458. PSC pipes of NP4 type conforming to IS: 784 may also be used for
H.P. culverts.
1.5. Provision of bridge and culvert with respect to Catchment Area: It is generally
found that when catchment area is upto 1 sq. km (100 hectares) a culvert is required and
for catchment area more than 1 sq. km (100 hectares), a small bridge will be necessary.
ARTICLE 2
SITE SELECTION & INVENTORY
2.1. Selection of Site: Normally selection of site for culverts and small bridges is
guided by road alignment. However where there is choice, select a site:
(i) Which is situated on a straight reach of stream, sufficiently down stream of bends;
(ii) Which is sufficiently away from the confluence of large tributaries as to be beyond
their disturbing influence;
(iii) Which has well defined banks;
(iv) Which make approach roads feasible on the straight; and
(v) Which offers a square crossing.
2.2. Existing Drainage Structures: If, there is an existing road or railway bridge or culvert
over the same stream within 500 m from the selected site, the best means of ascertaining
the maximum discharge is to calculate it from data collected by personal inspection of the
existing structure. Intelligent inspection and local inquiry will provide very useful
information, namely, marks indicating the maximum flood level, the afflux, the tendency
to scour, the probable maximum discharge, the likelihood of collection of brushwood
during floods, and many other particulars. It should be seen whether the existing structure
is too large or too small or whether it has other defects. The size and other parameters of
existing structures must be recorded. Also whether the road has been overtopped must
be enquired and recorded.
2.3. Inventory should also include taking notes on channel conditions from which the silt
factor and the co-efficient of rugosity can be computed.
ARTICLE 3
Collection of DESIGN DATA
3.1. In addition to the information obtained by inventory of an existing structure, the design
data described in the following paragraphs have to be collected. What is specified here
is sufficient only for small bridges and culverts. For larger structures, detailed instructions
contained in IRC 5 the Standard Specifications & Code of Practice for Bridges – General
Features of Design and IRC SP 54 Project Preparation Manual for Bridges should be
3.2. Catchment Area: When the catchment, as seen from the "topo" (G.T.) sheet, is
less than 1.25 sq. km in area, a traverse should be made along the watershed. Larger
catchments can be read from the 1 cm = 500 m topo maps of the Survey of India by
marking the watershed in pencil and reading the included area by placing a piece of
transparent square paper over it.
Catchment area can be delineated using topo-sheets of appropriate scale developed by
Survey of India. For flat areas where ground elevation difference is very less, catchment
area and other physiographic parameters can be estimated by using digital elevation
model (DEM) and GIS tools. Catchment area may also be calculated by software’s like
BHUVAN developed by ISRO /or by Google Earth. images in Arch-GIS or 3D-Civil
platform which can be used to delineate even small catchment areas.
3.3. Cross-sections: For a sizable stream, at least three cross-sections should be taken
at right angles to the river alignment, namely, one at the selected site, one upstream and
another downstream of the site, all to the horizontal scale of not less than 1 cm to 10 m
or 1/1000 and with an exaggerated vertical scale of not less than 1 cm to 1 m or 1/100.
Approximate distances, upstream and downstream of the selected site of crossing at
which cross-sections should be taken are given in Table 3.1.
Table 3.1
Catchment Area
1. Upto 3.0 sq.km
Distance (u/s and d/s of the crossing) at
which cross-sections should :be taken
100 m
2. From 3.0 to 15 sq. km
300 m
3. Over 15 sq. km
500 m
The cross-section at the proposed site of the crossing should show level at close intervals
and indicate outcrops of rocks, pools, etc. Often an existing road or a cart track crosses
the stream at the site selected for the bridge. In such a case, the cross-section should not
be taken along the center line of the road or the track as that will not represent the natural
shape and size of the channel. The cross-section should be taken at a short distance on
downstream of the selected site.
3.4. In the case of very small streams (catchments of 40 hectares or less) one cross
section may do but it should be carefully plotted so as to represent truly the normal size
and shape of the channel on a straight reach.
3.5. Highest Flood Level: The highest flood level should be ascertained by judicial local
observation, supplemented by local enquiry, and marked on the cross-sections. Design
HFL corresponding to design flood of a given return period can be found from stagedischarge Curve
3.6. Longitudinal Section: The longitudinal section should extend upstream and
downstream of the proposed site for the distances indicated in Table 3.1 and should show
levels of the bed, the low water level and the highest flood level.
3.7. Velocity Observation: Attempts should be made to observe the velocity during an
actual flood and, if that flood is smaller than the Design peak flood, the observed velocity
should be suitably increased. The velocity thus obtained is a good check on the accuracy
of that calculated theoretically. Simplest way is to use a float to find surface velocity (Vs)
. Mean flow velocity is 0.8Vs.
3.8. Trial Pit Sections:
3.8.1 Where the rock or some firm undisturbed soil stratum is not likely to be far below
the alluvial bed of the stream, a trial pit should be dug down to such rock or firm soil. But
if there is no rock or undisturbed firm soil for a great depth below the stream bed level,
then the trial pit may be taken down roughly 2 to 3 meter below the lowest bed level. The
location of each trial pit should be shown in the cross-section of the proposed site. The
trial pit section should be plotted to show the kind of soils passed through. However depth
of trial pit in soils shall be minimum 2 m for culverts and 3 m for small bridges.
For more detailed investigation procedure given in IRC: 78 may be referred to.
3.8.2. For Pipe culverts, one trial pit is sufficient. The result should be inserted on the
cross-section. '
3.9 Return period or frequency of peak run off
Return periods or frequency of peak run off for design of small bridges and culverts shall
be considered as under:
Small bridges – 100 years
Culverts -- 25 years to 50 years depending on importance of road
3.10 Iso-Pluvial Maps: Iso-pluvial maps prepared jointly by CWC, RDSO, IMD & MORTH
and published by CWC for different regions titled “Flood Estimation Report” give 24 hr.
rainfall value of different return periods e.g. 25,50 and 100 yr. return period. Table-4.1
can be used to convert 24 hr. rainfall of given frequency (to be read from Iso-Pluvial maps)
to rainfall corresponding to time of concentration to be determined for the given
ARTICLE 4
4. 1 . Although records of rainfall exist to some extent, actual records of floods are seldom
available in such sufficiency as to enable the engineer accurately to infer the worst flood
conditions for which provision should be made in designing a bridge. Therefore, recourse
has to be taken to theoretical computations. In this Article some of the most popular
empirical formulae are mentioned.
4.2. Dickens Formula
Q = CM3/4
…… (4.1)
the peak run-off in m3/s and M is the catchment area in sq. km
11 - 14 where the annual rainfall is 60 - 120 cm
14-19 where the annual rainfall is more than 120 cm
22 in Western Ghats
Ryve's Formula : This formula was devised for erstwhile Madras Presidency.
Q = CM2/3
…… (4.2)
Run-off in m3/s and M is the catchment area in sq. km
6.8 for areas within 25 km of the coast
8.5 for areas between 25 km and 160 km of the coast
10.0 for limited areas near the hills
4.4. Ingli's Formula: This empirical formula was devised for erstwhile Bombay
๐ ธ=
๐ ๐ ๐ ๐ ด
……. (4.3)
√๐ ด+๐ ๐
Q = maximum flood discharge in m3/s
M = the area of the catchment in sq.km
4.5. These empirical formulae involve only one factor viz. the area of the catchment and
all the so many other factors that affect the run-off have to be taken care of in selecting
an appropriate value of the co-efficient. This is extreme simplification of the problem and
cannot be expected to yield accurate results.
4.6. A correct value of C can only be derived for a given region from an extensive
analytical study of the measured flood discharges vis-a-vis catchment areas of streams
in the region. Any value of C will be valid only for the region for which it has been
determined in this way. Each basin has its own singularities affecting run-off. Since actual
flood records are seldom available, the formulae leave much to the judgment of the
engineer. Also, since the formulae do not consider rainfall, they are unreliable. Many other
similar empirical formulae are in use but none of them encompasses all possible
conditions of terrain and climate.
4.7. Rational Formulae for Peak Run-off from Catchment
4.7. 1. In recent years, hydrological studies have been made and theories set forth which
comprehend the effect of the characteristics of the catchment on run-off. Attempts also
have been made to establish relationships between rainfall and run-off under various
circumstances. Some elementary account of the rationale of these theories is given in the
following paragraphs.
4.7.2. Main factors: The size of the flood depends on the following major factors.
Distribution in time and space
Nature of Catchment
Permeability of the soil and vegetable cover
Initial state of wetness
4.7.3. Relation between the intensity and duration of a storm: Suppose in an
individual storm, F cm of rain falls in T hours, then over the whole interval of time T, the
mean intensity ๐ ผ will be F/T cm per hour. Now, within the duration T, imagine a smaller
time interval t (Fig. 4.1). Since the intensity is not uniform through-out, the mean intensity
reckoned over the time interval t (placed suitably within T) will be higher than the mean
intensity i.e. ๐ ผ taken over the whole period. Mean intensity i.e.I=F/T in cm/hr is less than
Ic which is the critical intensity of rainfall in cm/hr corresponding to time of concentration,
tc in hrs
It is also known that the mean intensity of a storm of shorter duration can be higher than
that of a prolonged one.
In other words, the intensity of a storm is some inverse function of its duration. It has
been reasonably well established that
๐ ผ
๐ +๐ ถ
……. (4.5a)
๐ ก+๐
Where c is a constant
Analysis of rainfall statistics has shown that for all but extreme cases, c = 1 [5] * when
time is measured in hours and precipitation in cm.
๐ ผ
๐ +1
……. (4.5b)
๐ ก+1
๐ +1
๐ ๐ ๐ = ๐ ผ(
๐ น
๐ ก+1
๐ +1
๐ ก+1
……. (4.5c)
……. (4.5d)
Thus, if the total precipitation F and duration T of a storm are known then the intensity
corresponding to t, which is a time interval within the duration of the storm can be
* Refers to the number of the publication in the Bibliography.
4.7.4. For an appreciation of the physical significance of this relationship, some typical
cases are considered below.
๐ น
๐ +1
๐ ก+1
Take an intense but brief storm which drops (say) 5 cm of rain in 20 minutes. The average
intensity comes to 15 cm per hour. For a short interval t of, say 6 minutes, within the
duration of the storm the intensity can be as high as
=18.2 cm per hour
Storms of very short duration and 6 minute intensities within them (and, in general, all
such high but momentary intensities of rainfall) have little significance in connection with
the design of culverts except in built-up areas where the concentration time can be very
short (see para 4.7.5.1) due to the rapidity of flow from pavements and roofs.
Next consider a region where storms are of medium size and duration. Suppose 15 cm
of rainfalls in 3 hours. The average intensity works out to 5 cm per hour. But in time
interval of one hour within the storm the intensity can be as much as
๐ ผ+1
=10 cm per hour
For the purpose of designing waterway of bridge such a storm is said to be equivalent of
a "one hour rainfall of 10 cm".
Lastly, consider a very wet region of prolonged storms, where a storm drops, say, 1 8 cm
of rain in 6 hours. In a time interval of one hour within the storm the intensity can be as
high as
๐ ผ+1
=10.5 cm per hour
Thus such a storm is equivalent of a "one hour storm of 10.5 cm".
4.7.5 "One-hour rainfall" for a region for designing waterway of bridges/culverts:
Since a bridge should be designed for peak run-off resulting from the severest storm (in
the region) that occurs once in 100 years, following procedure may be adopted for design
of waterway.. Let the total precipitation of that storm be F cm and duration T hours.
Consider a time interval of one hour somewhere within the duration of the storm. The
precipitation in that hour could be as high as
๐ น
๐ +1
๐ น
(1 +
Hence the design of the bridge will be based on a "one-hour rainfall of say I0 cm", where
๐ น
(1 +
... (4.8)
Suppose Fig. 4.1 represents the severest storm experienced in a region. If t represents
one hour, then the shaded area ADBC will represent l0.
It is convenient and common that the storm potential of a region for a given period of
years should be characterized by specifying the "one-hour rainfall" I0 of the region for the
purpose of designing the waterways of bridges in that region.
I has to be determined from F and T of the severest storm. That storm may not necessarily
be the most prolonged storm. The correct procedure for finding I0 is to take a number of
really heavy and prolonged storms and work out / from F and T of each of them. The
maximum of the values of / thus found should be accepted as "one hour rainfall" of the
region for designing bridges.
I0 of a region does not have to be found for each design problem. It is a characteristic of
the whole region and applies to a pretty vast area subject to the same weather conditions.
I0 of a region should be found once for all and should be known to the local engineers.
Heaviest rainfall in mm/hour in current years for a particular place may be obtained from
Meteorological Department of the Government of India.
Start with I0 and then modify it to suit the concentration time (see next para) of the
catchment area in each specific case. This will now be explained.
Since Design flood for bridges is 100 year and that of culvert is 25 to 50 year, rainfall
corresponding to those return periods only should be used for estimating design peak
flood. Design rainfall can be found from Intensity-duration curves for different return
periods provided such curve are available from IMD for the region in which bridge/culvert
is located. However, such curves are rarely available. Iso-Pluvial maps, as mentioned in
para 3.10, can also be used to convert 24 hour rainfall to rainfall corresponding to time
of concentration to be determined for the given catchment.
4.7.5.1. Time of concentration (tc): The time taken by the run-off from the farthest point
on the periphery of the catchment (called the critical point) to'' reach the site of the culvert
is called the "concentration time". In considering the intensity of precipitation it was said
that the shorter the duration considered the higher the intensity will be. Thus safety would
seem to lie in designing for a high intensity corresponding to a very small interval of time.
But this interval should not be shorter than the concentration time of the catchment under
consideration, as otherwise the flow from distant parts of the catchment will not be able
to reach the bridge in time to make its contribution in raising the peak discharge.
Therefore, when examining a particular catchment, only the intensity corresponding to
the duration equal to the concentration period (tc) of the catchment, needs to be
4.7.5.2. Estimating the concentration time of a catchment (tc): The concentration time
depends on (1) the distance from the critical point to the structure; and (2) the average
velocity of flow. The slope, the roughness of the drainage channel and the depth of flow
govern the later. Complicated formulae exist for deriving the time of concentration from
the characteristics of the catchment. For our purpose, however, the following simple
relationship [11] will do
tc = the concentration time in hours
L = the distance from the critical point to the structure in km.
H = the fall in level from the critical point to the structure in m.
L and H can be found from the survey plans of the catchment area and t calculated from
Equation (4.9).
Plate 1 contains graphs from which tc can be directly read for known values of L and H.
4.7.6. The critical or design intensity: The critical intensity for a catchment is that
maximum intensity which can occur in a time interval equal to the concentration time tc of
the catchment during the severest storm (in the region) of a given frequency Ic.
Since each catchment has its own tc it will have its own Ic
If we put t = tc in the basic equation (4.5d) and write Ic for the resulting intensity, we get
Ic =
๐ น
๐ +1
๐ ก๐ +1
Combination this with equation (4.8), we get
4.7.7 Calculation of run-off: A precipitation of Ic cm per hour over an area of A
hectares, will give rise to run-off
Q = 0.028 A Ic m3/s
To account for losses due to absorption etc. introduce a co-efficient P.
Q = 0.028 PAIC
= run-off in m3/s
= area of catchment in hectares
= critical intensity of rainfall in cm per hour
= co-efficient of run-off for the catchment characteristics
The principal factors governing P are: (i) porosity of the soil, (ii) area, shape and size of
the catchment, (iii) vegetation cover, (iv) surface storage viz. existence of lakes and
marshes, and (v) initial state of wetness of the soil. Catchments vary so much with regard
to these characteristics that it is evidently impossible to do more than generalize on the
values of P. Judgment and experience must be used in fixing P. Also see Table 4. 1 for
Table 4.1 Maximum Value of P in the Formula Q = 0.028 PAIc
Steep, bare rock and also city pavements
Rock, steep but wooded
Plateaus, lightly covered
Clayey soils, stiff and bare
-dolightly covered
Loam, lightly cultivated or covered
-dolargely cultivated
Sandy soil, light growth
-docovered, heavy brush
4.798. Relation between intensity and spread of storm: Rainfall recording stations are
points in the space and therefore the intensities recorded there are point intensities.
Imagine an area round a recording station. The intensity will be highest at the center and
will gradually diminish as we go farther away from the center, till at the fringes of the area
covered by the storm, intensity will be zero. The larger the area considered the smaller
would be the mean intensity. It is, therefore, logical to say that the mean intensity is some
inverse function of the size of the area.
If I is the maximum point intensity at the center of the storm then the mean intensity
reckoned over an area "a" is some fraction "f " of I. The fraction f depends on the area “a”
and the relation is represented by the curve in Fig. 4.2 which has been constructed from
statistical analysis [5] .
In hydrological theories it is assumed that the spread of the storm is equal to the area of
the catchment. Therefore in Fig. 4.2 the area “a” is taken to be the same as the area of
the catchment.
The effect of this assumption can lead to errors which, on analysis have been found to be
limited to about 12 per cent [5]
4.7.9. The run-off formula: Introducing the factor ƒ in the Equation 4.11 we get,
... (4.12)
Q = 0.028 P ƒ AIc
In the equation 4.14(a), I0 measures the role played by the clouds of the region and λ
that of the catchment in producing the peak run-off.
It should be clear from the foregoing discussion that the components of λ, are function
of A, L and H of the catchment.
4.7..10 Resume of the Steps for Calculating the Run-Off
Step 1:
Step 2:
Note down A in hectares, L in km and H in metre from the survey maps
of the area.
Estimate Io for the region, preferably from rainfall records failing that from
local knowledge.
Step 4:
Calculate Q = A I o λ
4.7. 1 11. Example: Calculate the run-off for designing a bridge across a stream.
Given Catchment: L = 5 km; H = 30 metre; A = 10 sq. km= 1000 hectare. Loamy soil
largely cultivated.
Rainfall: The severest storm that is known to have occurred in 20 years resulted in 1 5
cm of rain in 2.5 hours.
4.7.12. Run-off curves for small catchment areas (Plate 2) : Suppose the catchment
areas A in hectares and the average slope S of the main drainage channel are known.
Assuming that the length of the catchment is 3 times its width, then both L and H [as
defined in para (4.7.5.2)], can be expressed in terms of A and S and then tc calculated
from equation (4.9).
Also for small areas, ƒ may be taken equal to one, then vide para 4.7.9.
Q = P I0 A
( ๐ ก๐ +1 )
For I0 = 1 cm, the equation becomes,
Q = PA
( ๐ ก๐ +1 )
Hence Q can be calculated for various values of P, A and S. This has been done and
curves plotted in Plate 2.
Plate 2 can be used for small culverts with basins upto 1500 hectares or 1 5 sq. km. The
value of run-off read from Plate 2 are of "One Hour rainfall", I0, of one cm. These values
have to be multiplied by the I0 of the region. An example will illustrate the use of this Plate.
4.7.13. Example: The basin of a stream is loamy soil largely cultivated, and the area of
the catchment is 10 sq. km. The average slope of the stream is 10 per cent. Calculate the
run-off (I0, the one hour rainfall of the region is 2.5 cm).
Use Plate 2. For largely cultivated loamy soil P = 0.3 vide the Table in set in Plate 2.
Enter the diagram at A = 10 sq. km = 1000 hectares; move vertically up to intersect the
slope line of 10 per cent. Then, move horizontally to intersect the OO line join the
intersection with P = 0.3 and extend to the run-off (q) scale and read.
q = 10.2 m3/s
Multiply with I0.
Q= 10.2x2.5 = 25.5 m3/s
4.7.14. In conclusion: The use of empirical formulae should be done with due diligence.
The average designer who cannot rely so much on his intuition and judgment should go
by the rational procedure outlined above.
The data required for the rational treatment, viz., A, L and H can be easily read from the
survey plans. As regards I0 it should be realized that this does not have to be calculated
for each design problem. This is the storm characteristic of the whole region, with pretty
vast area, and should be known to the local engineers.
Complicated formulae, of which there is abundance, have been purposely avoided in this
Article. Indeed, for a terse treatment, the factors involved are so many and their interplay
so complicated that recourse need be taken to such treatment only when very important
structures are involved and accurate data can be collected. For small bridges, the simple
formulae given here will give sufficiently accurate results.
ARTICLE 5
5.1. In a stream with rigid boundaries (bed and banks) the shape and the size of the crosssection is significantly the same during a flood as after its subsidence. If the HFL is plotted
and the bed slope is measured, it is simple to calculate the discharge.
5.2. But a stream flowing in alluvium, will have a larger cross sectional area when in flood
than that which may be surveyed and plotted after the flood has subsided. During the
flood the velocity is high and, therefore, an alluvial stream scours its bed, but when the
flood subsides, the velocity diminishes and the bed progressively silts up again. From this
it follows that before we start estimating the flood conveying capacity of the stream from
the plotted cross-section, we should ascertain the depth of scour and plot on the crosssection the average scoured bed line that is likely to prevail during the high flood.
5.3. The best thing to do is to inspect the scour holes in the vicinity of the site, look at the
size and the degree of incoherence of the grains of the bed material, have an idea of the
probable velocity of flow during the flood, study the trial bore section and then judge what
should be taken as the probable average scoured bed line.
5.4. Calculation of Velocity: Plot the probable scoured bed line. Measure the crosssectional area A in m2 and the wetted perimeter P in m. Then calculate the hydraulic mean
depth, R by the formula.
๐ ด
(in m)
Next, measure the bed slope S from the plotted longitudinal section of the stream. Velocity
can then be easily calculated from one of the many formulae. To mention one, viz., the
Manning's formula:
๐ ผ
(R2/3, S1/2)
V = Mean velocity of flow() (to be undone)
the hydraulic mean depth
S = the energy slope which may be taken equal to the bed slope, measured over a
reasonably long reach, say 500 m or more
the rugosity co-efficient
For values of n, see Table 5.1. Judgment and experience are necessary in selecting a
proper value of n for a given stream.
Table 5.1 Rugosity Co-efficient, n
Perfect Good
Natural Streams
( 1 ) Clean, straight bank, full stage, no rifts or deep
(2) Same as ( 1 ), but some weeds and stones
(3) Winding, some pools and shoals, clean
(4) Same as (3), lower stages, more ineffective slope
and sections
(5) Same as (3), some weeds and stones
(7) Sluggish river reaches, rather weedy or with
very deep pools
(8) Very weedy reaches
(6) Same as (4), stoney sections
5.5. Calculation of Discharge
Q = A.V.
is a function of the size, shape and roughness of the stream and is called its
conveyance factor. Thus, the discharge carrying capacity of a stream depends on its
conveyance factor and slope.
5.6. When the cross-section is not plotted to the natural scale (the same scale horizontally
and vertically), the wetted perimeter cannot be scaled off directly from the section and
has to be calculated. Divide up the wetted line into a convenient number of parts, AB, BC
and CD, etc. (Fig. 5.1). Consider one such part, say PQ, let PR and QR be its horizontal
and vertical projections.
Then PQ = √๐ ๐ 2 + ๐ ๐ 2 V Now, PR can be measured on the horizontal scale of the given
cross-section and QR on the vertical. PQ can then be calculated. Similarly, the length of
each part is calculated. Their sum gives the wetted perimeter.
Fig. 5.1
5.7. If the shape of the cross-section is irregular as happens when a stream rises above
its banks and shallow overflows are created (Fig. 5.2) it is necessary to subdivide the
channel into two or three sub-sections. Then R and n are found for each sub-section and
their velocities and discharges computed separately.
Fig. 5.2
Where further elaboration is justified, corrections for velocity distribution, change of slope,
etc. may be applied. Books on Hydraulics give standard methods for this.
5.8. Velocity Curves: To save time in computation, curves have been plotted in Plate 3.
Given R, S and n, velocity can be read from this plate.
5.9. Better Measure than Calculate Velocity: It is preferable to observe the velocity
during a high flood. When it is not possible to wait for the occurrence of high flood, the
velocity may be observed in a moderate flood and used as a check on the theoretical
calculations of velocity. In making velocity observations, the selected reach should be
straight, uniform and reasonably long.
5.10. The flood discharge should be calculated at each of the three cross-sections, which
as already explained in para 3.3 should be plotted for all except very small structures. If
the difference in the three discharges, thus, calculated is more than 10 per cent the
discrepancy has to be investigated.
ARTICLE 6
6.1. Estimated Flood Discharge from Flood Marks on an Existing Structure
6.1.1. Having collected the necessary information from inspection as mentioned in para
2.2, the discharge passed by an existing structure can be calculated by applying an
appropriate formula. In Article 15 some formulae for calculating discharges from flood
marks on existing bridges are discussed. Worked out examples have been included in
Article 17.
6.1.2. Distinct water mark on bridge piers and other structures can be easily found
immediately following the flood. Sometimes these marks can be identified years
afterwards but it is advisable to survey them as soon after the flood is possible.
Turbulence, standing wave and slashing may have caused a spread in the flood marks
but the belt of this spread is mostly narrow and a reasonably correct profile of the surface
line can be traced on the sides of piers and faces of abutments. This is perhaps the most
reliable way of estimating a flood discharge because in the formulae discussed
in Article- 15 the co-efficient involved have been accurately found by experiments.
6.2. Fixing Design Discharge
6.2.1. The recommended rule: Flood discharges can be estimated in three different
ways as explained in Para 4.1 to 6.1.2. The values obtained should be compared. The
highest of these values should be adopted as the design discharge Q, provided it does
not exceed the next highest discharge by more than 50 per cent. If it does, restrict it to
that limit. That is 1.5 times of second highest.
6.2.2. Sound economy: The designer is not expected to aim at designing a structure of
such copious dimensions that it should pass a flood of any possible magnitude that can
occur during the lifetime of the structure. Sound economy requires that the structure
should be able to pass easily floods of a specified frequency and that extraordinary and
rare floods should pass without causing excessive damage to the structure or the road.
6.2.3. The necessity for this elaborate procedure for fixing Q arises for sizeable structures.
As regards small culverts, Q may be taken as the discharge determined from the run-off
ARTICLE 7
7.1. The section of a stream, having rigid boundaries, is the same during the flood and
after its subsidence. But this is not so in the case of streams flowing within, partially or
wholly, erodible boundaries. In the latter case, a probable flood section has to be evolved
from the theoretical premises for the purpose of designing a bridge; it is seldom possible
to measure the cross-section during the high flood.
7.2. Wholly Erodible Section. Lacey's Theory: Streams flowing in alluvium are wide
and shallow and meander a great deal. The surface width and the normal scoured depth
of such streams have to be calculated theoretically from concepts which are not wholly
rational. The theory that has gained wide popularity in India is "Lacey's Theory of Flow in
Incoherent Alluvium". The salient points of that theory, relevant to the present subject, are
outlined here.
7.3. A stream, whose bed and banks are composed of loose granular material, that has
been deposited by the stream and can be picked-up and transported again by the current
during flood, is said to flow through incoherent alluvium and may be briefly referred to as
an alluvial stream. Such a stream tends to scour or silt up till it has acquired such a crosssection and (more particularly) such a slope that the resulting velocity is "non-silting and
non-scouring". When this happens the stream becomes stable and tends to maintain the
acquired shape and size of its cross-section and the acquired slope. It is then said: "to
have come to regime" and can be regarded as stable.
7.4. Lacey's Equation: When an alluvial stream carrying known discharge, Q has come
to regime, it has a regime wetted perimeter P, a regime slope S, and regime hydraulic
mean depth R, In consequence, it will have a fixed area of cross-section A and a fixed
velocity V. For these regime characteristics of an alluvial channel, Lacey suggests [18]
the following relationships. It should be noted that the only independent entities involved
are Q and Ksf. The Ksf is called silt factor and its value depends on the size of the grains
of the alluvium. Its value is given by the formula:
... (7.1a)
Ksf =1.76 √๐ ๐
where dm is the weighted mean diameter of the particles in mm. Table 7.1 gives values
of Ksf for different bed materials.
(Typical method of determination of weighted mean diameter of particles (d m) as given
in Appendix-2 of IRC:5 is reproduced in Plate 4).
(a) Regime Cross-Section
P = 4.8 Q ½
... (7.1b)
Table 7.1 Silt Factor Ksf in Lacey's Equations [18] = 1.76 √๐ ๐
Type of bed material
Coarse silt
Silt/fine sand
Medium sand
Coarse sand
Fine bajri and sand
Heavy sand
0.081 to 0.158
0.233 to 0.505
1.29 to 2.00
0.5 to 0.6
0.8 to 1.25
2.0 to 2.42
7.5. The Regime Width and Depth: Provided a stream is truly regime , it is destined
tocome to regime according to Lacey. It will then be stable and have a section and slope
conforming to his equations. For wide alluvial streams the stable width W can be taken
equal to the wetted perimeter P of Equation (7.1a).
That is
W = P = 4.8 Q ½ (applicable for wide stream only)
….. (7.2a)
Also, the normal depth of scour D on a straight and unobstructed part of a wide stream
may be taken as equal to the hydraulic mean radius R in Equation (7. 1 c). Hence,
ARTICLE 8
8. 1. The General Rule for Alluvial Streams: The linear waterway of a bridge/culverts
a wholly alluvial stream should normally be kept equal to the width required for stability,
viz., that given by Equation (7.2a).
8.2. Unstable Meandering Streams: A large alluvial stream, meandering over a wide
belt, may have several active channels separated by land or shallow sections of nearly
stagnant water. The actual (aggregate) width of such streams may be much in excess of
the regime width required for stability. In bridging such a stream it is necessary to provide
training works that will contract the stream. The cost of the latter, both initial and recurring,
has to be taken into account in fixing the linear waterway.
8.3. In the ultimate analysis it may be found in some such cases, that it is cheaper to
adopt a linear waterway for the bridge/culverts somewhat in excess of the regime width
given by Equation (7.2a). But as far as possible, this should be avoided. When the
adopted linear waterway exceeds the regime width it does not follow that the depth will
become less than the regime depth D given by Equation (7.2b). Hence, such an increase
in the length of the bridge/culverts does not lead to any countervailing saving in the depth
of foundations. On the contrary, an excessive linear waterway can be detrimental in so
far as it increases the action against the training works.
8.4. Contraction to be Avoided: The linear waterway of the bridge/culverts across an
stream should not be less than the regime width of the stream. Any design that envisages
contraction of the stream beyond the regime width, necessarily has to provide for much
deeper foundation. Much of the saving in cost expected from decreasing the length of the
bridge/culverts may be eaten up by the concomitant increase in the depth of the
substructure and the size of training works. Hence, except where the section of the stream
is rigid, it is generally troublesome and also futile from economy
consideration to attempt contracting the waterway.
8.5. Streams not Wholly Alluvial: When the banks of a stream are high, well defined,
and rigid (rocky or some other natural hard soil that cannot be affected by the prevailing
current) but the bed is alluvial, the linear waterway of the bridge/culverts should be made
equal to the actual surface width of the stream, measured from edge to edge of water
along the HFL on the plotted cross-section. Such streams are later referred to as quasialluvial.
8.6. Streams with Rigid Boundaries: In wholly rigid streams the rule of para 8.5
applies, but some reduction in the linear waterway may, across some streams with
moderate velocities, be possible and may be resorted to, if in the final analysis it leads to
tangible savings in the cost of the bridge.
8.7. As regards streams that overflow their banks and create very wide surface widths
with shallow side sections, judgment has to be used in fixing the linear waterway of the
bridge. The bridge/culverts should span the active channel and detrimental afflux
avoided. See also Article 18.
ARTICLE 9
9.1. Alluvial Streams
9.1.1. What is the significance of the Normal Scour Depth? If a constant discharge were
passed through a straight stable reach of an alluvial stream for an indefinite time, the
boundary of its cross-section should ultimately become cosine curve.
This will happen when regime conditions come to exist. The depth in the middle of the
stream would then be the normal regime depth.
In nature, however, the flood discharge in a stream does not have indefinite duration. For
this reason the magnitude and duration of the flood discharge carried by it would govern
the shape of the flood section of any natural stream. Some observers have found that
curves representing the natural stream sections during sustained floods have sharper
curvature in the middle than that of an ellipse. In consequence, it is believed that Lacey's
normal depth is an under estimate when applied to natural streams subject to sustained
floods. However, pending further research, Lacey's equations may be
9.1.2. As discussed later in Article 1 1 , the depth of foundations is fixed in relation to the
maximum depth of scour, which in turn is inferred from the normal depth of scour. The
normal depth of scour for alluvial streams is given by Equation (7.2b), so long as the
bridge/culverts does not contract the stream beyond the regime width W given by
Equation (7.2a).
9.1.3. If the linear waterway of the bridge for some special reason, is kept less than the
regime width of the stream, then the normal scour depth under the bridge will be greater
than the regime depth of the stream (Fig. 9.1).
W = the regime width of the stream
the designed waterway; when the bridge is assumed to cause contraction L is
less than W
D = The normal scour depth when L = W
DI = The mean depth of scour under the bridge with L less than W
According to Clause 106.9.3.1 of IRC 5
Db = discharge in m3/s per m width
ksf = silt factor for material obtained upto deepest anticipated scour.
= 1 .76 √๐ ๐ , ๐ ๐ being the weighted mean diameter of particles in mm.
dsm = mean depth of scour in m.
Fig. 9.1
The value of Db shall be total design discharge divided by the effective linear waterway
between abutments or guide bunds.
This formulae take into account the effect of contraction and, therefore, no further
modification arc needed. When the bed is protected by apron and curtain wall, the scour
considerations will be different as discussed in Article-20.
9.2. Quasi-Alluvial Streams
9.2. 1 . Some streams are not wholly alluvial : A stream may flow between banks which
are rigid in so far as they successfully resist erosion, but its bed may be composed of
loose granular material which the current can pick-up and transport. Such a stream may
be called quasi-alluvial to distinguish it, on the one hand, from a stream with wholly rigid
boundaries and, on the other, from a wholly alluvial stream. Since such a stream is not
free to erode its banks and flatten out the boundaries of its cross-section as a wholly
alluvial stream does, it does not acquire the regime cross-section which Lacey's equations
9.2.2. It is not essential that the banks should be of rock to be in erodible. Natural mixtures
of sand and clay may, under the influence of elements, produce material hard enough to
defy erosion by the prevailing velocity in the stream.
Across a stream section, the natural width of which is nowhere near that prescribed by
Lacey's theory, it is expected to find that the banks, even though not rocky are not friable
enough to be treated as incoherent alluvium for the application of Lacey's Theory. Such
cases have, therefore, got to be discriminated from the wholly alluvial streams and treated
on a different footing.
9.2.3. In any such case the width W of the section, being fixed between the rigid banks,
can be measured. But the normal scour depth D corresponding to the design discharge
Q has to be estimated theoretically as it cannot be measured during the occurrence of
high flood.
9.2.4. When the stream width is large compared to depth : In Article-5, for calculating
the discharge of the stream from its plotted cross-section, the probable scoured bed line
(para 5.3) was drawn.
When the stream scours down to that line it should be capable of passing the discharge
calculated there, say q m3 /s. But the discharge adopted for design, Q, may be anything
upto 50 per cent more than q (see para 6.2.1). Therefore, the scour bed line will have to
be lowered further.Suppose the normal scour depth for Q is D and that for q is d, then,
D=d(๐ )3/5
... (9.2)
Since d is known, D can be calculated. This relationship depends on the assumption that
the width of the stream is large as compared, with its depth, and therefore, the wetted
perimeter is approximately equal to the width and is not materially affected by variations
in depth. It also assumes that the slope remains unaltered.
area x velocity
R P C R2/3 S1/ 2
K R 5/3
where K is a constant.
Hence, R varies as Q 3/5 . Since in such streams R is very nearly equal to the depth,
therefore, D varies as Q 3/5 . Hence, the equation (9.2).
From the above relationship it follows that if Q is 1 50 per cent of q, D will be equal to 127
per cent of d.
9.2.5. Alternatively, the normal depth of scour of wide streams may be calculated as
under. If the width of the stream is large as compared with its depth, then W may be taken
as P and D as R.
area x velocity
(PR) V- (WD) V, where Vis the mean velocity
D= ๐ ๐
Now W is the known fixed width of the stream. If the velocity V has actually been observed
(para 5.9), then D can be calculated from the above equation. For mean velocity, refer
relevant clause in IRC:6.
9.2.6. Suppose the velocity has not been actually measured during a flood, but the slope
S is known.
Knowing Q, W and S, D can be calculated from this equation.
For quickness, velocity curves in Plate 3 can be used. Assume a value of R and fix a
suitable value of the rugosity co-efficient n appropriate for the stream. Corresponding to
these values and the known slope, read the velocity from Plate 3. Now calculate the
discharge (= VRW). If this equals the design discharge Q, then the assumed value of R
is correct. Otherwise, assume another value of R and repeat. When the correct value of
R has been found, take D equal to R. (See the worked out Example in Article-16).
9.2.7. The procedure described above can be applied if either the slope of the stream or
the actual observed velocity is known. If either of these are not known, the following
procedure for approximate calculation of the normal scour depth can be applied.
Suppose the wetted perimeter of the stream is P and its hydraulic mean depth R. If Q is
its discharge, then,
area x velocity
(PR) [CR2/3 S1/2]
.... (9.6a)
Now, if this stream, carrying the discharge Q, had been wholly alluvial, with a wetted
perimeter P1, and hydraulic mean depth R1 for regime conditions, then,
(P1R1) [CR2/3 S1/2]
.... (9.6ab)
Also, for a wholly alluvial stream Lacey's Theory would give:
Now substituting values of P1 and R1 from equations (9.6c) and (9.6d) in (9.6e), we get
If the width W of the stream is large compared with its depth D, then writing W for P and
D for R in equation (9.6f).
Thus, if the design discharge Q, the natural width W, and the silt factor Ksf are known,
the normal scour depth D can be calculated from Equation (9.7).
The above reasoning assumes that the slope at the section in the actual case under
consideration is the same as the slope of the hypothetical (Lacey's) regime section,
carrying the same discharge. This is not improbable where the stream is old and its bed
material is really incoherent alluvium. But if there is any doubt about this, the actual slope
must be measured and the procedure given in para 9.2.6 applied.
9.2.8. When the stream is not very wide: If the width of the stream is not very large as
compared with its depth, then the methods given above will not give accurate enough
results. In such a case draw the probable scoured bed line on the plotted cross-section,
measure the area and the wetted perimeter and calculate R.
Corresponding to this value of R and the known values of S and n, read velocity from
Plate 3. If the product of this velocity and the area equals the design discharge, the
assumed scoured bed line is correct. Otherwise, assume another line and repeat the
process. Then measure D.
9.2.9. Effect of contraction on normal scour depth: If, for some special reason, the
linear waterway L of a bridge across a quasi-alluvial stream is kept less than the natural
unobstructed width W of the stream (Fig. 9.1), then the normal scour depth under the
bridge D1 will be greater than the depth D ascertained above for the unobstructed stream.
Covered by the relationship:
Because Db, of L case will be more than Db, of W case.
9.3. Scour in Clay and Bouldary Strata: There are no rational methods for
assessment of scour in clay or bouldary strata. Guidelines for calculating silt factor for
bed materials consisting of gravels and boulders as given in Appendix-I of IRC:78-2000
may be adopted and are reproduced in paras 9.3. 1 and 9.3.2.
9.3. 1. Scour in clay: Scour in clay is generally less than scour in sand. Normally in field
we get a mixture of sand and clay at many places. For the purpose of assessment
following definition of sand and clay can be given.
Sand -
Where แถฒ is equal to or more than 15° even if c (Cohesion of soil) is more
than 0.2 kg/cm2
(Silt factor Ksf will be calculated as per provisions of para 7.4 or Table
Where แถฒ is less than 15° & c (Cohesion of soil) is more than 0.2 kg/cm2
Scour in sand of above definition can be calculated by the formulae given earlier. In clay
instead of silt factor (Ksf) clay factor (Ksfc) is adopted Ksfc = F (1+√๐ )
c= Cohesion in kg/cm2 and
F= 1.5 แถฒ for แถฒ ≥ 10° < 15°
... (9.9)
= 1.75 for แถฒ ≥ 5°< 10°
= 2.0 for แถฒ <50°
Scour depth (dsm) = 1.34 (Db2 / Ksfc) 1/3
Db = discharge per unit width
9.3.2. Bouldary strata: There is no rational method to assess scour in bouldary strata of
boulders or pebbles. In the absence of any formula Ksf may be determined as per Clause
703.2.2 of IRC: 78 and adopted. If, say, average size of pebbles is db
It is, however, better to investigate depth of foundations adopted in past for similar
foundation and decide depth on the basis of precedence. Protection work around
foundations in the form of curtain wall and apron or garland blocks should be provided,
when the foundation is laid on bouldary strata.
ARTICLE 10
10.1. In considering bed scour, we are concerned with alluvial and quasi-alluvial streams
only and not with streams which have rigid beds.
1 0.2. In natural streams, the scouring action of the current is not uniform all along the
bed width. It is not so even in straight reaches. Particularly at the bends as also round
obstructions to the flow, e.g., the piers of the bridge, there is deeper scour than normal.
In the following paragraphs, rules for calculating the maximum scour depth are given. It
will be seen that the maximum scour depth is taken as a multiple of the normal scour
depth according to the circumstances of the case.
10.3. In order to estimate the maximum scour depth, it is necessary first to calculate the
normal scour depth. The latter has already been discussed in detail. To summarise what
has been said earlier, the normal scour depth will be calculated as under:
(i) Alluvial Streams. Provided the linear waterway of the bridge is not less than the
regime width of the stream, the normal scour depth D is the regime depth as calculated
from Equation (7.2b).
(ii) Streams with Rigid Banks but Erodible Bed. Provided the linear waterway of the
bridge is not less than the natural unobstructed surface width of the stream, the normal
scour depth d is calculated as explained in Article 9.
10.4. Rules for finding Maximum Scour Depth. The rules for calculating the maximum
scour depth from the normal scour depth are:
Rule (1) : For average conditions on a straight reach of the stream and when the bridge
is a single span structure, i.e. it has no piers obstructing the flow, the maximum scour
depth should be taken as 1.27 times the normal scour depth,
Rule (2) : For bad sites on curves or where diagonal current exist or the bridge is multispan structure, the maximum scour depth should be taken as 2 times the normal scour
depth, modified for the effect of contraction when necessary.
10.5. The finally adopted value of maximum scour depth must not be less than the depth
(below HFL) of the deepest scour hole that may be found by inspection to exist at or near
the site of the bridge.
The following example will illustrate the application of the rules in para 10.4 above.
10.6. Example, A bridge is proposed across an alluvial stream (K sf= 1.2) carrying a
discharge of 50 m3/s. Calculate the depth of maximum scour when the bridge consists of
(a) 3 spans of 6 m and (b) 3 spans of 8 m
Regime surface width of the stream
W = 4.8Q 1/2 = 4.8 x 50 1/2 = 33.94 m
Regime depth
Maximum scour depth
when span (3x6 m), Db the discharge per metre width is
50/18, i.e., 2.778 cumecs
dsm = 1 .34 (2.7782/l .2)1/3 = 2.49 m
Maximum depth of scour for pier
= 2dsm = 2x2.49 = 4.98 m
Maximum depth of scour for abutment
= 1 .27 d sm = 1.27x2.49 = 3. 16 m
When span is 3 x 8 m, Db the discharge per metre width is
50/24, i.e., 2.083 cumecs
d sm = 1 .34 (2.083 2 /l .2) I/3 = 2.055 m
Maximum depth of scour for pier
= 2dsm = 2x2.055 = 4.11 m
Maximum depth of scour for abutment
= 1.27dsm = 1.27 x 2.055 = 2.61 m
10.7 For small bridges across alluvial channel having multiple spans, the foundation
levels for abutments should be kept the same as that of pier for following reasons:
(i) In case of small spans, the scour hole around pier could extend up to abutment.
(ii) Abutment foundation at higher level may create a surcharge effect over the
foundation of adjacent pier.
(iii) In case of outflanking of the bridge the abutment in any case has to be designed for
scour all around condition.
ARTICLE 11
11.1. The following rules should be kept in view while fixing the depth of bridge
Rule (1) In Soil. The embedment of foundations in soil shall be based on assessment of
anticipated scour. Foundations may be taken down to a comparatively shallow
depth below the bed surface provided good bearing stratum is available and
foundation is protected against scour. The minimum depth of open foundations
shall be upto stratum having adequate bearing capacity but not less than 2.0m
below the scour level or protected scour level.
Rule (2) In Rocks. When a substantial stratum of solid rock or other material not erodible
at the calculated maximum velocity is encountered at a level higher than or a little
below that given by Rule (1) above, the foundations shall be securely anchored
into that material. This means about 0.6 m into hard rocks with an ultimate crushing
strength of 10 MPa or above and 1 .5 m in all other cases.
Rule (3) All Beds. The pressure on the foundation material must be well within the safe
bearing capacity of the material.
These rules enable one to fix the level of the foundations of abutments and piers.
11.2 The above rules are applicable for open foundations only. For deep foundations
like well, and pile foundations, wherever adopted depending upon site requirements depth
of foundations shall be worked out as per IRC: 78.
ARTICLE 12
12.1. As a rule, the number of spans should be as small as possible, since piers obstruct
flow. Particularly, in mountainous regions, where torrential velocities prevail, it is better to
span from bank to bank using no piers if possible.
12.2. Length of Span: In small structures, where open foundations can be laid and solid
abutments and piers raised on them; it has been analyzed that the following approximate
relationships give economical designs.
S = 1 .5 H
For Masonry arch bridges
For RCC Slab Bridges
S = Clear span length in metres.
H = Total height of abutment or pier from the bottom of its foundation to its top in metres.
For arched bridges it is measured from foundation to the intrados of the key stone.
12.3. Vertical Clearance: After fixing the depth of foundations Df, the vertical clearance
is added to it to get H. The minimum vertical clearance shall be provided as per Table
Discharge in
Upto 0.30
upto 3.0
Above 3 and upto 30
Above 30 and upto 300
Table 12.1
Minimum vertical clearance in
For openings of culverts having arched decking, the clearance below the crown of the
intrados of arch shall not be less than 1/10 of the maximum depth of water plus 1/3 of the
rise of arch intrados.
Further to keep the free board of approaches not less than 1750 mm ,the vertical
clearance in slab/box cell bridges may be increased suitably.
In designing culverts for roads across flat regions where streams are wide and shallow
(mostly undefined dips in the ground surface), and in consequence the natural velocities
of flow are very low, the provision of clearance serves no purpose. Indeed it is proper to
design such culverts on the assumption that the water at the inlet end will pond up and
submerge the inlet to a predetermined extent. This will be discussed in Article 19.
In case of structure over artificial channels or canals, etc. the minimum vertical clearance
should be taken 600 mm above the Full Supply Level.
12.4. The Number of Spans:
124. 1. If the required linear waterway L is less than the economical span length it has to
be provided in one single span.
1 2.4.2. When L is more than the economical span length (S) the number of spans (N)
required is tentatively found from the following relation:
L = NS
12.4.3. Since N must be a whole number (preferably odd) S has to be modified suitably.
In doing so it is permissible to adopt varying span lengths in one structure to keep as
close as possible to the requirements of economy and to cause the least obstructions to
the flow.
12.5. To facilitate inspection and carrying out repairs, the minimum vent height of culverts
should normally be 1500 mm. The vent size of irrigation culverts may be decided
considering the actual requirements and site condition. For pipe culverts the minimum
diameter of pipes should be 1.20 m..
12.6 If a large number of small bridges and culverts are required to be provided in a
project, it is important that the span lengths or box sections should be standardized, so
that the repetitive use of false work or use of precast method of construction. may be
appropriately adopted such design would make substantial reduction in overall
construction period of the project.
ARTICLE 13
13.1. Details of small bridges and culverts of probable spans and heights conforming to
latest IRC codes and guidelines are incorporated with a view to cut short the time in
preparation of estimates and design of culverts and attain uniform standards and quality
control in the work.
13.2. Geometric Standards
13.2.1. IRC standards: Standards contained in IRC:73 and IRC:86 are adopted for
Geometric Standards. The overall widths adopted for culverts and small bridges for 2lane carriageway are as follows.
NH and SH MDR
12 m Minimum
8.4 m
13.2.2. Design loads for 2-Iane roadway: Design loading for culverts and small bridges
should be as below:
Village Road and ODR (Rural Roads)
NH, SH and MDR
2-lanes IRC Class A or 70R
whichever gives worst effect
Depending upon the carriageway
width refer IRC-6
1 3.2.3. Width of roadway: The width of a culvert and small bridge (along the direction
of flow) should be such that the distance between the outer faces of the parapets will
equal the full designed width of the formation of the road. Any proposed widening of the
road formation in the near future should also be taken into account in fixing the width of
the structure. In case of high banks, the length of culvert should be judiciously decided to
avoid high face walls.
1 3.2.4. In small bridges, the width (parallel to the flow of the stream) should be sufficient
to give a minimum clear carriageway of 4.25 m for a single-lane bridge and 7.5 m for a
two-lane bridge between the inner faces of the kerbs or wheel guards. Extra provision
should be made for footpaths, etc., if any are required. Width of bridges and culverts shall
be at least equal to width of road way in the section.
1 3.2.5. Siting of structures and gradients: Culverts and small bridges must be sited on
the straight alignment of roads. If the Nalla is crossing the road at angles other than right
angle,either skew culverts and small bridges should be provided or, if economical, the
Nalla should be suitably trained. The gradient of road may be provided on the bridges
and culverts . If the bridges are situated at change of gradient (hump), the profile of
vertical curve should be given in wearing coat. Alternatively, the profile could be given in
the deck itself. It shall be ensured that the bearing surface of deck slab on the
abutment/pier cap shall be horizontal.
13.3. Design:
13.3.1. Road top level: For maintaining the geometric standards of the road, culverts and
small bridges should be constructed simultaneously or prior to the earthwork for road as
otherwise there would be the following disadvantages.
(1) Practically, every culvert and small bridge becomes a hump on the road and
geometric of the road is affected.
(2) Duplicate work of consolidation of approaches giving rise to extra cost.
13.3.2. Minimum span and clearance : From the consideration of maintenance of
culverts, it is desirable that the span of slab culvert is kept minimum 2 m and height 1.5
m and diameter of pipes 1.2 m. Culverts of small span or diameter are found to get choked
due to silting and also cause difficulty in cleaning.
13.3.3. Pipe culverts: Pipe culverts shall conform to IS category NP4. The cushion
between the top of the pipe and the road level shall not be less than 600 mm. First class
bedding consisting of compacted granular material can be used for height of fill upto 4 m
and concrete cradle bedding upto a maximum height of fill upto 8 m.
For small size culverts, RCC pipe culverts with single pipe or up to 6 pipes placed side by
side (with minimum clear distance of 600mm) depending upon the discharge may be
used as far as possible, as they are likely to prove comparatively cheaper than slab
RCC pipes in two rows one above the other have also been used for small bridges on
cost considerations, specially for providing waterway in breached section of roads.
13.3.4. RCC slab: RCC slab culverts and small bridges should be adopted where the
founding strata is rocky or of better bearing capacity. In case where adequate cushion is
not available for locating pipe culvert RCC slab culvert should be adopted. RCC slab
culverts/bridges are also useful for cattle crossing during dry weather.
13.3.5. RCC box cell structures: In a situation where bearing capacity of soil is low,
RCC Box type culvert should be preferred.
13.3.6. Balancing culverts: Balancing culvert are to be located at points on L section
of the road where down gradients meet. These balancing culverts balance the discharge
from either side of the road. Observation of the road alignment during rains also gives a
good idea about location of balancing culverts.
13.4 Numbering of Culverts and Small Bridges:
For details reference may be made to "Recommended Practice for Numbering Bridges
and Culverts", IRC:7.
13.5. General Design Aspects and Specifications: The type design of pipe culverts and
RCC slab culverts and slab bridges given here are based on following general aspects.
Coursed rubble stone masonry for substructure and parapet walls is generally found to
be economical in comparison to mass concrete substructure. The masonry below or
above the ground level should be as per IRC:40. If bricks having minimum crushing
strength of 7 Mpa are available, these can also be used for culverts.
13.5.1. Parapet wall and railing : For culverts, where parapet walls are provided they
shall be of plain concrete M15 grade or brick or store masonry with 450 mm top width. In
case of pipe culverts no parapet walls are needed and guard stones would be adequate
except for culverts on hill roads. Guard stones provided shall be of size 200x200x600
mm. Railings as given in Standard Drawings of MORT&H may also be provided for
culverts and small bridges. Railings or parapets shall have a minimum height above the
adjacent roadway or footway safety kerb surface of 1.1 m less one half the horizontal
width of the top rail or top of the parapet. Crash barriers may be provided when they are
found functionally required. Crash barriers when provided shall conform to provisions in
IRC:5 and while adopting MORT&H standard drawings, the design of deck slab shall be
checked for provision of crash barriers.
1 3.5.2. Wearing coat: Normally, the wearing surfaces of the road
carried over the
culverts. For low category road which do not have bituminous surfaces, concrete wearing
coat of average 75 mm or bituminous wearing coat is provided. On the small bridges
wearing coat is provided as per IRC:5.
1 3.5.3. Approach slab: Approach slab can be dispensed with in case of culverts. For
Small bridges approach slab as per IRC:5 shall be provided.
13.5.4. Deck slab: Grade of concrete shall be as per IRC 112. M 25 concrete for moderate
and M30 concrete for severe conditions of exposure and high strength deformed bars
conforming to IS: 1786 are specified for the deck slabs.
13.5.5. Expansion joint: For small bridges, premoulded bituminous sheet (like, shalitex
board) of 20 mm thickness are required to be provided.
13.5.6. Pier/abutment cap/coping: The minimum thickness of reinforced cap over solid
PCC/RCC substructure shall be 200 mm and that in case of masonry substructure shall
not be less than 500 mm. The minimum grade of concrete shall be M 25 and M 30 for
moderate and severe conditions of exposure respectively. However, the coping over the
returns may be of M 15 grade and thickness not less than 100 mm.
13.5.7. Section of pier abutment and returns: The abutment and pier sections should
be so designed as to withstand safely the worst combination of loads and forces as
specified in the IRC:6.
13.5.8. Top width of pier/abutment: In respect of masonry and concrete
piers/abutments minimum width at top of pier and abutments for slab bridges just below
the caps shall be as per Table 13.1. Tar paper bearings shall be provided between
abutment/peir cap and RCC slab for spans upto 10 m.
Span (in m)
Table 13.1
Minimum width at top
of abutment/pier (mm)
If the velocity flow is more than 4.5 m/s and river carries abrasive particles, it is advisable
to design section of foundation and pier considering their effect. A sacrificial layer of
brick/stone masonry of suitable thickness and height shall be provided irrespective of total
height of substructure.
In the case of arch bridges, the top width of abutments and piers should be adequate to
accommodate skew decks and to resist the stresses imposed under the most unfavorable
conditions of loading.
13.5.9. Return walls or wing walls: Wing walls are generally at 45° angle to the
abutment and are also called as splayed wing walls. Walls parallel to road are called as
return walls.
Where embankment height exceeds 2 m, splayed return walls may be preferred. The
length of straight return should normally be 1.5 times the height of the embankment.
Where the foundations of the wing walls can be stepped up, having regard to the soil
profile, this should be done for the sake of economy. Quite often short return walls meet
the requirements of the site and should be adopted.
The top width of wing walls and returns shall not be less than 450 mm.
Layout of wing walls for skew bridges/culverts shall be prepared keeping in
view the height and normal distance from the road to the wing wall
Slope of top of wing wall to be such that height of wall matches with the slope
of embankment
13.5.10. Weep holes and water spouts: Adequate number of weep holes at spacing not
exceeding 1 m in horizontal and vertical direction should be provided to prevent any
accumulation of water and building up of the hydrostatic pressure behind the abutment
and wing walls. The weep holes should be provided at about 1 50 mm above low water
level or ground level whichever is higher. Weep holes shall be provided with 100 mm dia
AC pipes for structures in plain/reinforced concrete, brick masonry and stone masonry.
For brick and stone masonry structures, rectangular weep holes of 80 mm wide and 150
mm height may also be provided. Weep holes shall extend through the full width of the
concrete/masonry with slope of about 1 vertical to 20 horizontals towards the drainage
In case of stone masonry, the spacing of weep holes shall be adjusted to suit the height
of the course in which they are formed. The sides and bottom of the weep holes in the
interior shall be made up with stones having fairly plain surface.
For spans more than 5 m one water spout of 100 mm dia. in the center of the slab located
on either side of the deck shall be provided. The spacings of drainage spouts shall not
exceed 1 0 m.
In case of one side camber, the number of drainage spouts shall be doubled and location
suitably adjusted.
13.5.1 1. Foundation concrete: Foundation concrete shall not be less than M 15 grade.
If the foundation level is below water table, 10 per cent extra cement is to be added in
concrete. The minimum thickness of footing shall be 300 mm. For foundation resting on
rock a leveling course of at least 150 mm thickness in M 15 grade of concrete shall be
13.5.12. Arches: The type of superstructure depends on the availability of the
construction materials and its cost. An evaluation of the relative economics of RCC slabs
and masonry arches should be made and the latter adopted where found more
The masonry arches may be either of cement concrete blocks of M 15 or dressed stones
or bricks in 1:3 cement mortar. The crushing strength of concrete, stone or brick units
shall not be less than 105 kg/cm2. Where stone masonry is adopted for the arch ring, it
shall be either coursed rubble masonry or ashlar masonry.
13.5.13. Raft foundation: Raft foundations are found to be quite suitable for small
bridges and culverts where the founding strata is soft and has SBC upto 10 t/m2. The
following aspects are to be kept in consideration.
(1) Raft foundations are suitable for all types of structures other than pipe culverts.
(2) Protection needs to be provided in the form of apron.
(3) Cut-off should be done first, i.e., before the raft. Immediately, after the raft is complete,
aprons on U/s and D/s should be completed.
(4) Details of raft foundation are given in Article 21.
13.6. Quality Control
13.6.1. Although, the work of culverts and small bridges is simple it is necessary to have
quality control in the work of stone/brick masonry and concrete in deck slab, bar bending,
etc. Reference may be made to "Guidelines on Quality Systems for Road Bridges",
IRC: SP:47.
1 3.6.2. Specifications should be in accordance with "Specification for Road and Bridge
Works" of Ministry of Road Transport and Highways published by Indian Roads Congress.
13.7. Setting out of culverts and small bridges: Setting out of culverts and small bridges
should be done from 4 masonry/ concrete pillars, two in the direction of road and two
along the stream, all placed along two center lines. The top of pillars in the direction of
road should be at the proposed top level of deck slab. Two lines, one along the direction
of stream and the other along the center line of road should be inscribed on one of the
pillars and all distances should be measured with respect to these lines. The pillars should
be placed sufficiently away from the zone of excavation.
13.8. Masonry Work
13.8.1. All masonry work shall conform to IRC:40. The mortar mix in case of cement sand
shall be 1:3, 1:4 or 1:5, whereas, in case of cement lime sand it shall be 1.0:0.5:4.5.
13.8.2. Brick proposed to be used shall be of minimum compressive strength of 7 MPa.
However, for rivers with velocity of 4.5 m/s and carrying highly abrasive particles, this
shall be increased to 10 MPa.
13.8.3. Brick and stone masonry shall conform to IRC:40.
13.9. Concrete
13.9.1. According to IRC:112 the minimum grade of plain concrete is M 15 of concrete
and that of RCC is M 20. The size of reinforcing steel to be used for RCC slabs and the
grading of aggregates are specified in relevant codes and specifications. It is advisable
to use power driven concrete mixer. Similarly, vibrators should also be made available.
Furthermore, precast concrete cover blocks must be provided to ensure bottom cover to
reinforcement. Water cement ratio must be limited to 0.45 maximum. In case of use of
Plasticiser w/c ratio can be restricted to 0.4. Size of coarse aggregate will be 20 mm for
RCC and upto 40 mm for plain concrete. Wherever feasible, precast construction may be
adopted, which ensures quality and speed of construction.
13.10. Bar bending: Lengths of bars and numbers are given in standard drawings.
Cutting of bars from available stock must be done carefully. Generally, tendency of cutting
bars of required lengths and discarding pieces of shorter lengths give rise to greater
wastages. Normally staggered overlaps to the extent of 25 per cent may be provided.
Calculated quantities of steel are increased suitably to account for overlaps, its length
conforming to IRC:112 . Steel chairs should be provided for maintaining correct position
of top bars.
ARTICLE 14
14.1. Abutment and Wing Wall Sections: For RCC slab culverts designed for IRC
single lane of class 70 R loading or 2-lanes of IRC class A loading, the abutment, pier
and wing wall sections shall be designed as per IRC 112..
The base widths of the abutment and the pier depend on the bearing capacity of the soil.
The pressure at the toe of the abutment should be worked out to ensure that the soil is
not overstressed.
The pier sections should be made preferably circular in the case of skew crossings.
1 4.2. Filling behind the abutments, wing walls and return walls shall confirm to IRC:78 as
reproduced in Appendix "”A”.
14.3. Unreinforced Masonry Arches: Plate 6 shows the details of arch ring of segmental
masonry arch bridges without footpaths for spans 6 m and 9 m.
The section of abutment and pier for masonry arch bridges will have to be designed taking
into account the vertical reaction, horizontal reaction and the moment at springing due to
dead load and live load. Table 14.1 gives the details of horizontal reaction, vertical
reaction and moment at springing for arch bridges of span 6 m and 9 m and Table 14.2
gives the influence line ordinates for horizontal reaction, vertical reaction and moment at
springing for a unit load placed on the arch ring.
Table 14.1 Vertical Reaction, Horizontal Reaction and Moment at Springing Due to
Dead Load of Arch Ring Masonry, Fill Material and Road Crust for One Meter of
Arch Measured Along the Transverse Direction (i.e. Perpendicular to the
Direction of Traffic) for Right Bridges
SI. No.
Horizontal Reaction
Vertical Reaction
Moment at
(Tonne Metres)
(+) 0.30
(+) 0.47
Notes: 1. Unit weight of arch ring masonry, fill materials and the road crust is assumed
as 2.24 t/m3.
2. Positive sign for moment indicates tension on the inside of arch ring.
Table 14.2 influence Line Ordinates for Horizontal Reaction (H) Vertical Reaction at
Support (VA) and (VB) and Moment at Springing (MA) and (MB) for Unit Load, say
1 Tonne Located along the Arch Axis at an Angle 9 Degrees from the Radius OC.
Rise of Arch is One Quarter of Span (Fig. 14.1)
SI. θ Degree
H in
VA in
VB in
(tonnes-m) (tonnes-m)
(3) 15
(4) 25
(5) 35
(6) 45
(7) 53°8'
(b) Effective Span 9 m
(a) Effective Span 6 m
(1) 0
(2) 5
Note: Positive sign for moment indicates tension on the inside of arch ring
14.4. RCC Slabs
14.4.1. The details of RCC slabs to be used for culverts and bridges at right crossings
and skew crossings shall be designed as per IRC 112.
14.5. Box Cell Structures: The details for single cell box upto 8 m opening, for double
cell upto 3 m opening of each cell and triple cell upto 3 m opening of each cell with and
without earth cushion for varying bearing capacity may be designed as per IRC 112.
14.6. RCC Pipe Culverts: The details of pipe culverts of 1.2 m dia, with single or double
pipes having cement concrete or granular materials in bed are given in Plates 8- 11
ARTICLE 15
15.1. The formulae for discharge passing over broad crested weirs and drowned orifices
have been developed ab initio in this section. These formulae are very useful for
computing flood discharges from the flood marks left on the piers and abutments of
existing bridges and calculating afflux in designing new bridges. It is necessary to be
familiar with the rationale of these formulae to be able to apply them intelligently.
15.2. Broad Crested Weir Formulae applied to Bridge Openings: In Fig. 15.1, X-X is
the water surface profile, and Z-Z the total energy line. At Section 1, the total energy.
The area of flow at Section 2,
= BC x linear waterway
=(1-in) HL
Where L is the linear waterway. From Eq. (15.2) Velocity at Section 2
v = (2gn H) ½
Therefore, the discharge through the bridge
Q = av
= (l-n) HL(2gnH) ½
To account for losses in friction, a coefficient Cw may be introduced. Thus,
Q = Cw (l-n) HL(2gnH) ½
= Cw √2๐ LH 3/2 (n ½ - n 3/2)
The depth BC adjusts itself so that the discharge passing through the section is
maximum. Therefore, differentiating
----- =0
On exit from the bridge, some of the velocity head is reconverted into potential head due
to the expansion of the section and the water surface is raised, so that Dd is somewhat
greater than BC, i.e. greater than 66.7 per cent of H. In fact, observations have proved
that, in the limiting condition,
Dd can be 80 per cent of Du. Hence, the following rule:
"So long as the afflux (Du -Dd) is not less than
depends on Du and is independent of Dd
Dd, the weir formula applies, i.e., Q
The coefficient Cw may be taken as under:(1) Narrow Bridge opening with or without floors
(2) Wide bridge opening with floors
The fact that the downstream depth Dd has no effect on the discharge Q, nor on the
upstream depth Du when the afflux is not less than 4 Dd is due to the formation of the
"Standing Wave"
(3) Wide bridge opening with no bed floors
15.3. The Orifice Formulae: When the downstream depth, Dd is more than 80 per cent
of the upstream depth Du, the weir formula does not hold good, i.e. the performance of
the bridge opening is no longer unaffected by Dd.
In Fig. 15.2, X-X is the water surface line and Z-Z the total energy line.
Apply Bernoulli’s Equation to points 1 and 2, ignoring the loss of head (h) due to entry
and friction.
In the field it is easier to work in terms of h = Du- Dd instead of h'. But h is less than n as
on emergence from the bridge the water surface rises, due to recovery of some velocity
energy as potential head. Suppose eu2/2g represents the velocity energy that is
converted into potential head.
15.4. In Conclusion: Let us get clear on some important points
(1) In all these formulae Dd is not affected in any way by the existence of the bridge. It
depends only on the conveyance factor and slope of tail race. Dd has, therefore, got to be
actually measured or calculated from area - slope data of the channel as explained
already in Article 7.
(2) The Weir Formula applies only when a standing wave is formed, i.e., when the afflux
(h = Du - Dd) is not less than ¼ Dd.
(3) The Orifice Formulae with the suggested values of CQ and e should be applied when
the afflux is less than ¼ Dd.
15.5. Examples have been worked out in Articles 1 6 and 1 7 to show how these formulae
can be used to calculate aflux and discharge under bridges.
ARTICLE 16
16.1. The afflux at a bridge is the heading up of the water surface caused by it. It is
measured by the difference in levels of the water surfaces upstream and downstream of
the bridge (Fig. 16.1).
16.2. When the waterway area of the openings of a bridge is less than the unobstructed
natural waterway area of the stream, i.e., when the bridge contracts the stream, afflux
occurs. Contraction of the stream is normally not done, but under some circumstances it
is taken recourse to, if it leads to ponderable economy. Also, in the case of some alluvial
streams in plains the natural stream width may be much in excess of that required for
regime. When spanning such a stream, it has to be contracted to, more or less, the width
required for stability by providing training works.
16.3. Estimating afflux is necessary to see its effect on the 'clearance' under the bridge,
on the regime of the channel upstream of the bridge; and on the design of training works.
16.4. For calculating afflux we must know (1), the discharge Q, (2) The unobstructed width
of the stream W, (3) the linear waterway of the bridge L, and (4) the average depth
downstream of the bridge Dd.
16.5. The downstream depth Dd is not affected by the bridge: it is controlled by the
conveyance factor and slope of the channel below the bridge. Also, the depth, that
prevails at the bridge site before the construction of the bridge, can be assumed to
continue to prevail just downstream of the bridge after its construction. Thus, Dd is the
depth that prevails at the bridge site before its construction. To estimate afflux, we must
know Dd. In actual problems, Dd is either given or can be calculated from the data
1 6.6. Example: A bridge, having a linear waterway of 25 m, spans a channel 33 m
wide carrying a discharge of 70 m3 /s. Estimate the afflux when the downstream depth is
1 m.
Dd= 1 m; W = 33m; L = 25m
h= 33๐ ข -1
… (16.2)
Subtracting for h from (16.2) in (16.1) and rearranging
u= 0.0617 u3 + 1.386 u=1.68 m/sec
Alternatively, assume that h is more than ¼ Dd
and apply the Weir formula
Q= 1.706 Cw LH 3/2
Substituting for u in (16.1)
h=0.263 m
70=1.706 x 0.94 x 25 x H 3/2
H = 1.45m
H = Du + u2/2g = Du (approx.)
Or: Du = 1.45 m (approx.)
Q=W Du u
... 70 = 33 x 1.45 u
... u = 1.46; u2/2g = 0.1086 m
H= Du + u2/2g
1.45 = Du + 0.1086
Du = 1.3414 m
h= Du - Dd = 1.3414 – 1.0=0.3414 m
Adopt h = 0.3414 m. Since h is actually more than ¼ Dd, therefore, the value of afflux
arrived by the Weir Formula is to be adopted.
16.7. Example: The unobstructed cross-sectional area of flow of a stream of 90 m 2 and
the width of flow is 30 m. A bridge of 4 - spans of 6 m clear is proposed across it. Calculate
the afflux when the discharge is 280 m3/s.
w = 30 m; L = 24 m, Dd= 30= 3.00 m
๐ ฟ
The depth before the construction of the bridge is the depth downstream of the bridge
its construction. Hence, Dd= 3.00 m
= 0.8
By the Orifice Formula the discharge through the bridge
Now, the discharge just upstream of the bridge
280 = (3 + h)30u
…… .(16.4)
Putting for h from (16.4) in (16.3) and rearranging
u = 2.33 + .02195 u3
L = 2.81 m/sec
Putting for u in (1 6.4)
h = 0.32 m < 1/4 Dd
16.8. Example: A bridge of 3 spans of 8 m each is proposed across a stream, whose
unobstructed width is 36 m, slope 1/2000 and discharge 400 m3/sec. Calculate the
afflux (n=0.03) (Fig. 16.2).
We have first to find Dd,
Q = AV = (RP) V = RWV
... RV =
= 11.11
Knowing n = 0.03; S = 1/2000, read velocity for various values of R from Plate 3 and
select that pair whose product is 11.11. Thus, we get.
R = 5.1
V = 2.18
Take Dd = R =5.1 m
Now, W = 36rn, L = 24 M, Dd = 5.1 m
๐ ฟ
= 0.67 Therefore, C0 = 0.865; e = 0.95
By the Orifice Formula, the discharge through the bridge
ARTICLE 17
17.1. Calculating Discharge by the Weir Formulae
Example: The unobstructed width of a stream is 40 m. The linear waterway of a bridge
across is 27 m. In a flood, the average depth of flow downstream of the bridge was 3.0
m and the afflux was 0.9 m. Calculate the discharge (Fig. 17.1).
Since h is more than 0.25 Dd, therefore, the Weir Formula will apply
w = 40 m; L = 27 m, h = 0.9 m
Let the velocity of approach be u m/sec. The discharge at a section just upstream of the
Q = u x 3.9 x 40 = 156 u
u 2/3 -0.0222 u2 = 1.70
or u = 2.45 m/sec
Putting the value of u in (17.1 a) or (17.1b) we get Q
Q – 156 x 2.45
= 382 m3/sec
๐ ฟ
Try the Orifice Formula
= 0.67
... C0 = 0.865; e = 0.95
u = 2.36
Substituting for u in (17.1c) and (17. Id) we get Q
Q= 156 x 2.36 = 368.16 m3/sec
This result is about the same as given by the first method. In fact, the Orifice Formula,
with the recommended value of Co and e gives nearly correct results even where the
conditions are appropriate for the Weir Formula. But the converse is not true.
17.2. Calculating Discharge by the Orifice Formula
Example: The unobstructed width of a stream is 30 m and the linear waterway of the
bridge across is 22 m. During a flood the average depth of flow downstream of the bridge
was 1 .6 m and the afflux 0.10 m. Calculate the discharge (Fig. 17.2).
Given: W = 30 m, L = 22 m, h 0.1m, Depth of flow =1.6 m. Let velocity of approach be u
m/s. The discharge at a section just upstream of the bridge will be.
Q = u x 1.7 x 30
Contraction =
๐ ด
๐ ฟ
= 0.73
Corresponding to this C0 = 0.87 and e = 0.90
The discharge under the bridge, by the Orifice Formula
Substituting for u in (1 7.2a) and (1 7.2b) to get Q
= 1.51 x 1.7 x 30
= 77.01 cu. m/sec
17.3. The Border Line Cases: An example will now follow to illustrate what results are
obtained by applying the Weir Formula and Orifice Formulae to cases which are on the
border line, i.e., where the afflux is just ¼ Dd.
17.4. Example: A stream whose unobstructed width is 35 m is spanned by a bridge
whose linear waterway is 30 m. During a flood the average downstream depth was 2.6 m
and the afflux was 0.65 m. Calculate the discharge (Fig. 17.3).
u = 3.27 m/s
Put for u in (17.3a) or (17.3b)
Q= 1 13.75 x 3.2 =371.96 m3/s
By the Orifice Formula
Ce = 0.90
๐ ด
๐ ฟ
= 0.85
Contraction =
e = 0.44
If u is the velocity of approach, the discharge just upstream of the bridge.
... (17.3c)
Q = 35 x3.25u = 1 13.75u
The discharge under the bridge by the Orifice Formula
Q = 0.906 x 4.43 x 30 x 2.6 (0.65 + 0.0735 u2) ½
= 310.98 (0.65 + 0.0735 u2) ½
Equating values of Q from (17.3c) and (17.3d) and Squaring and rearranging
113.75 u = 310.98 x (0.65 + 0.0735 u2) ½
... u = 3.27 m/s
Substituting for u in (1 7.3c) and (1 7.3d), we get Q
Q= 1 13.75 x 3.27 =371.96 m3/s
ARTICLE 18
18.1. In plains where the ground slopes are gentle and the natural velocities of flow in
streams are low, the flood water may spill over one or both the banks of the stream at
18.2. Height of Approach Roads: Consider the case where main channel carries the
bulk of the discharge and a small fraction of it flows over the banks somewhere upstream
of the bridge. If the overflow strikes high ground at a short distance from the banks, it can
be forced back into the stream and made to pass through the bridge. This can be done
by building the approach roads of the bridge solid and high so that they intercept the
overflow. In this arrangement, the linear waterway of the bridge must be ample to handle
the whole discharge without detrimental afflux. Also, the top level of the approach road
must be high enough to prevent overtopping. If the velocity of the stream is V(m/s), the
๐ ฃ2
water surface level, where it strikes the road embankment, will be 19.6 (m) higher than HFL
in the stream at the point, where the overflow starts. This arrangement is, therefore,
normally feasible where the stream velocity is not immoderately high.
18.3. Subsidiary or Relief Culverts: Sometimes, however, the overflow spreads far and
away from the banks. This is often the case in alluvial plains, where the ground level falls
continuously away from the banks of the stream. In sue h cases, it is impossible to force
the overflow back into the main stream. The correct thing to do is to pass the overflow
through relief culverts at suitable points in the road embankment. These culverts have to
be carefully designed. They should not be too small to cause detrimental ponding up of
the overflow, resulting in damage to the road or some property, nor, should they be so
big as to attract the main current.
18.4. Dips and Breaching Sections in Approach Roads: It is sometimes feasible as
well as economical to provide permanent dips (or alternatively breaching sections) in the
bridge approaches to take excessive overflows in emergencies. The dips or breaching
sections have to be sited and designed so that the velocity of flow through them does not
become erosive, cutting deep channels and ultimately leading to the shifting of the main
current. However, since the state highways, national highways, and expressways are to
be designed as all-weather roads, dips and breaching sections may be considered only
for rural roads and major district roads.
18.5. Retrogression of Levels: Suppose water overflows a low bank somewhere
upstream of the bridge and after passing through a relief culvert, rejoins the main stream
somewhere lower down. When the flood in the main channel subsides, the ponded up
water at the inlet of the subsidiary culvert gets a free fall. Under such conditions deep
erosion can take place. A deep channel is formed, beginning at the outfall in the mains
stream and retrogressing towards the culvert. This endangers the culvert. To provide
against this, protection has to be designed downstream of the culvert so as to dissipate
the energy of the falling water on the same lines as is done on irrigation falls. That is a
suitable cistern and baffle wall should be added for dissipating the energy and the issuing
current should be stilled through a properly designed expanding flume.
ARTICLE 19
19.1. Feasibility of Pipe and Box Culverts Flowing Full
19.1.1. Some regions along plain consist of vast flat without any deep and defined
drainage channels in it. When the rain falls, the surface water moves in some direction in
a wide sheet of nominal depth. So long as this movement of water is unobstructed, no
damage may occur to property or crops. But when a road embankment is thrown across
the country intercepting the natural flow, water ponds up on one side of it. Relief has then
to be afforded from possible damage from this ponding up by taking the water across the
road through causeways or culverts.
19.1.2. In such flat regions the road runs across wide but shallow dips and, therefore, the
most straightforward way of handling the surface flow is to provide suitable dips (i.e.,
causeways) in the longitudinal profile of the road and let water pass over them.
1 9.1.3. There may, however, be cases where the above solution is not the best. Some
of its limitations may be cited. Too many causeways or dips detract from the usefulness
of the road. Also, the flow of water over numerous sections of the road, makes its proper
maintenance problematic and expensive. Again, consider the case of a wet cultivated or
waterlogged country (and flat plains are quite often swampy and waterlogged) where the
embankment has necessarily got to be taken high above the ground. Frequent dipping
down from high road levels to the ground produces a very
undesirable road profile. And, even cement concrete slabs, in dips across a waterlogged
country, do not rest evenly on the mud underneath them. Thus, it will appear that
constructing culverts in such circumstances should be a better arrangement than
providing dips or small causeways.
19.1.4. After we have decided that a culvert has to be constructed on a road lying across
some such country, we proceed to calculate the discharge by using one of the run off
formulae, having due regard to the nature of terrain and the intensity of rainfall as already
explained in Article-4. But the natural velocity of flow cannot be estimated because (i)
there is no defined cross-section of the channel from which we may take the area of crosssection and wetted perimeter and (ii) there is no measurable slope in the drainage line.
Even where we would calculate or directly observe the velocity, it may be so small that
we could not aim at passing water through the culvert at that velocity,
because the area of waterway required for the culvert (A+๐ ) is prohibitively large. In such
cases the design has to be based on an increased velocity of flow through the culvert and
to create the velocity the design must provide for heading up at the inlet end of the culvert.
Economy, in design being the primary consideration, the correct practice, indeed is to
design a pipe or a box culvert on the assumption that water at the inlet end may head
upto a predetermined safe level above the top of the inlet opening. This surface level of
the headed up water at the upstream end has to be so fixed that the road bank should
not be overtopped, nor any property in the flood plain damaged.
Next, the level of the downstream water surface should be noted down. This will depend
on the size of the slope of the leading out channel and is normally, the surface level of
the natural unobstructed flow at the site, that prevails before the road embankment is
After this we can calculate the required area of cross-section of the barrel of the culvert
by applying the principles of hydraulics discussed in this Article.
19.1.5. The procedure set out above is rational and considerable research has been
carried out on the flow of water through pipe and box culverts, flowing full.
19.1 .6. In the past, use was extensively made of empirical formulae which gave the
ventway area required for a culvert to drain a given catchment area. Dun's Drainage Table
is one of the class and is purely empirical. This table is still widely used, as it saves the
trouble of hydraulic calculations.
But it is unfortunate that recourse is often taken rather indiscriminately to such short cuts,
even where other more accurate and rational procedure is possible and warranted by the
expense involved. Dun's Table or other in that class, should NOT be used until suitable
correction factors have been carefully evolved from extensive observations (in each
particular region with its own singularities of terrain and climate) of the adequacy or
otherwise of the existing culverts vis-a-vis their catchment area.
19.1.7. Considerations of economy require that small culverts, in contrast with relatively
larger structures across defined channels, need not be designed normally to function with
adequate clearance for passing floating matter. The depth of a culvert should be small
and it does not matter if the opening stops appreciably below the formation level of the
road. Indeed, it is correct to leave it in that position and let it function even with its inlet
submerged. This makes it possible to design low abutments supporting an arch or a slab,
or alternatively, to use round pipes or square box barrels.
19.1.8. High headwall should not be provided for retaining deep over-fills. Instead of this
the length of the culverts should be increased suitably so that the road embankment, with
its natural slopes, is accommodated without high retaining headwalls.
19.1.9. Where masonry abutments supporting arches or slabs are designed for culverts
functioning under "head", bed pavements must be provided. And, in all cases, including
pipe and box culverts, adequate provision must be made at the exit against erosion by
designing curtain walls. Where the exit is a free fall, a suitable cistern and baffle wall must
be added for the dissipation of energy and stilling of the ensuring current.
19.2. Hydraulics of the Pipe and Box Culverts Flowing Full
19.2.1. The permissible heading up at the inlet: It has been explained already that
where a defined channel does not exist and the natural velocity of flow is very low, it is
economical to design a culvert as consisting of a pipe or a number of pipes of circular or
rectangular section functioning with the inlet submerged. As the flood water starts heading
up at the inlet, the velocity through the barrel goes on increasing. This continues till the
discharge passing through the culvert equals the discharge coming towards the culvert.
When this state of equilibrium is reached the upstream water level does not rise any
For a given design discharge the extent of upstream heading up depends on the vent way
of the culvert. The latter has to be so chosen that the heading up should not go higher
than a predetermined safe level. The criterion for safety being that the road embankment
should not be overtopped, nor any property damaged by submergence. The fixing of this
level is the first step in the design.
19.2.2. Surface level of the tail race: It is essential that the HFL in the outfall channel
near the exit of the culvert should be known. This may be taken as the HFL prevailing at
the proposed site of the culvert before the construction of the road embankment with
some allowance for the concentration of flow caused by the construction of the culvert.
19.2.3. The operating head when the culverts flow full: In this connection the cases
that have to be considered are illustrated in Fig. 19.1. In each case the inlet is submerged
and the culvert flows full. In case (a) the tail race water surface is below the crown of the
exit and in case (b) it is above that. The operating head in each case is marked "H". Thus,
we see that: "When the culvert flows full, the operating head, H, is the height of the
upstream water level measured from the surface level in the tail race or from the crown
of the exit of the culvert whichever level is higher".
19.2.4. The velocity generated by "H" : The operating head "H" is utilized in (i) supplying
the energy required to generate the velocity of flow through the culvert (ii) Forcing water
through the inlet of the culvert, and (iii) overcoming the frictional resistance offered by the
inside wetted surface of the culvert.
If the velocity through the pipe is v, the head expended in generating is
๐ ฃ2
As regards the head expended at the entry it is customary to express it as a fraction
Ke of the velocity head
kƒ of the
๐ ฃ2
๐ ฃ2
. Similarly, the head required for overcoming the friction of
. From this it follows that:
H = [l + KQ + Kƒ]
๐ ฃ2
... (19.1)
From this equation we can calculate the velocity v, which a given head H will generate
in a pipe flowing full, if we know Ke and Kf.
19.2.5. Values of Ke and Kf : Ke principally depends on the shape of the inlet. The
following values are commonly used:
Ke = 0.08 for bevelled or
Bell - mouthed entry
= 0.505 for sharp edged entry
... (19.2)
As regards Kf it is a function of the Length L of the culvert, its hydraulic mean radius R,
and the co-efficient of rugosity n of its surface.
The following relationship exists between Kf and n:
For cement concrete circular pipes or cement plastered masonry culverts of rectangular
section, with the co-efficient of rugosity n = 0.01 5, the above equation reduces to:
The graphs in Fig. 19.2 are based on Equation 1 9.4. For a culvert of known sectional
area and length, Kf can be directly read from these graphs.
19.2.6. Values of Ke and Kf modified through research: Considerable research has
recently been carried out on the head lost in flow through pipes. The results have
unmistakably demonstrated the following: The entry loss co-efficient Ke depends not only on the shape of the entry but also on the
size "entry and the roughness of its wetted surface. In general, Ke, increases with an
increase in the e of the inlet.
Also Kf the friction loss co-efficient, is not independent of Ke. Attempts to make the entry
efficient repercuss adversely on the frictional resistance to flow offered by the wetted
surface of the barrel. In other words, if the entry conditions improve (i.e. if Ke decreases),
the friction of the barrel increases (i.e. Kf increases). This phenomenon can be explained
by thinking of the velocity distribution inside the pipe. When the entry is square and sharp
edged, high velocity lines are concentrated nearer the axis of the barrel, while the bellmouthed entry' gives uniform distribution of velocity over the whole section of the barrel.
From this it follows that the average velocity being the same in both cases, the velocity
near the wetted surface of the pipe will be lower for square entry than for bellmouthed
entry. Hence, the frictional resistance inside the culvert is smaller when the entry is square
than when it is bell-mouthed. Stream lining the entry is, therefore, not an unmixed
Consequently, it has been suggested that the values of Ke, and Kf should be as given in
Table 19.1
Table 19.1 Values of Ke and KfI9 l
Entry and
Circular pipes
Square entry
Bevelled entry
1.107 R0.5
Ke =
0.00394 L/R1 -2
Rectangular culverts
Bevelled entry
Square entry
0.572 R0.3
0.00394 L/R1. 2
0.0035 L/R1. 25
0.0035L/R 1.25
19.2.7. Design calculations: We have said that
H = (l +Ke + Kf)
๐ ป
i.e. v = 4.43 (1+๐ พ
๐ +๐ พ๐
๐ ป
Q = A x 4.43 (1+๐ พ
๐ +๐ พ๐
Suppose we know the operating head H and the length of the barrel L, and assume that
the diameter of a round pipe or the side of a square box culvert is D.
From D calculate the cross-sectional area A and the hydraulic mean radius R of the
Now from R and L compute Ke and Kf using appropriate functions from Table 19.1. Then,
calculate Q from Equation (19.5). If this equals the design discharge, the assumed size
of the culvert is correct. If not, assume a fresh value of D and repeat.
19.2.8. Design chart (Plate 24): Equation (19.5) may be written as
Q = ๏ ฌ √2๐ ๐ ป
…… (19.6)
๐ ด
๏ ฌ = (1+๐ พ +๐ พ
๏ ฎ๏ ฎ๏ ฎ๏ ฎ๏ ฎ
๐ )1/2
(๏ ฑ๏ น๏ ฎ๏ ท)
It is obvious that all components of ๏ ฌ in Equation (19.7) are functions of the cross-section,
length, roughness, and the shape of the inlet of the pipe. Therefore, ๏ ฌ represents the
conveying capacity of the pipe and may be called the 'Conveyance Factor'. The
discharge, then depends on the conveyance factor of the pipe and the operating head. In
Plate 23, curves have been constructed from equation (1 9.7) from which Q can be
directly read for any known values of ๏ ฌ and H.
Also, in the same Plate, Tables are included from which X can be taken for any known
values of (i) length, (ii) diameter in case of circular pipes or sides in case of rectangular
pipes, and (iii) conditions of entry, viz., sharp-edged or round. The material assumed is
cement, concrete and values of Ke and Kf used in the computation are based on functions
in Table 19.1.
The use of Plate 24 renders the design procedure very simple and quick. Examples will
now follow to illustrate.
19.2.9. Example data:
(1) Circular cement concrete pipe flowing full with bevelled entry
(2) Operating head = 1 m
(3) Length of the pipe = 25 m
(4) Diameter = 1 m
Find the discharge.
For L=25 m and D=1 m, the conveyance factor
๏ ฌ =0.618
See, in Plate 24, the Table for circular pipes with rounded entry.
Now refer to the curves in the same Plate. For ๏ ฌ = 0.6 1 8 and H= 1 m
Q=2.72 m3 /sec
19.2.10. Example: Design a culvert consisting of cement concrete circular pipes with
bevelled entry and flowing full, given: (Fig. 19.3).
=20 m
R.L. of ground in metres
H.F.L of tail race in metres
Permissible heading up at inlet R.L.
Length of culvert
Since we shall try pipes of diameters exceeding 0.8 m, the culvert will function as
Assumed value of D = (1) 1 m; (2) 1.5 m;
H= 1.8 -D = (1)0.8 m; (2) 0.3 m;
From Plate 24, Q = (1) 2.54 m3/s; (2) 3.5 m3/s
Number of pipes
Require 10/Q = (1)3.93; (2)2.85
Hence, 4 pipes of 1 metre diameter will suit.
Discharge per pipe
Say 4 Say 3
Improved Intake to Increase Culvert Capacity
Culverts In dips and low height roads are often provided below ground. The portion of
culvert lying under ground eventually silt up and its conveying capacity is drastically
reduced. Improved intake design with proper design of inlet and outlet transitions
connecting the culvert with the channel like a siphon in canal cross-drainage works will
increase the conveying capacity of culvert. It is advisable to provide the inlet level of
culverts at least 150 mm below lowest bed level at the section, so that water will always
pass through the culvert.
“Scour in Culverts”
Where the bed is unprotected, scour depth in the abutments should be found as in the
case of bridges on alluvial fine and coarse soil to find the depth of foundation. When the
bed is rigid as in case of box culverts, curtain/ cut-off walls must be provided both
upstream and downstream to protect against scour. Flexible stone pitching/stone gabions
laid over geo-synthetic filter of length 3 to 4 times estimated scour depth should be
provided downstream to arrest erosion of bed and banks where outlet velocity is high.
ARTICLE 20
20.1. Floor Protection Works:
In case structures founded on erodible soil are protected against scour by floor
protection works, the following is considered as sound practice.
20.1.1. For structures where adoption of shallow foundations becomes economical by
restricting the scour, floor protection may be provided. The floor protection will comprise
of rigid flooring with curtain walls and flexible apron so as to check scour, washing away
or disturbance by piping action, etc. Usually performance of similar existing works is the
best guide for finalizing the design of new works. However, the following minimum
specification for floor protection shall be followed while designing new structures subject
to the general stipulation that post protection works velocity under the structures does not
exceed 2 m/s and the intensity of discharge is limited to 2m3/m. In case it does not
satisfies, design of floor protection work need to be done as per IRC 89
20.1.2. Suggested Specifications:
20.1.2.1. Excavation for laying foundation and protection works should be carried out as
per specifications under proper supervision. Before laying the foundation and protection
works the excavated trench should be thoroughly inspected by the Engineer-in-Charge
to ensure that:
(a) There are no loose pockets, unfiled depressions left in the trench.
(b) The soil at the founding level is properly compacted to true lines and level.
(c) All concrete and other elements are laid in dry bed.
20.1.2.2. Rigid flooring: The rigid flooring should be provided under the bridge and it
should extend for a distance of at least 3 m on upstream side and 5 m on downstream
side of the bridge.
However, in case the splayed wing walls of the structure are likely to be longer, the
flooring should extend upto the line connecting the end of wing walls on either side of the
The top of flooring should be kept 300 mm below the lowest bed level.
Flooring should consist of 150 mm thick flat stone/bricks on edge in cement mortar 1 :3
laid over 300 mm thick cement concrete M20 grade laid over a layer of 1 50 mm thick
cement concrete M15 grade. Joints at suitable spacings (say 20 m) may be provided.
20.1.2.3. Curtain walls: The rigid flooring should be enclosed by curtain walls (tied to the
wing walls) with a minimum depth below floor level of 2 m on upstream side and 2.5 m on
downstream side. The curtain wall should be in cement concrete M20 grade or brick/stone
masonry in cement mortar 1:3. The rigid flooring should be continued over the top width
of curtain walls. In this context, relevant provision in "Guidelines for design and
construction of river training and control works for road bridges", IRC: 89 is also referred.
20.1.2.4. Flexible apron: Flexible apron 1 m thick comprising of loose stone boulders
(Weighing not less than 40 kg) should be provided beyond the curtain walls for a minimum
distance of 3 m on upstream side and 6 m on downstream side. Where required size
stones are not economically available, cement concrete blocks or stones in wire crates
may be used in place of isolated stones. In this context, relevant provision in IRC: 89 is
also referred.
20.1.2.5. Wherever scour is restricted by provision of flooring/flexible apron, the work of
flooring/apron etc., should be simultaneously completed along with the work on
foundations so that the foundation work completed is not endangered.
20.2. Maintenance:
20.2. 1. The bridge structures are more susceptible to damages during monsoon. It is
generally observed that following factors contribute mainly to damage.
(a) Choking of vents
(b) Wash outs of approaches
(c) Dislodgement of wearing course and cushion
(d) Scour on D/S (downstream)
(e) Silting on U/S (upstream)
(f) Collection of debris on approaches in cutting
20.2.2. To minimize the occurrence of above phenomena, it is necessary to take
adequate steps as below:
(1) The vents should be thoroughly cleaned before every monsoon.
(2) The bridge vents should be cleared after the first monsoon flood as the flood carries
maximum debris with it.
(3) Keep approaches almost matching with existing bank, i.e., cutting or embankment
should be minimum to avoid wash outs of approaches.
(4) Disposal of water through side gutters shall be properly planned so that it does not
damage the cross-drainage work proper.
(5) The wearing coat with cushion should be sufficiently stable and it should not get
dislodged during floods.
(6) In the event of approaches being in cutting there is a tendency of whirling of water at
the approaches. This leads to collection of debris in the approaches. After the floods
recede, huge heap of debris is found on the approaches. This should be quickly cleared.
ARTICLE 21
21.1. Raft foundation is preferred when the good foundable strata is not available within
a reasonable depth. Thus, the sandy layer or sand and silty foundations warrant provision
of raft foundation. While providing raft foundation, some important points should be kept
in view.
21.1.1. Raft top should be kept 300 mm below the lowest bed level. This will ensure
protection to raft and also would avoid silting tendency on U/S and scouring tendency on
D/S. The raft will also not be subjected to stresses due to temperature variations.
21.1.2. U/S and D/S aprons should be provided in accordance with IRC 89 to protect the
bridge from scouring or undermining. The width of U/S and D/S aprons should be 1.5 dsm
and 2.0 dsm respectively (Fig. 21.1).
21.1.3. The depth of cut-off wall should be 30 cm below the scour level. The normal scour
๐ ท2
depth is worked out by the formula dsm = 1 .34 x (๐ พ ๐ ) (Refer Equation 9.1).
(Scour Depth need not be increased by any factor as in case of open foundations as
stipulated in IRC: 78).
Fig. 21.1 Scour Depth and Apron Width for Raft
21.1.4. Longitudinal cut-off walls should be provided on U/S and D/S side and they should
be connected by cross cut off walls. Longitudinal cut-off walls safeguard the bridge from
scour whereas the cross-cut-off walls keep the longitudinal cut-off walls in position and
also protect the bridge from scouring particularly due to out flanking.
21.1.5. The raft is generally as wide as the deck but in certain cases may be narrower
than the deck (Fig. 21.2).
21.1.6. Pressure relief holes may be provided in the raft to relieve the raft from possible
uplift pressure from below. The holes need to be carefully packed with graded filter
material to prevent outflow of soil particles of the foundation strata along with the flow of
water (Fig. 21.3).
ARTICLE 22
22.1. Generally, the black cotton (B.C.) soil is of expansive nature. As it comes in contact
with water, the montmorillonite group cells expand. This phenomenon leads to heavy
pressure on structure and the structure may develop cracks and fail. It is, therefore,
necessary to safeguard the structure from the ill-effects of the damaging nature of the
soil. It is desirable to cut the contact of expansive soil and the foundation structure. This
can be achieved by providing a sandy media all around the foundation. Such nonexpansive layer not only cuts the all around contact between soil and foundation but also
absorbs energy of swelling and shrinking of foundation soil below the layer of sand and
keeps the foundation safe.
22.2. Since expansive soils have low SBC, ground improvement is required. The same
can be done by introducing a granular layer of suitable thickness below the pipe which
would help in load dispersion to a wider area and thus reduce the stress on the soil below
the granular layer.. Such layer, improves Safe Bearing Capacity (SBC) of the strata to a
considerable extent and safeguards the foundation from the adverse effects of the
expansive soil also (Fig. 22.1).
Fig. 22.1 Hume Pipe Culvert in BC Soil
ARTICLE 23
23.1. Where to Provide Box Structures: Box structures are hydraulically efficient
structures where thickness of walls and slab are small and there is least obstruction to
flow. Box cell structures are most suitable for soils having SBC 8-12 T/m2. Box cell
structures are also provided to keep the deck level lower. The box being a continuous
structure, the thickness of top slab is less than that of a simply supported span of same
length. Also, expansion joints at the deck ends are dispensed with.
When the river or Nalla has sandy bed and/or purely clayey strata, the independent
foundations are likely to be deeper and this may enhance the cost of culverts and small
bridges. Under these circumstances box culverts are found to be a better solution.
Several such box cell structures have shown a good in service performance. Purely sandy
soil or clayey strata may be at few places but mixed soils are available in several cases.
Where แถฒ value of mixed soil is less than 15°, it may be treated as a clayey soil. Similarly,
where safe bearing capacity of soil is found to be less than 10 t/m2, box culverts are most
suitable for such type of soils.
23.3. Foundation: Where there is a clayey stratum, top soil below box may be replaced
by a layer of granular soil of suitable thickness like sandy murum and stone dust etc. or
Where there is murum and mixed soil having <j) more than 1 5°, there is no need of
providing sandy layer.
The box cell structures Shall be designed as per IRC 112.
Box cell structures are to be provided with curtain walls and apron and these must be
completed before floods. The best practice is to lay foundations of curtain wall and apron
first and then lay box.
( 1 ) "Highway Practice in the United States of America", Public Roads Administration,
Federal Works Agency, Washington 25, D.C. ,1949.
(2) "Engineering Hydraulics, Proceedings of the Fourth Hydraulic Conference", Iowa
Institute of Hydraulics Research, June 12-15, 1949, Edited by Hunter Rouse, John Wiley
& Sons, Inc., New York, 1950.
(3) "Elements of Applied Hydrology", Don Johnstone and William P. Cross, The Ronald
Press Company, New York.
(4) "Rainfall and Run-off', Edgar E. Foster, The Macmillian Company, New York,.
(5) "The Flood Estimation and Control", B.D. Richards, Chapman and Hall Ltd., London.
(6) "A Treatise on Applied Hydraulics", Herbert Addison, 3 rd Edition, Chapman and Hall
Ltd., London.
(7) "Irrigation Pocket Book", Compiled by Robert Burton, Bockley, E & F N Spoon Ltd.,
(8) "Canals and Related Structures, Design", Supplement No. 3, Part 2, Engineering
Design of Volume X, Design and Construction Reclamation Manual", United States
Department of Interior, Bureau of Reclamation.
(9) University of Iowa Studies, Bulletin No. 1, "Flow of Water Trough Culvert", David L.
Yarid, Floyd A. Nagler and Sherman M. Woodward, The University of Iowa, Iowa.
(10) United States Department of Agriculture, Technical Bulletin No. 442, "Bridge Piers
as Channel Obstructions", David L. Yarnell, United States Department of Agriculture,
Washington D.C.
(11) "California Culvert Practice", State of California, Department of Public Works,
Division of Highways, Sacramento.
(12) "Highway Design and Construction", Arthur G. Bruce, and John Clarkeson, 3rd
Edition, International Text-book Company, Scranton, Pennsylvania.
(13) "Hydraulics for Engineers and Engineering Students", F.C. Lea, 4th Edition, Edward
Arnold &Co.
(14) "Military Engineer Services Handbook, Roads, Volume III", Sixth Edition,
Government of India, Central Publication Branch, Calcutta.
(15) "Standard Specification and Code of Practice for Roads Bridges - Section I – General
Features of Design ", IRC:5
(16) '"Standard Specifications and Code of Practice for Road Bridges - Section II - Loads
and Load Combinations", IRC:6
(17) "Recommended Practice for Numbering Culverts, Bridges and Tunnels", IRC:7.
(18) "Standard Specifications and Code of Practice for Road Bridges, Section IV - Brick,
Stone and Block Masonry ", IRC:40
(19) "Standard Specifications and Code of Practice for Road Bridges, Section VII Foundations & Substructure", IRC:78
(20) "Guidelines for Design and Construction of River Training and Control Works for
Road Bridges", IRC: 89
(21) "Code of Practice for Concrete Road Bridges", IRC:112
(22) "Project Preparation Manual for Bridges ", IRC:SP 54
(23) "Regime Flow in Incoherent Alluvium", G. Lacey, Central Board of Irrigation and
Power, Paper No. 20.
(24) "The Hydraulic of Culverts", by F.T. Mavis, The Pennsylvania State College
Engineering Experimental Research Station Series Bulletin No. 56. | {"url":"https://studylib.net/doc/25918381/guidelines-for-the-design-of-small-bridges-and-culvert-dr...","timestamp":"2024-11-05T05:30:06Z","content_type":"text/html","content_length":"179644","record_id":"<urn:uuid:978e4ec1-5c57-4430-ab83-8b71bc6b521a>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00435.warc.gz"} |
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tl;dr – It is possible to construct a winning strategy at the lottery by choosing the numbers that other people rarely select. We discuss this and prove it on a small example.
There are many things I don’t like with so-called math reasoning and lotteries, and I wanted to write about it for a very long time. So, on the one hand we have the classic scammers who try to sell
you the “most probable numbers” (or alternatively the “numbers that are due”). Of course, neither strategy is mathematically valid (because the draws are independent). On the other hand, many
“educated” and “rational” people argue that, given that the expected value of a lottery ticket is negative (because the probability of wining a prize at the lottery is very low), smart people should
never buy lottery tickets.
Comic by Zach Wiener, http://www.smbc-comics.com
Now what if we could find a (mathematically correct!) strategy to make the expected value of our ticket positive? The idea is to choose the numbers that other players choose the least often, so that
when we win a prize, it will be divided among fewer other players. But will it be enough to make a significant difference?
The example
Let’s consider a lottery where players have to choose 6 numbers out of 19. The total number of players is 10000. The favorite numbers of the players are 1, 2, 3, 4, 5 and the least favorite are 15,
16, 17, 18, 19. They are respectively selected 2 times more often and 2 times less often than the other numbers 6 – 14. The company who runs the lottery decides to give the players back 90 percent
of the amount of the tickets (thus ensuring a 10% profit) depending on the number of numbers they have chosen that also are in the right combination:
• 0 or 1 correct number: 0%
• 2 correct numbers: 42%, shared with other winners
• 3 correct numbers: 10%, shared with other winners
• 4 correct numbers: 3%, shared with other winners
• 5 correct numbers: 4%, shared with other winners
• 6 correct numbers (the jackpot): 40%, shared with (the eventual) other winners
Then we compute the expected value for each ticket that was bought. You can find the R code I used on my GitHub page. I plotted the expected gains against an indicator of the rarity of the
combination chosen by each player (the harmonic mean of the inclusion probabilities):
Expected gains wrt a measure of the frequency of the combination chosen
As we predicted, the expected gains are higher if you chose an “unpopular” combination. But what impresses me most is the order of magnitude of the effect. It is indeed possible to find a combination
that yields a positive expected value (points on the left that are above the red line)!
Further work
I have no idea how all this works when we change the parameters of the problem: numbers to choose from (49 in France for example), number of players, choices of the players (inclusion probabilities
of the numbers), payoffs, etc. I bet that the shape of the curve remains the same, but I wonder how high the expected value can get for the rarest combinations, and if it is always possible to find a
winning strategy. I might try to work on an analytical solution when I find some time because I believe it involves some sampling theory.
Finally, a question to all people who never play the lottery because the expected value is negative, would you start buying tickets now that you know there exists a strategy with positive expected
Comic by Zach Wiener, http://www.smbc-comics.com
PS: Henri pointed out chapter 11 of Jordan Ellenberg’s “How not to be wrong” which deals with interesting mathematical facts about the lottery, including a similar discussion as this post. Be sure to
check it out, it’s really great! | {"url":"https://nc233.com/tag/lottery/","timestamp":"2024-11-05T18:48:19Z","content_type":"text/html","content_length":"48458","record_id":"<urn:uuid:da8ae7c2-d414-4ffc-930e-84193ffe1dc7>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00058.warc.gz"} |
Chaining/Taping On Level Ground And Sloping GroundChaining/Taping On Level Ground And Sloping GroundChaining/Taping On Level Ground And Sloping Ground
The process of measuring the distance between any two points by using a chain is called chaining. If used tape instead of a chain then the process is called taping. Generally, we use tape nowadays
due to its accuracy and portability. Chains help us to measure the distance between any two points, which ultimately means it supports us to measure the area of any ground, as the surface of grounds
can be divided into many points.
Level grounds are also called uneven ground which is the contradiction of plane surface. It is a very cozy job to measure distance on a plane surface. But it gets complexed when the surface is uneven
or on a certain level.
Instruments required for taping on level grounds are:
· Tape
· Ranging Rod (minimum 3)
· Peg
· Arrows (minimum 10)
If we want to measure the distance between two points AB. Let us consider the distance between A and B is L. And let L be the distance greater than the total length of the tape. If we are using tape
of 30m then L is greater than 30m. So to measure L, we divide L into different segments. One segment could be in the range of 10-20m.
Even though we have 30m tape, we do not utilize the whole tape and only use a maximum of 20m. It is because if we use the whole 30m of the tape then there is a sagging effect. This means the tape
does not get stretched to its required potential as its own weight causes it to sag. So we use a maximum of 20m of our tape to measure distance in each segment.
Then the procedure to measure distance on level grounds are:
Step 1: Install peg in the two endpoints. Here, we install one peg at
mark A and another peg at mark B.
Step 2: Fix two ranging rod at the end stations (points) i.e. one rod at
station A and another at station B.
Step 3: 1^st person stands about 2m behind the ranging rod at the station
A. 2^nd person stands at point B with the ranging rod. 3^rd person carries a ranging rod and moves forward from point A. 3^rd person starts to move the ranging rod to and fro under the command of 1^
st person until the ranging rod lies on line AB and mark that point as C. Then the third person installs an arrowhead at point C.
The ranging rod will lie on the line AB when this ranging rod will coincide or blocks the ranging rod at B when viewed by 1^st person at station A.
Step 4: This process is repeated several times and required
intermediate points are determined. Let us divide AB into 3 segments pointed as points A, C, D, and B.
Step 5: Now, we use the tape or chain. For this, we have preferred
tape. We measure the distance between those three segments using tape each segment at one time, starting from point A.
Important Tip: While measuring any segment, say AC we measure or use tape from the nearest ground level as possible. It is done to avoid the sag effect or the inaccurate measurement of distance due
to the inclined installation of the ranging rod. Also, while measuring, we should make sure that we measure the perpendicular distance between any two points.
Let the recorded measurements of AC, CD, and DB be L1, L2, and L3 respectively. And consider them as forward length.
Step 6: Now, again we start to measure different segments. But this
time we start to measure from point B. For this, again we divide the length AB into several segments. But this time we start from point B. Now, the 1^st person starts at point B and starts ranging.
Let the intermediate points be D’, and C’.
We cannot use the previous intermediate points C and D while measuring from B, because we have to provide an independent check. If we had used points C and D, then it would provide us a dependent
check and it would not make sense as we just measured the same distance segments two times.
Let the recorded lengths of BD', D'C', and C'A be L1' L2' AND L3' respectively. Consider those lengths as backward lengths.
Hence, total forward length(L)= L1+L2+L3
Total backward length(L’) =L1’+L2’+L3’
Average length= (L+L’)/2
Discrepancy= │ L – L’ │
The average length is the required measurement of length L, only if the precision is greater than or equal to (≥) 1 in 2000. If precision does not lie within this range, then the whole process is
carried out again, unless a precision is obtained. Then only the required length is obtained.
Precision=1/average length/discrepancy
What do you mean by precision is 1 in 2000 (1/2000)?
The precision is 1 in 2000. This means while measuring the distance of 2000m, the discrepancy should only be 1m. Or, while measuring the distance of 2000m, the fluctuating value (or change in value)
between forward and backward length should only be a maximum of 1m.
Chaining/Taping on sloping ground
By the direct method,
The whole process is the same as above i.e. taping on level ground. For measuring the sloping ground, the precision should be 1 in 1000.
Here the distance between point A and B is L1+L2+L3 while starting from point A. Again we determine the different length segments initiating from point B. Finally, we check the precision using
discrepancy and average length and obtain the required measurement.
Important Tip:
Q. How do we know the measurement reading we have taken is perpendicular value or exactly straight or accurate measurement?
Ans: suppose from the above figure, we have starting end of the tape is at A and we have to take perpendicular measurement with ranging rod at C. For this we fix, starting end of tape at A and
lengthen tape towards the ranging rod at C. To obtain an exactly perpendicular measurement, we move the tape's end to and fro (or up and down) at in ranging rod at C until the shortest distance is
measured at the tape in point C. This shortest distance in the tape is the exactly perpendicular measurement.
Q. What is the least count of engineering tape?
Ams: The least count of engineering tape is 2mm or 0.002m. Hence, while taking reading from tape in the field the third digit after the dot should always be a even number. For eg: 5.314m, 7.798m, etc
Here our reading is 5.314 means our length is 5.3m, 1cm and 4mm.
4 comments :
1. Helpful!!
2. Well explained....
3. Great article Lot's of information to Read...Great Man Keep Posting and update to People..Thanks 안전놀이터
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15 A copper wire when bent in the form of a square encloses an ... | Filo
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area of the unshaded portion within the rectangle. (Take ) 18 In the adjoining figure, is a square of side . and are two diagonals of the square. Two semicircle are drawn with AD
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Question 15 A copper wire when bent in the form of a square encloses an area of . same wire is bent into the form of a circle, find the area of the circle. with diameter . (b) In the given figure,
Text find the area of the unshaded portion within the rectangle. (Take ) 18 In the adjoining figure, is a square of side . and are two diagonals of the square. Two semicircle are drawn with AD
Updated On Dec 4, 2022
Topic All topics
Subject Mathematics
Class Class 9
Answer Type Video solution: 1
Upvotes 114
Avg. Video 6 min | {"url":"https://askfilo.com/user-question-answers-mathematics/15-a-copper-wire-when-bent-in-the-form-of-a-square-encloses-33313530373132","timestamp":"2024-11-12T02:16:28Z","content_type":"text/html","content_length":"195634","record_id":"<urn:uuid:9ebf54ff-6d6d-4728-a882-0075a3e36a90>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00445.warc.gz"} |
Constructing Histograms | Curious Toons
Table of Contents
Introduction to Histograms
What is a Histogram?
A histogram is a special type of bar graph that represents the distribution of numerical data by showing the frequency of data points within specified intervals, known as bins. Unlike regular bar
graphs, which can represent categorical data, histograms are specifically designed for continuous data. Each bar in a histogram corresponds to a bin, which covers a range of values. The height of the
bar indicates how many data points fall within that range. For instance, if we have the ages of students in a class, we might create bins like “10-12”, “13-15”, and so forth. The height of each bar
tells us how many students fall into each age group. This visual representation makes it easier to spot patterns, trends, and distributions in data. Histograms help us understand how frequently
values occur and can reveal important features such as central tendency (where most data points lie) and spread (how much the data varies). Overall, a histogram is a powerful tool in statistics and
data analysis that allows us to quickly grasp complex information at a glance.
Importance of Histograms in Data Representation
Histograms play a crucial role in data representation because they transform raw data into meaningful visual insights. One of their primary benefits is that they help us understand the underlying
distribution of a dataset. By showing how data points are spread out across different ranges, histograms enable us to identify patterns and trends that might be overlooked in a simple list of
numbers. For instance, we can quickly see if the data is normally distributed, skewed, or has any outliers. Additionally, histograms facilitate comparison between different datasets. For example, if
we have test scores from two different classes, overlaying their histograms can help us visually assess differences in performance and variability.
Moreover, histograms make it easier to communicate complex data to others, such as classmates or teachers who may not be familiar with statistical terms. By providing a clear and intuitive visual,
histograms enhance understanding and engagement with the data. Thus, mastering the art of constructing and interpreting histograms is an essential skill for anyone interested in data analysis,
statistics, or scientific research.
Understanding Data Distribution
Types of Data: Continuous vs. Discrete
When we talk about data, it’s essential to understand the two primary types: continuous and discrete. Continuous data refers to measurements that can take on any value within a given range. Think of
it like height, weight, or temperature. These measurements can be broken down into smaller increments, meaning you could have a height of 5.5 feet or even 5.563 feet. Continuous data can come from
things you can measure accurately, making it infinitely variable.
On the other hand, discrete data represents distinct and separate values, often counted rather than measured. For example, the number of students in a classroom or the number of cars in a parking lot
are discrete because you can only have whole numbers—no fractions or decimals involved. Understanding the difference between these types of data is crucial because it affects how we visualize and
interpret our data sets. In histograms, for example, continuous data is represented with smooth intervals, while discrete data might use distinct bars, highlighting their separate categories.
Frequency Distribution and Its Role
Frequency distribution is a fundamental concept in statistics that helps us summarize and organize data effectively. It refers to a way of recording how many times each value or range of values
(known as “bins”) occurs in a dataset. For instance, if we were to record the ages of students in our school, a frequency distribution would show how many students fall within specific age
ranges—like 13-14 years, 15-16 years, and so forth.
This organization makes it easier to understand the data and identify patterns or trends. Frequency distributions are essential for constructing histograms, as they provide the necessary counts for
each bin. In a histogram, the height of each bar represents the frequency of data points in that interval, allowing for quick visual analysis. By understanding frequency distributions, we can draw
insights from the data, like identifying where most students fall in terms of age or how scores are spread out in a test. Overall, frequency distribution serves as a foundational building block for
data analysis and visualization, making it an indispensable tool in statistics!
Steps to Construct a Histogram
Collecting and Organizing Data
Before we can create a histogram, the first step is to collect and organize our data. Data collection involves gathering information that you want to analyze, which can come from surveys,
experiments, or observations. For instance, if we’re studying the number of pets owned by students in our school, we could gather that data through a class survey where every student reports how many
pets they have at home.
Once we have our data, organizing it is crucial for clarity and understanding. This usually means making a list of values or using a table. It’s important to ensure that your data is clean — meaning
there are no duplicates or irrelevant entries. After organization, we can start to determine the range of our data—this tells us the smallest and largest values. Organizing the data allows us to see
patterns or trends that might not be obvious at first glance. It’s like preparing the ingredients before you cook a meal; everything needs to be in order so that when you start making your histogram,
you can clearly represent the information.
Choosing Appropriate Bins
After you’ve collected and organized your data, the next step is to choose appropriate bins for your histogram. Bins are intervals that group our data points, and their selection is crucial because
they can significantly affect how the information is presented. When determining your bins, consider the range of your data—this is the difference between the largest and smallest values.
For instance, if the data reflects pet ownership, and the maximum number of pets is 10, you might choose bins like 0-1, 2-3, 4-5, and so forth. It’s important to have equal-width bins that make sense
for your data. Too few bins can oversimplify the data, while too many can complicate it and make it hard to interpret. Aim for around 5 to 15 bins; this range often strikes a good balance. Remember,
the goal is to convey the underlying patterns of the data clearly, so think about what will best represent the story your data tells. With the right bins, your histogram will effectively illustrate
how the data is distributed.
Creating a Histogram
Using Graphing Software
In today’s digital age, graphing software is an incredibly helpful tool for constructing histograms effectively and efficiently. Programs like Excel, Google Sheets, or specialized statistical
software can automate the process of creating these visual representations of data. To begin, you should gather your data and organize it into a frequency table, which indicates how many values fall
within specific intervals or “bins.” Most graphing software will allow you to input these values easily.
After inserting your data, you can select the histogram option to generate a visual output. This software often provides customization features, enabling you to adjust bin sizes, colors, and labels
to enhance clarity and presentation. Unlike drawing by hand, software-generated histograms can be modified effortlessly, which is ideal for exploratory data analysis. You’ll also find that these
tools can handle large datasets and complex calculations in seconds, allowing you to focus more on interpreting the results. Mastering graphing software is a valuable skill not only for math class
but also for future studies and professional work, as it emphasizes precision and the effective communication of information.
Drawing Histograms by Hand
While digital tools are convenient, there’s something valuable about understanding how to draw histograms by hand. This process deepens your comprehension of data representation and visualizes how
bin sizes and frequency affect the overall shape of the histogram. First, start by organizing your data into a frequency table. Determine your bins—these could represent age ranges, test scores, or
any quantitative measure relevant to your data.
Once your table is set, outline your axes. The horizontal axis (x-axis) will represent the bins, while the vertical axis (y-axis) shows the frequency. Remember to label each axis clearly for better
understanding. Next, for each bin, draw a bar that corresponds to the frequency; the height of the bar will reflect how many data points fall into that specific range. Ensure that the bars are
touching, as this indicates that the data is continuous. This hands-on approach not only reinforces your understanding of how histograms represent data but also hones your skills in creating clear
and effective visualizations. Moreover, practicing this method will prepare you for situations where technology isn’t available, strengthening your foundational skills in data presentation.
Interpreting Histograms
Identifying Trends and Patterns
When we interpret histograms, one of the first things we should do is look for trends and patterns within the data. Histograms visualize how data points are distributed across different ranges or
“bins.” For instance, if we see that certain bins have much taller bars, it indicates that more data points fall within those ranges. This standout feature can signal trends, such as increased
frequency at specific intervals, helping us understand the central tendencies of the data, like peaks (modes), and any spread or gaps in the distribution. We might also observe patterns like normal
distributions—where the data clusters around a central point—or skewed distributions—which indicate that the data is not evenly spread. As you analyze a histogram, consider questions like: Where do
most values lie? Are there any outliers? The distribution shape can provide insights, not just about the data at hand, but also about possible underlying phenomena or behaviors. Recognizing these
trends and patterns can support deeper data analysis, enabling us to draw meaningful conclusions and make informed decisions based on our findings.
Common Misinterpretations and Errors
When working with histograms, it’s easy to make some common misinterpretations and errors. One frequent mistake is assuming that the height of the bars represents individual data points instead of
the frequency of values within each bin. This misunderstanding can lead us to think that taller bars signify larger individual values, rather than the actual number of occurrences in that range.
Additionally, students may misinterpret the width of the bins; uneven widths can misrepresent data distributions when they don’t account for the differing areas of the bars. Another common error
arises from overlooking outliers or special features, mistaking them for part of the overall trend. It’s essential to analyze histograms carefully, considering not just the visual aspects but the
data it represents. Lastly, students sometimes fail to recognize when a histogram is misleading—such as when the y-axis isn’t scaled properly, making differences appear exaggerated. By understanding
these misinterpretations and errors, you’ll sharpen your skills in analyzing histograms, ultimately leading to more accurate data interpretation and stronger conclusions.
As we conclude our exploration of constructing histograms, let’s take a moment to reflect on the significance of this tool in understanding the world around us. Histograms are not just bars on a
graph; they are a visual representation of data that allows us to decipher patterns, trends, and outliers in a seemingly chaotic set of numbers.
Have you ever considered how a simple histogram can reveal the heartbeat of a community? Whether it’s analyzing test scores, measuring rainfall, or tracking the growth of our favorite plants,
histograms empower us to make informed decisions and predictions. They invite us to ask questions: What story does the data tell? Are there unexpected peaks of interest, or valleys that deserve our
As you venture forward, remember that data is an integral part of our lives—shaping opinions, guiding policies, and influencing outcomes. Your ability to construct and interpret histograms is a key
skill that will enhance your analytical thinking. Embrace this knowledge, for mathematics is not just about numbers; it’s about uncovering the narratives that lie within. So, the next time you see a
histogram, ask yourself: What story is the data trying to tell me? | {"url":"https://curioustoons.in/constructing-histograms/","timestamp":"2024-11-09T18:59:55Z","content_type":"text/html","content_length":"108075","record_id":"<urn:uuid:b978878f-9308-47e2-8955-dd50fd21ab25>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00427.warc.gz"} |
On minimum bisection and related partition problems in graphs with bounded tree width
Electronic Notes in Discrete Mathematics 49 (): 481-488 (2015)
Minimum Bisection denotes the NP-hard problem to partition the vertex set of a graph into two sets of equal sizes while minimizing the number of edges between these two sets. We consider this problem
in bounded degree graphs with a given tree decomposition (T, X) and prove an upper bound for their minimum bisection width in terms of the structure and width of (T, X). When (T, X) is provided as
input, a bisection satisfying our bound can be computed in time proportional to the encoding length of (T, X). Furthermore, our result can be generalized to k-section, which is known to be APX-hard
even when restricted to trees with bounded degree.
Minimum Bisection
Minimum k-Section
Tree decomposition | {"url":"https://tore.tuhh.de/entities/publication/4ea71415-33e4-49c1-93e3-6b87d7041e6f","timestamp":"2024-11-02T05:14:40Z","content_type":"text/html","content_length":"894629","record_id":"<urn:uuid:d8dc3b77-3018-40b1-a19f-0e67dff36ad9>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00116.warc.gz"} |
Use the data from Appendix \(B\) to construct a scatterplot, find the value of the linear correlation coefficient \(r\), and find either the P-value or the critical values of \(r\) from Table A-5
using a significance level of \(\alpha=0.05 .\) Determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. (Save your work because the same
data sets will be used in Section 10-2 exercises.) Use all of the foot lengths and heights of the 19 males from Data Set 2 "Foot and Height" in Appendix \(\mathrm{B}\).
Short Answer
Expert verified
Calculate \(r\), compare with critical values or P-value < 0.05 to determine correlation.
Step by step solution
Gather Data
Collect the foot lengths and heights of the 19 males from Data Set 2 'Foot and Height' in Appendix B. Ensure all data points are correctly recorded.
Construct a Scatterplot
Plot the foot lengths on the x-axis and the corresponding heights on the y-axis. Each pair of values (foot length, height) should be represented as a single point on the scatterplot.
Calculate the Linear Correlation Coefficient
Use the formula for Pearson's correlation coefficient, \[ r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \]where \(n\) is the number of paired values,
\(x\) represents the foot lengths, and \(y\) represents the heights. Compute the necessary sums and substitute them into the formula to find the value of \(r\).
Find the Critical Values of r
Refer to Table A-5 to find the critical values of \(r\) for a significance level of \(\alpha = 0.05\) with \(n - 2\) degrees of freedom. For \(n = 19\), the degrees of freedom is \(17\). Find the
corresponding critical values \(r\text{critical}\).
Determine the P-value
Using the computed \(r\) from Step 3, find the P-value associated with this correlation coefficient. This can be done using statistical software or a P-value table.
Decision Making
Compare the computed \(r\) from Step 3 to the critical values from Step 4. If the computed \(r\) is outside the range of the critical values, or if the P-value is less than 0.05, there is sufficient
evidence to reject the null hypothesis and conclude that there is a linear correlation between foot length and height.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
scatterplot construction
To understand how two variables relate to each other, we first construct a scatterplot. A scatterplot is a type of graph that shows individual data points represented by dots. In this case, we are
examining foot lengths and heights of 19 males. We plot each foot length on the x-axis and the corresponding height on the y-axis.
First, gather all the data from the appendix and ensure it's recorded correctly. Then, use a graphing tool or software to construct the scatterplot.
Each point on the scatterplot corresponds to a pair of values (foot length, height). This visual representation helps us quickly observe any potential linear relationships between the two variables.
Pearson's correlation coefficient
After plotting the scatterplot, the next step is to measure the strength and direction of the linear relationship between our two variables - foot length and height. We do this using Pearson's
correlation coefficient, denoted as \( r \). The formula for Pearson's correlation coefficient is:
\[ r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \]
• \( n \) is the number of paired data points
• \( x \) represents the foot lengths
• \( y \) represents the heights
Compute the sums and substitute them into the formula to determine the value of \( r \).
The value of \( r \) ranges from -1 to +1. A value close to +1 indicates a strong positive linear correlation, while a value close to -1 indicates a strong negative linear correlation. A value around
0 suggests no linear relationship.
P-value determination
The next step is to determine the P-value associated with our computed correlation coefficient \( r \). The P-value helps us evaluate the significance of our results. It tells us the probability of
observing our data if the null hypothesis (no correlation) is true.
To find the P-value, you can use statistical software or refer to a P-value table that corresponds to \( r \) and the degrees of freedom (\( n - 2 \)). In this case, with \( n = 19 \), our degrees of
freedom is 17.
Consult the P-value table or enter your \( r \) value into statistical software to find the P-value. A particularly low P-value (typically less than 0.05) indicates that our results are statistically
significant, meaning there's a low probability that the observed correlation is due to chance.
critical value computation
To further confirm the significance of our correlation, we can compare our computed \( r \) against critical values. Critical values are thresholds that \( r \) must exceed to demonstrate a linear
Refer to Table A-5 to find the critical values for a significance level of \( \alpha = 0.05 \) and 17 degrees of freedom. Identify the two critical values \( r_{critical} \) for both tails of the
If our computed \( r \) is greater than the positive critical value or less than the negative critical value, we can reject the null hypothesis and conclude that a linear correlation exists.
hypothesis testing
Finally, we perform hypothesis testing to make our decision. Our null hypothesis (\( H_0 \)) states that there is no correlation between foot length and height. The alternative hypothesis (\( H_a \))
states that a correlation does exist.
To make our decision, we consider two factors:
• If our computed \( r \) is greater than the positive critical value or less than the negative critical value
• If the P-value is less than 0.05
If any of the conditions above hold true, we reject the null hypothesis. This suggests that there is sufficient evidence to support a linear correlation between foot length and height.
If not, we fail to reject the null hypothesis, concluding that there isn't enough evidence to assert a linear correlation between the variables.
This is how statistical tools like scatterplots, Pearson's correlation coefficient, P-values, critical value computation, and hypothesis testing work in unison to help us analyze and interpret data | {"url":"https://www.vaia.com/en-us/textbooks/math/essentials-of-statistics-6-edition/chapter-10/problem-32-use-the-data-from-appendix-b-to-construct-a-scatt/","timestamp":"2024-11-14T03:23:15Z","content_type":"text/html","content_length":"255092","record_id":"<urn:uuid:f3f4a975-1525-42c7-bd20-8964cc251e0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00323.warc.gz"} |
Testing a new resource system - Unvanquished
Testing a new resource system
We are currently trying out a new resource-based building system on our Development & Testing Server. The system itself is work in progress and could change at any time, but as our first test games
were a lot of fun and we find that it has a great potential it should be about time to explain to you how it currently works, so you can quickly adapt and discuss the system with us on a higher level
of understanding.
What’s a resource system?
Tremulous used a pool of build points that each team was given at the beginning of a match. Teams could spend the build points to build structures. When structures got destroyed, the build points
entered a queue and after a while became available to the teams again. There was a sudden death mechanism that disabled building alltogether after a timelimit was hit.
Resources are, just like Tremulous build points, spent to build structures but they are not returned if a structure gets destroyed. Think of them as an actual payment for building the structure, as
opposed to a mere limit of how many structures you can have simultaneously. Replacing and moving of structures is still possible and won’t be penalized: if a structure at n% health gets deconstructed
for either reason, you will get n% of its price back. The resource system I’m going to talk about describes the way that building resources, which we will still refer to as build points (BP), are
Resource Generating Structures (RGS)
Both teams are given an initial amount of build points. If they want to acquire more, they need to build Resource Generating Structures (RGS) that generate them over time. The human RGS is called
Drill, the alien one is called Leech. They, too, have a construction price. RGS have a spherical area of effect described by their range. If the areas of effect of two RGS intersect, they will have a
reduced efficiency, forcing teams to expand over the map.
Resource generation
Each RGS has an efficiency e. If a RGS doesn’t interfere with another one (that means that their distance is above a threshold), that efficiency will be at 100% (e = 1), otherwise it will be lower,
but positive. The mine rate of a RGS also depends on the level-wide (team independent) base rate r(t) (in BP/min), which itself depends on the time t (in minutes after the match begun). The current
generation rate of any RGS is simply e×r(t). r(t) has an initial value r(0) = R₀ which can be configured on the server. The base rate r(t) also has a half life period T (in min), which can also be
set on the server. This will replace the artifical sudden death, as teams will get a decreasing number of resources over time. The formula to calculate r(t) is R₀×2^-(t/T).
Let’s asssume we have a single RGS on the map. Since its area of effect does not overlap with another RGS, it will generate resources at 100% efficiency, so e = 1. Now let R0 be 15 BP/min and let T
be 15 min. If the RGS already exists at the beginning of the match (t = 0), it would generate at a rate of 15 BP/min at that moment. If the match has been running for 15 minutes (t = T), the RGS
would only generate 7.5 BP/min (no matter if it has been present since the beginning or has just been built). After 30 minutes (t = 2T), the RGS would generate 3.75 BP/min.
Interference: Idea and Approximation
One of the key ideas behind the RGS approach is that it should force the teams to expand and gain control of the map. This is accomplished by reducing the efficiencies of RGS that are built too close
to each other (another approach would’ve been to simply forbid placing of a RGS in the range of another one but we found this to be too artificial and rigid). Our idea was that a number of RGS
together would, no matter what teams they belonged to, generate at a rate that was proportional to the volume of the union of their spherical areas of effect (AOE). This alone would’ve been far too
complex to implement in a real time game but since we also need to assign every RGS an individual generation rate (because they might belong to different teams), this approach was out of question. We
decided to use an approximation of it though: We look at pairs of interfering RGS and calculate the intersection quota q. If q = 1 then the AOEs are identical. If q ≈ 0 then they barely touch. If q =
0.5 that would mean that 50% of the AOE of one RGS lays inside the AOE of the other one. Now for two interfering RGS, we independently halve the part of the efficiency that corresponds to the
overlapping area. So if the efficiency of one of the RGS was at 100% and 20% of its AOE overlapped with that of the other one, we would reduce the efficiency to 80% + 0.5 × 20% = 90%.
We only have two RGS on the map. They might belong to different teams or a single team. Now let the RGS range be 800 qu (quake units), so that their area of effect is a sphere with a radius of 800
qu. If the two RGS are at a distance of 1600 qu or higher, they would both mine at 100%. If they were both in the very same spot (which is not possible, of course), they would both mine at 50%
efficiency. If they were at a distance of 800 qu, parts of their AOE would overlap and we would find the relative size of the intersection area to be 5/16. So now we would half 5/16 of the initial
efficiency and get a new efficiency of (1 – 5/16) × 100% + 5/16 × 100% × 0.5 = 11/16 + 5/32 = 27/32. Both RGS together would now mine at a total efficiency of 2 × (27/32) = 54/32 which is exactly the
volume of the union of their AOEs divided by the volume of a single AOE. Note that our approximation is always exact if each RGS interfers with not more than another one.
Approximation side-effects
The pairwise approximation appproach yields exact results for pairs of RGS but produces an error when more than two RGS interfere. This error will always penalize the cluttering of RGS (meaning
building many of them close by). If, hypothetically, n RGS are built in the very same position, the total efficiency would be n/2^(n-1). So one or two RGS will yield 100% efficiency but three RGS
will only generate at 3/4 of the speed of a single RGS. If you build 10 RGS in the same spot the total efficiency would drop to less than 2%. While this effect is, in theory, an error, it has a
number of positive side effects:
• The number of calculations that is necessary for n interfering RGS is n×(n-1) and thus its complexity is in O(n^2). The fact that players have a motivation not to let RGS AOEs interfer will help
keep the server load low.
• You can choose to build two RGS closeby so that one takes over completely if the other one gets destroyed. Since this advantage comes at double price and build time, I consider it a balanced
decision that can make strategic building more interesting. At the same time the penalty will prevent mindless ‘more is better’ spamming of RGS.
• RGS can also be used as a weapon against campers: Just build a lot of RGS next to the enemy camp and they won’t be able to generate any more resources there.
Get involved!
If you want to help in the development of the resource system, give your feedback on the forums and help us testing by participating in our development games, which we will announce on the forums and
on IRC. | {"url":"https://unvanquished.net/testing-a-new-resource-system/","timestamp":"2024-11-08T03:58:38Z","content_type":"text/html","content_length":"40066","record_id":"<urn:uuid:330edd3c-ef3e-48cb-8069-2670240282d5>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00440.warc.gz"} |
GSEB Solutions Class 9 Maths Chapter 10 Circles Ex 10.1
Gujarat Board GSEB Solutions Class 9 Maths Chapter 10 Circles Ex 10.1 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 9 Maths Chapter 10 Circles Ex 10.1
Question 1.
Fill in the blanks:
1. The centre of a circle lies in ______ of the circle. (exterior/interior).
2. A point, whose distance from the centre of a circle is greater than its radius lies in ______ of the circle. (exterior/interior).
3. The longest chord of a circle is a ______ of the circle.
4. An arc is a ______ when its ends are the ends of a diameter.
5. A segment of a circle is the region between an arc and,_____ of the circle.
6. A circle divides the plane, on which it lies, in _____ parts.
1. The centre of a circle lies in the interior of the circle.
2. A point, whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle.
3. The longest chord of a circle is a diameter of the circle.
4. An arc is a semicircle when its ends are the ends of a diameter.
5. A segment of a circle is the region between an arc and the chord of the circle.
6. A circle divides the plane, on which it lies, in three parts.
Question 2.
Write True or Fiase: Give reasons for your answers.
1. The line segment joining the centre to any point on the circle is a radius of the circle.
2. A circle has an only a finite number of equal chords.
3. If a circle is divided into three equal arcs, each is a major arc.
4. A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
5. The sector is the region between the chord and its corresponding arc.
6. A circle is a plane figure.
1. True because all points on the circle are equidistant from the centre.
2. False because there are infinitely many points on the circle. So for each point on the circle, a point can be determined on the circle at a given distance from that point resulting in greatly
many equal chords.
3. False because for each arc, the remaining arc will have a greater length.
4. True because of the definition of diameter.
5. False by virtue of its definition.
6. True as it is a part of a plane.
Question 3.
1. A round pizza is cut into 4 pieces. Each piece represents ______
2. Circles with the same centre and different radii are called.
3. A chord of a circle divides the circle into two parts known as _____
4. A constant distance between the centre of a circle and any point on a circle is called
5. Diameter = 2 x ______
6. The boundary of a circle is called ______
1. sector
2. concentric circles
3. segment
4. radius
5. radius
6. circumference
Leave a Comment | {"url":"https://gsebsolutions.com/gseb-solutions-class-9-maths-chapter-10-ex-10-1/","timestamp":"2024-11-05T13:18:10Z","content_type":"text/html","content_length":"246270","record_id":"<urn:uuid:9e71ffd1-f9fc-4795-9c12-4d767687c1ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00201.warc.gz"} |
OpenFF Fragmenter
OpenFF Fragmenter
The main purpose of openff-fragmenter is to fragment molecules for quantum chemical (QC) torsion drives.
openff-fragmenter is still pre-alpha. It is not fully tested and the API is still in flux.
Currently two fragmentation schemes are supported:
• a Wiberg Bond Order (WBO) sensitive fragmentation scheme [1] (recommended)
• the fragmentation schema detailed by Rai et al [2] reffered to in this package as the ‘Pfizer’ scheme.
WBO Sensitive Fragmentation
The assumption when fragmenting molecules is that the chemistry is localized and that removing or changing remote substituents (defined as substituents more than 2 bonds away from the central bond
that is being driven in the torsion drive) will not change the torsion potential around the bond of interest. However, that is not always the case. openff-fragmenter uses the Wiberg Bond Order (WBO)
as a surrogate signal to determine if the chemistry around the bond of interest was destroyed during fragmentation relative to the bond in the parent molecule.
The WBO is a measure of electronic population overlap between two atoms in a bond. It can be quickly calculated from an empirical QC calculation and is given by:
\[W_{AB} = \sum_{\mu\in{A}}\sum_{\nu\in{B}}|D_{\mu\nu}|^2\]
Where \(A\) and \(B\) are atoms \(A\) and \(B\) in a bond, \(D\) is the density matrix and \(\mu\) and \(\nu\) are occupied orbitals on atoms \(A\) and \(B\) respectively.
openff-fragmenter calculates the WBO of the parent molecules, then fragments according to a set of rules and then recalculates the WBO of the fragments. If the WBO for the fragment of the bond of
interest changes more than a user’s specified threshold, openff-fragmenter will add more substituents until the WBO of the bond of interest is within the user specified threshold.
Chaya D Stern, Christopher I Bayly, Daniel G A Smith, Josh Fass, Lee-Ping Wang, David L Mobley, and John D Chodera. Capturing non-local through-bond effects when fragmenting molecules for quantum
chemical torsion scans. bioRxiv, 2020. URL: https://www.biorxiv.org/content/early/2020/08/28/2020.08.27.270934, arXiv:https://www.biorxiv.org/content/early/2020/08/28/2020.08.27.270934.full.pdf,
Brajesh K. Rai, Vishnu Sresht, Qingyi Yang, Ray Unwalla, Meihua Tu, Alan M. Mathiowetz, and Gregory A. Bakken. Comprehensive assessment of torsional strain in crystal structures of small
molecules and protein–ligand complexes using ab initio calculations. Journal of Chemical Information and Modeling, 59(10):4195–4208, 2019. PMID: 31573196. URL: https://doi.org/10.1021/
acs.jcim.9b00373, arXiv:https://doi.org/10.1021/acs.jcim.9b00373, doi:10.1021/acs.jcim.9b00373. | {"url":"https://docs.openforcefield.org/projects/fragmenter/en/stable/index.html","timestamp":"2024-11-09T04:25:51Z","content_type":"text/html","content_length":"18266","record_id":"<urn:uuid:2a604451-ade5-48bb-9860-71790a1f6324>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00315.warc.gz"} |
emerald cut ratio calculator
Emerald Cut Ratio Calculator
Calculating the ratio for an emerald-cut gemstone is crucial for achieving the desired aesthetic appeal. To simplify this process, we’ve developed an accurate and user-friendly Emerald Cut Ratio
How to Use:
Using the calculator is straightforward. Input the length and width values of your emerald-cut gemstone in the designated fields. Click the “Calculate” button to obtain the precise ratio, which plays
a key role in determining the gem’s visual balance.
The formula for calculating the emerald cut ratio is:
Example Solve:
Let’s consider an example where the length of the emerald-cut gemstone is 7 mm, and the width is 5 mm.
Q: Why is the emerald cut ratio important?
A: The emerald cut ratio directly influences the visual appearance of the gemstone, ensuring it has the desired elongated or squarer look.
Q: Can I use this calculator for other gem cuts?
A: No, this calculator is specifically designed for emerald-cut gemstones. Different cuts may have different ratio calculation methods.
Q: Are there specific ideal ratios for emerald-cut gemstones?
A: While there’s no one-size-fits-all ideal ratio, common preferences range between 1.30 and 1.60, depending on personal aesthetics.
In conclusion, our Emerald Cut Ratio Calculator provides a convenient and precise way to determine the ratio for emerald-cut gemstones. Whether you’re a jeweler or a gem enthusiast, this tool can aid
in achieving the perfect visual balance for your gemstone. | {"url":"https://calculatordoc.com/emerald-cut-ratio-calculator/","timestamp":"2024-11-13T21:21:15Z","content_type":"text/html","content_length":"91393","record_id":"<urn:uuid:04be973b-d858-4fce-9ad1-4ca795ceb261>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00385.warc.gz"} |
De Moivre's Theorem
De Moivre's Theorem resources
Complex Numbers Test 01 (DEWIS)
Seven questions on complex numbers. Testing modulus, multiplication, division, Argand diagram, polar form, De Moivre's theorem. DEWIS resources have been made available under a Creative Commons
licence by Rhys Gwynllyw & Karen Henderson, University of the West of England, Bristol.
Polar Form and De Moivre's Theorem - Numbas
3 questions. Finding modulus and argument of complex numbers. Use De Moivre's Theorem to find powers of complex numbers. Numbas resources have been made available under a Creative Commons licence by
the School of Mathematics & Statistics at Newcastle University. | {"url":"https://www.mathcentre.ac.uk/topics/complexnumbers/demoivres-theorem/","timestamp":"2024-11-05T07:07:47Z","content_type":"text/html","content_length":"8512","record_id":"<urn:uuid:c5347b32-f44e-4d1e-ba8b-f165724d0176>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00235.warc.gz"} |
Integration problem of ti-89
integration problem of ti-89 Related topics: algebra operations
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Author Message
drunsdix42 Posted: Thursday 17th of Apr 17:37
Hello everyone! . Ever since I have started integration problem of ti-89 at college I never seem to be able to learn it well. I am quite good at all the other
branches, but this particular chapter seems to be my weakness . Can some one aid me in learning it easily ?
Registered: 24.09.2002
From: NJ
Back to top
oc_rana Posted: Saturday 19th of Apr 11:03
Hey. I think I can assist . Can you explain some more on what your troubles are? What exactly are your problems with integration problem of ti-89? Getting a good
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Why not try this out? It could be just be the answer for your problems .
Registered: 08.03.2007
From: egypt,alexandria
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Paubaume Posted: Sunday 20th of Apr 09:27
Hello ! Algebrator is a very wise piece of software. I use it frequently to solve questions. You must try it. It may solve your issue .
Registered: 18.04.2004
From: In the stars... where you
left me, and where I will wait
for you... always...
Back to top
Rishy_Locer Posted: Monday 21st of Apr 12:02
Yeah! That's a great alternative to the costly private tutors and costly online training . The single page formula list offered there has served me in every
Intermediate algebra internal that I have taken up in the past. Even if you are an intermediate in Algebra 2, the Algebrator is very useful since it offers both
simple and challenging exercises and drills for practice.
Registered: 12.02.2003
Back to top
fveingal Posted: Wednesday 23rd of Apr 08:15
It can be ordered right here – https://softmath.com/. I’ve heard from someone that they even offer an unconditional money back guarantee, so go ahead and order a
copy, I’m sure you’ll love it .
Registered: 11.07.2001
From: Earth
Back to top | {"url":"https://softmath.com/algebra-software-5/integration-problem-of-ti-89.html","timestamp":"2024-11-11T04:15:33Z","content_type":"text/html","content_length":"41501","record_id":"<urn:uuid:6a559168-4012-40d4-968b-f3f321a4d5c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00570.warc.gz"} |
Built-In Training
Train deep learning networks for sequence and tabular data using built-in training functions
After defining the network architecture, you can define training parameters using the trainingOptions function. You can then train the network using the trainnet function. Use the trained network to
predict class labels or numeric responses, or forecast future time steps.
You can train a neural network on a CPU, a GPU, multiple CPUs or GPUs, or in parallel on a cluster or in the cloud. Training on a GPU or in parallel requires Parallel Computing Toolbox™. Using a GPU
requires a supported GPU device (for information on supported devices, see GPU Computing Requirements (Parallel Computing Toolbox)). Specify the execution environment using the trainingOptions
Deep Network Designer Design and visualize deep learning networks
Train and Test Network
trainingOptions Options for training deep learning neural network
trainnet Train deep learning neural network (Since R2023b)
testnet Test deep learning neural network (Since R2024b)
predict Compute deep learning network output for inference
minibatchpredict Mini-batched neural network prediction (Since R2024a)
scores2label Convert prediction scores to labels (Since R2024a)
confusionchart Create confusion matrix chart for classification problem
sortClasses Sort classes of confusion matrix chart
Featured Examples | {"url":"https://au.mathworks.com/help/deeplearning/built-in-training-for-sequences.html?s_tid=CRUX_lftnav","timestamp":"2024-11-06T11:39:50Z","content_type":"text/html","content_length":"83898","record_id":"<urn:uuid:f42e71c4-7501-40e8-bf14-77cc334ef522>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00665.warc.gz"} |
Addressing formula of (i,j)th element of a m*n matrix in Column-major order - Quescol
Addressing formula of (i,j)th element of a m*n matrix in Column-major order
In this article we will learn how to calculate address of (i,j)th element of a matrix m*n in Column major order. Below is the explanation of it.
Column Major Order Formula
The Location/Address of element A[i, j] can be obtained by evaluating expression:
LOC (A [i, j]) = base_address + w * [m * j + i]
Base_Address = Address of the first element in the array.
w(Size of element)= Word size means a number of bytes occupied by each element of an Array. It depends on the data type, e.g., 4 bytes for an int in many systems.
m = Number of rows in the array.
i = is the row index
j = is the column index
Here we have taken a 2-D array A [2, 4] which has 2 rows and 4 columns.
1. Column Index Multiplied by Row Count: The term (M * j) finds the starting position of the jth column. Since it’s column-major, to get to the start of the jth column, we skip j columns, each of
which has m elements.
2. Row Index Addition: Adding the row index i moves down to the ith element in that column.
3. Scaling by Element Size: Multiplying by w(size of element) adjusts the address according to the actual memory size of each element in the matrix.
4. Adding Base Address: Finally, adding the Base_Address of the matrix gives the absolute memory address of the element A[i, j].
Problem to solve on Column Major Order
Now to calculate the base address of any index using column major order we can use process given below.
Consider a 4×3 (m=4, n=3) matrix stored in column-major order starting at base address 2000, and assume each element is an integer type taking 4 bytes. To find the address of the element in the
second row and third column (A[1, 2]):
It can be calculated as follow:
Given base_address = 2000, w(Size)= 4, m(number of rows)=4, i(row index)=1, j(column index)=2
LOC (A [i, j]) = base_address + W [m * j + i ]
LOC (A[1, 2]) = 2000 + 4 *[4 * 2 + 1]
= 2000 + 4 * [8 + 1]
= 2000 + 4 * 9
= 2000 + 36
= 2036 | {"url":"https://quescol.com/data-structure/addressing-formula-in-column-major-order","timestamp":"2024-11-13T10:51:01Z","content_type":"text/html","content_length":"81366","record_id":"<urn:uuid:a768c6eb-f6fc-4ceb-9ebe-4492ddf8c37c>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00374.warc.gz"} |
NCERT Exemplar Class 9 Maths Chapter 1 Number Systems - Tennelli IndustriesNCERT Exemplar Class 9 Maths Chapter 1 Number Systems - Tennelli Industries
The equipment is expected to produce $240,000 in cash inflows and $160,000 in cash outflows annually. The company uses straight-line depreciation, and has a 40% tax rate. Determine the annual
estimated net income and net cash inflow. Problem D Span Fruit Company has used relevance in accounting a particular canning machine for several years. The company is considering buying a
technologically improved machine at a cost of $232,000.
EXERCISE 1–8 (LO Net Income, Shares Issued
Problem F Slow to Change Company has decided to computerize its accounting system. The company has two alternatives—it can lease a computer under a three-year contract or purchase a computer
outright. Following are the asset, liability, and equity items of Dumont Inc. at January 31, 2015, after its first month of operations. The following list shows the various ways in which the
accounting equation might be affected by financial transactions. You have also translated word sentences into algebraic equations and solved some word problems.
Unit 2: Accounting Principles and Practices
Write a memorandum to your instructor summarizing the key points of the article. Be sure to include a copy of the article used for this assignment. Compute the net present value of each of the five
proposals. Do you recommend replacing the old vans? Support your answer with computations and disregard all factors not related to the preceding data. Management currently has other opportunities
that will yield 18%.
1. If the company leases the trucks, the lease contract will run for six years.
2. Management will not make any investment unless at least an 18% rate of return can be earned.
3. Problem B Graham Company currently uses four machines to produce 400,000 units annually.
4. Exercise B Zen Manufacturing Company is considering replacing a four-year-old machine with a new, advanced model.
EXERCISE 1–12 (LO Correcting Financial Statements
Business decision case B Slick Company is considering a capital project involving a $225,000 investment in machinery and a $45,000 investment in working capital. The machine has an expected useful
life of 10 years and no salvage value. The annual cash inflows (before taxes) are estimated at $90,000 with annual cash outflows (before taxes) of $30,000. The company uses straight-line
Using the net present value method, show whether the company should sell the equipment. Prepare a schedule to support your conclusion. Calculate the net present value of out-of-pocket costs for the
purchase alternative. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License .
NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1
Below are the December 31, 2023, year-end accounts balances for Mitch’s Architects Ltd. This is the business’s second year of operations. Now we’ll develop a strategy you can use to solve any word
problem. This strategy will help you become successful with word problems. We’ll demonstrate the strategy as we solve the following problem.
If any of the items are not to be recorded, leave the row blank. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don’t
understand, look them up in a dictionary or on the Internet. If we take control and believe we can be successful, we will be able to master word problems. The height or length of an object or the
distance between two distinct excel accounting and bookkeeping objects can be determined with the help of trigonometric ratios.
Annual before-tax net cash inflow from the machine is expected to be $7,000. Calculate the unadjusted rate of return. Alternate problem A Mark’s Manufacturing Company is currently using three
machines that it bought seven years ago to manufacture its product. Each machine produces 10,000 units annually. | {"url":"https://tennelli.com/ncert-exemplar-class-9-maths-chapter-1-number/","timestamp":"2024-11-06T08:37:25Z","content_type":"text/html","content_length":"414874","record_id":"<urn:uuid:87741428-5cb2-4ade-a046-6f2594e15456>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00240.warc.gz"} |
MATH 0342
This course prepares students to enroll in MATH 1342, Elementary Statistics Mathematics, when a grade of ā Cā or better is earned. Topics in this course include the real number system, solving
linear equations and inequalities, graphing linear equations and inequalities, solving systems of linear equations, exponents and polynomials, and factoring polynomials. This course will not transfer
to a senior college; however, it will count for Non-degree credit from Weatherford College. Attendance and tutorials required.
Either a C or better in MATH 0301 or placement by TSI instrument. | {"url":"https://catalog.wc.edu/mathematics-math/math-0342","timestamp":"2024-11-07T15:56:06Z","content_type":"text/html","content_length":"12998","record_id":"<urn:uuid:0b8e378b-298d-48e4-bdd1-db6922c12ae7>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00480.warc.gz"} |
perplexus.info :: Just Math : Fancy quadrilateral
Since there are no constraints given except the two radii, if we find one working example quadrilateral, we need only use this one to measure the distance.
There is a subset of of q's that can have an inclrcle (ic).
Likewise there is a subset of q's that can have a circumcircle (cc).
We look in the intersection of those subsets.
E.g, squares are candidates, but the ratio 12/7 does not equal sqrt(2).
Looking at more members found in the intersection: all rombi have an ic, but non-square rombi cannot have a cc! This brings us back to squares.
(It is mentioned in subsequent comments that some isosceles trapezoids are found in the intersection.)
Things get better with the kites, a superset of rombi. All kites have a cc, and the right kites (kites having a pair of opposed right angles) have a cc, so that's a good place to look to satisfy the
12/7 constraint. The right right kite will also have the convenience of bilateral symmetry, with the distance to be measured lying along the fold.
Edited on October 27, 2019, 10:03 am
Posted by Steven Lord on 2019-10-25 15:46:38 | {"url":"http://perplexus.info/show.php?pid=11914&cid=61582","timestamp":"2024-11-13T16:09:43Z","content_type":"text/html","content_length":"13108","record_id":"<urn:uuid:d37e9929-48b7-452d-884b-0813682660e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00049.warc.gz"} |
2019 AMC 10A Problems/Problem 14
The following problem is from both the 2019 AMC 10A #14 and 2019 AMC 12A #8, so both problems redirect to this page.
For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$
It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures: $[asy] unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw
((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+
(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0)); draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/
12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/
2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6)); draw((-1/3,1-d)--(-1/3,-1-d),Arrows); draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot
((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d)); draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt
(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt
(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d)); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/
2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d));
dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); [/asy]$
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$, since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two
We proceed by contradiction. Assume a configuration of four lines exists such that there exist only two intersection points. Let these intersection points be $A$ and $B$. Consider two cases:
Case 1: No line passes through both $A$ and $B$
Then, since an intersection point is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$. Then, since there can be no additional intersections, the 2
lines that pass through $A$ cant intersect the 2 lines that pass through $B$, and so 2 lines passing through $A$ must be parallel to 2 lines passing through $B$. Then the two lines passing through
$B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2: There is a line passing through $A$ and $B$
Then there must be a line $l_a$ passing through $A$, and a line $l_b$ passing through $B$. These lines must be parallel. The fourth line $l$ must pass through either $A$ or $B$. Without loss of
generality, suppose $l$ passes through $A$. Then since $l$ and $l_a$ cannot coincide, they cannot be parallel. Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.
All possibilities have been exhausted, and thus we can conclude that two intersections is impossible. Our answer is given by the sum $0+1+3+4+5+6=\boxed{\textbf{(D) } 19}$.
Video Solution 1
~Education, the Study of Everything
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_8","timestamp":"2024-11-08T12:23:11Z","content_type":"text/html","content_length":"55043","record_id":"<urn:uuid:6d483332-4222-4e62-8685-1061153e4a47>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00728.warc.gz"} |
An object with a mass of 6 kg is hanging from a spring with a constant of 4 (kg)/s^2. If the spring is stretched by 9 m, what is the net force on the object? | HIX Tutor
An object with a mass of #6 kg# is hanging from a spring with a constant of #4 (kg)/s^2#. If the spring is stretched by #9 m#, what is the net force on the object?
Answer 1
What happens when you stretch the string and release, is that,the restoring force tries to pull the object upwards with the force #(F= K×x=4×9=36 N)#
So, net force acting on the object becomes #(mg-F)# i.e #(6×10-36) N# or #24 N# downwards.
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Answer 2
The net force on the object can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The
formula for Hooke's Law is ( F = -kx ), where ( F ) is the force exerted by the spring, ( k ) is the spring constant, and ( x ) is the displacement from the equilibrium position.
• Mass of the object, ( m = 6 ) kg
• Spring constant, ( k = 4 ) (kg)/s(^2)
• Displacement from equilibrium position, ( x = 9 ) m
Using Hooke's Law, we can find the force exerted by the spring:
[ F = -kx = -(4)(9) = -36 \text{ N} ]
Since the force exerted by the spring is opposite to the direction of displacement, we take it as negative.
To find the net force on the object, we also need to consider the force due to gravity. The force due to gravity can be calculated using the formula ( F = mg ), where ( m ) is the mass of the object
and ( g ) is the acceleration due to gravity (( g \approx 9.8 ) m/s(^2) on Earth).
[ F = (6)(9.8) = 58.8 \text{ N} ]
The net force on the object is the vector sum of the forces due to the spring and gravity:
[ \text{Net force} = F_{\text{spring}} + F_{\text{gravity}} = -36 + 58.8 = 22.8 \text{ N} ]
So, the net force on the object is 22.8 N.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Loïc Baumann's Blog
The new release of Team Foundation Server 2010 went out few weeks ago. As I was lazy, I installed a quick VM to test it out, I put 512 Meg of RAM because I was cheap and planned to setup the basic
Well, it didn’t go very well at first as the new Team Foundation Server Administration Console rejected my configuration: not enough RAM, 1024 is at least needed!
Well, I was a bit stubborn on this and thought that 512MB could be fine for my own VPC evaluation (yes, cheap and stubborn, what a guy!), so I tried to find if there was a way to overcome this
My buddy Reflector helped me to find out what could be done and here’s the solution to my problem:
If there’s an error that pops up, you can make the verification mechanism ignores it by setting an environment variable (that’s right, a Windows environment variable) corresponding to the criteria
you want to bypass.
In my case the criteria was called “VMEMORY”, the formulae is not that hard, create a variable TFS_IGNORE_VMEMORY (with a value of 1 for instance, but it doesn’t matter), and you’re good to go!
Nice, of course if not recommended to bypass this with prod server, but for evaluation purpose I guess it’s all right.
So, during the check, here’s the error TF255157: The amount of RAM available on this computer is not sufficient to complete this process. You must have the following amount of RAM available: 1024 MB.
Then you can check the log to get the criteria that’s failing:
Create a System environment variable:
Restart your Team Foundation Configuration Wizard for the process to see the environment variable!
And this time, everything’s working like a charm! | {"url":"http://loicbaumann.fr/en/2010/02/19/under-the-limits/","timestamp":"2024-11-05T10:15:35Z","content_type":"text/html","content_length":"46223","record_id":"<urn:uuid:72ba1f5a-043c-4694-a3e2-50cf533de6dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00115.warc.gz"} |
In the literature, although many studies are describing the structural break in the linear regression model with time-series data, studies investigating this issue with cross-sectional data are
limited. In this study, the performance evaluation of some approaches used to determine the structural break in a linear regression equation based on cross-sectional data was performed. In this
context, firstly, the structural break problem is defined. Then, the theoretical expositions of some well-known methods which determine the structural break are given. The methods which used to
determine the structural breaks may show performance differences under the effect of some factors. The performances of selected methods were evaluated with a simulation study in the context of
the difference of constant terms, the difference of slope coefficients, location of break-point, sample size and homogeneity of error variances. The results of the simulation showed that the
performances become different in terms of some structural features from the suggested methods for determination of structural break.
Keywords: Structural Break, Linear Regression Models, Cross-Sectional Data, Break-Point
JEL Classifications: C40, C530.
DOI #: 10.33818/ier.450804
Makalenin Tam Metni
Bu sayıdaki tüm makaleler
^1 Zümre Özdemir Güler, Research Asistant of Econometrics, Karamanoğlu Mehmet Bey University, Karaman, TURKEY, 70200, (email: zumreozd@kmu.edu.tr), Tel: +903382262000, Fax: +903382262023.
^2 Mehmet Akif Bakır, Professor of Statistics, Gazi University, Ankara, TURKEY, 06500, (email: mabakir@gazi.edu.tr), Tel: +903122021470, Fax: +903122122279. | {"url":"https://ead.org.tr/ozetler/450804.html","timestamp":"2024-11-06T05:36:02Z","content_type":"application/xhtml+xml","content_length":"5482","record_id":"<urn:uuid:41066132-5cca-4703-9b6d-ddd665dc79d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00255.warc.gz"} |
IPMAT Rohtak 2019 Question Paper
If the speed of boat in still water on Saturday was 27 km/hr and the speed of boat in still water on Wednesday was $$66\dfrac{2}{3}\%$$ more than that of Saturday and time taken to travel upstream on
Wednesday is $$16\dfrac{1}{3}$$ times than time taken by it to travel downstream on Saturday, then find the speed of stream (in kmph) on Saturday?
If the time taken by boat to travel upstream on Monday is $$27\dfrac{1}{5}$$ hrs. more than the time taken by it to travel downstream on the same day, then find the speed of boat in still water on
Monday ?(speed of boat in still water is same in upstream as in downstream)
If the time taken by boat to travel upstream on Wednesday is 6/7 times than the time taken to travel downstream on Monday and the speed of boat in still water on Monday is 15 kmph then find the
speed of boat in still water on Wednesday? ( speed of boat in still water is different for different days)
The speed of boat in still water on Saturday was 21 km/hr. and that on Sunday was $$28\dfrac{4}{7}\%$$ more than that on Saturday, if the time taken by boat to travel upstream on Saturday is $$2\
dfrac{1}{2}$$ times the time taken to travel downstream on Sunday, then find the time taken by the boat to cover a distance of 125 km upstream on Saturday?
If the time taken by boat to travel upstream on Friday is 30 hours more than the time taken by it to travel downstream on Wednesday and the speed of boat in still water on Friday is 17 kmph, then
find the upstream speed of boat on Wednesday? (speed of boat in still water is different on different days)
A Container contains X litres of Milk. A thief stole 50 litres of Milk and replaced it with the same quantity of water. He repeated the same process further two times. And thus, Milk in the container
is only X-122 litres. Then what is the quantity of water in the final mixture?
Veena has to pay Rs. 2460 to Sita, 5 Months later at 6% SI per annum, and Gita has to pay Sita same amount at 7.5% SI per annum after certain months. If both took the same amount of loan from Sita
then Gita paid loan after how many months?
Use the figure below to answer questions 8 through 9.What is the area of the shaded figure?
Use the figure below to answer questions 8 through 9.
What is the ratio of the area of Circle M and the area of Circle K?
If x% of y is equal to z then what percentage of z is x? | {"url":"https://cracku.in/ipmat-rohtak-2019-question-paper-solved?page=1","timestamp":"2024-11-10T02:42:48Z","content_type":"text/html","content_length":"164502","record_id":"<urn:uuid:601a97e3-e3f8-49de-952a-889c32472078>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00765.warc.gz"} |
Conic Section Eccentricity Calculator - GEGCalculators
Conic Section Eccentricity Calculator
The eccentricity of a conic section is a measure of its shape. It quantifies how stretched or compressed the conic is compared to a circle. For ellipses, it ranges from 0 to 1, with 0 indicating a
perfect circle. For hyperbolas, it’s greater than 1, while parabolas have an eccentricity of 1, giving them a unique open shape.
Conic Section Eccentricity Calculator
Conic Section Eccentricity (e) Shape
Circle 0 Perfectly round
Ellipse 0 < e < 1 Stretched or compressed circle
Parabola e = 1 Unique open curve
Hyperbola e > 1 Stretched, open curve with two branches
How do you find the eccentricity of a conic section? The eccentricity of a conic section can be found using the formula:
Eccentricity (e) = c / a
where ‘c’ is the distance from the center to one of the foci, and ‘a’ is the distance from the center to a vertex.
What is the eccentricity of a conic? The eccentricity of a conic section determines its shape. It is a measure of how “stretched” or “squished” the conic is compared to a perfect circle. Eccentricity
values range between 0 and 1 for ellipses, equal 1 for parabolas, and greater than 1 for hyperbolas.
What is the formula for the eccentricity ratio? There is no standard formula for an “eccentricity ratio.” Eccentricity is typically expressed as a single value for a conic section.
What if eccentricity is 1 in a conic section? If the eccentricity of a conic section is 1, it represents a parabola. In a parabola, the distance between the focus and the directrix is equal, creating
a unique, open shape.
How do you find the eccentricity of an ellipse if foci are given? To find the eccentricity of an ellipse if the foci are given, you can use the formula:
Eccentricity (e) = distance between foci (2c) / length of major axis (2a)
How do you find the eccentricity of a hyperbola? To find the eccentricity of a hyperbola, you can use the formula:
Eccentricity (e) = c / a
where ‘c’ is the distance from the center to one of the foci, and ‘a’ is the distance from the center to a vertex.
What is the eccentricity of ellipse, parabola? The eccentricity of an ellipse is a value between 0 and 1, while the eccentricity of a parabola is always equal to 1.
What eccentricity is and how it is measured? Eccentricity is a measure of how “stretched” or “squished” a conic section is compared to a perfect circle. It is measured as the ratio of the distance
between the foci (or focus and directrix in the case of a parabola) to the length of the major axis.
What does 1 eccentricity mean? An eccentricity of 1 means that the conic section is a parabola. In a parabola, the distance between the focus and the directrix is equal, giving it a unique, open
What is the formula for minimum eccentricity? There is no specific formula for finding the minimum eccentricity of a conic section. The eccentricity depends on the specific parameters (e.g., the
semi-major axis and semi-minor axis for an ellipse or the distance from the vertex to the focus for a parabola) of the conic.
How do you find the eccentricity of a parabola? To find the eccentricity of a parabola, you can use the formula:
Eccentricity (e) = 1
This is because the eccentricity of a parabola is always equal to 1.
Is the eccentricity of a parabola always 1? Yes, the eccentricity of a parabola is always equal to 1.
Why is the eccentricity of an ellipse between 0 and 1? The eccentricity of an ellipse is between 0 and 1 because it represents how “squished” or “stretched” the ellipse is compared to a perfect
circle. When the eccentricity is 0, the ellipse is a perfect circle, and as it approaches 1, the ellipse becomes more elongated.
What is the eccentricity and distance between foci? The eccentricity of a conic section is the ratio of the distance between its foci to the length of its major axis.
What is the minimum eccentricity that an ellipse can have? The minimum eccentricity that an ellipse can have is 0, which occurs when the ellipse is a perfect circle.
What if the eccentricity of a hyperbola is 2? If the eccentricity of a hyperbola is 2, it indicates that the hyperbola is highly stretched and elongated. In practice, the eccentricity of hyperbolas
is typically greater than 1 but not equal to 2.
How do you find the eccentricity of the ellipse or hyperbola? You can find the eccentricity of an ellipse or hyperbola using the formula:
Eccentricity (e) = c / a
where ‘c’ is the distance from the center to one of the foci, and ‘a’ is the distance from the center to a vertex for an ellipse, or ‘a’ is the distance from the center to a point on one of the
branches for a hyperbola.
How do you find the eccentricity of a vertex? The eccentricity of a conic section is not found by directly calculating the eccentricity of a vertex. Instead, you find the eccentricity using the
distances between the foci and the geometry of the conic section.
What is the eccentricity of an ellipse, hyperbola, parabola, and circle?
• Ellipse: Eccentricity ranges from 0 to 1.
• Hyperbola: Eccentricity is greater than 1.
• Parabola: Eccentricity is always equal to 1.
• Circle: Eccentricity is always equal to 0.
What is the eccentricity of the general equation of an ellipse? The eccentricity (e) of the general equation of an ellipse is determined by the values of ‘a’ and ‘b’ in the equation:
x^2/a^2 + y^2/b^2 = 1
The eccentricity can be calculated as:
e = √(1 – (b^2/a^2))
What is the eccentricity of an ellipse calculator? An eccentricity calculator is a tool or software that helps calculate the eccentricity of an ellipse based on its geometric properties, such as the
lengths of its semi-major and semi-minor axes.
Is eccentricity always a number? Yes, eccentricity is always a numerical value, and it is a dimensionless quantity.
How do you find the eccentricity of an object? The eccentricity of an object is typically not calculated unless the object’s shape can be described by a conic section (ellipse, hyperbola, parabola,
or circle). For other objects, eccentricity may not be a relevant parameter.
What is an example of eccentricity? An example of eccentricity is the shape of planetary orbits. The eccentricity of an orbit determines how elongated or stretched the orbit is, with values close to
0 indicating nearly circular orbits and values close to 1 indicating highly elliptical orbits.
What is the eccentricity of a graph? The term “eccentricity” in the context of a graph refers to a different concept. It is a measure of how far a vertex is from the farthest vertex in the graph. It
is not related to conic sections.
What is another word for eccentricity? Synonyms for eccentricity include peculiarity, quirkiness, oddity, singularity, idiosyncrasy, and uniqueness.
What is the average eccentricity? There is no concept of average eccentricity unless you are referring to a specific set of conic sections with varying eccentricities.
What is the maximum eccentricity? The maximum eccentricity for a conic section depends on the type of conic:
• For an ellipse, the maximum eccentricity is 1.
• For a hyperbola, there is no theoretical maximum, but it can be any value greater than 1.
• For a parabola, the eccentricity is always 1.
• For a circle, the eccentricity is always 0.
What is the smallest value of eccentricity? The smallest value of eccentricity is 0, which occurs in the case of a perfect circle.
What is the unit of eccentricity? Eccentricity is a dimensionless quantity and does not have units.
What object has an eccentricity of 1? Objects with an eccentricity of 1 include parabolic reflectors and the paths of objects in free fall under gravity.
Which conic section has no eccentricity? A perfect circle has no eccentricity, as its eccentricity value is always 0.
Can you have negative eccentricity? Eccentricity is defined as a positive value, so it is not typically expressed as negative. It represents how “stretched” or “squished” a conic section is compared
to a perfect circle.
What does it mean if an object has an eccentricity greater than 1? If an object has an eccentricity greater than 1, it represents a hyperbolic shape. In the context of conic sections, eccentricity
values greater than 1 are associated with hyperbolas.
Can an ellipse have an eccentricity of 1? No, an ellipse cannot have an eccentricity of 1. The eccentricity of an ellipse always falls between 0 and 1, with 0 indicating a perfect circle and values
close to 1 indicating a highly elongated ellipse.
What happens to the eccentricity of an ellipse as the foci are moved closer together? As the foci of an ellipse are moved closer together, the eccentricity of the ellipse decreases. When the foci
coincide at the center, the eccentricity becomes 0, and the ellipse becomes a perfect circle.
Is eccentricity the ratio of distance? Yes, eccentricity is essentially the ratio of distances. It is the ratio of the distance between the foci (or focus and directrix for a parabola) to the length
of the major axis of a conic section.
Do all ellipses must also have the same eccentricity? No, all ellipses do not have to have the same eccentricity. The eccentricity of an ellipse can vary depending on its specific dimensions and
What if the eccentricity of an ellipse is 3/7? If the eccentricity of an ellipse is 3/7, it means that the ellipse is moderately stretched compared to a perfect circle but not highly elongated.
What is the eccentricity of a completely flat ellipse? A completely flat ellipse, also known as a degenerate ellipse, has an eccentricity of 0. In this case, it degenerates into a line segment.
What can the eccentricity of the hyperbola never be equal to? The eccentricity of a hyperbola can never be equal to 0 because hyperbolas are characterized by their eccentricity values being greater
than 1.
What is the eccentricity of a rectangular hyperbola? A rectangular hyperbola is a special case of a hyperbola with an eccentricity of exactly 2.
What is the eccentricity of a hyperbola always? The eccentricity of a hyperbola is always greater than 1. It can vary, but it is never less than 1.
What is the eccentricity of a parabola (e=1)? The eccentricity of a parabola is always equal to 1.
What is the eccentricity of all conics? The eccentricity of all conics can vary. It is 0 for circles, between 0 and 1 for ellipses, 1 for parabolas, and greater than 1 for hyperbolas.
What is the eccentricity of each vertex? Eccentricity is not typically associated with individual vertices of a conic section. It is a property of the entire conic.
What is the difference in eccentricity of a circle and a flat ellipse? The eccentricity of a circle is always 0, indicating a perfectly round shape. In contrast, a flat or degenerate ellipse has an
eccentricity of 0, but it represents a line segment, not a closed curve.
What is the general equation of a conic with eccentricity? The general equation of a conic section with eccentricity ‘e’ depends on the type of conic:
• For an ellipse: (x^2/a^2) + (y^2/b^2) = 1 with eccentricity given by e = √(1 – (b^2/a^2))
• For a hyperbola: (x^2/a^2) – (y^2/b^2) = 1 with eccentricity given by e = √(1 + (b^2/a^2))
• For a parabola: y^2 = 4ax with eccentricity e = 1
What is the formula for eccentricity of a structure? The formula for eccentricity in the context of structures or mechanics may differ from the geometric eccentricity used for conic sections. The
specific formula would depend on the parameters and context of the structural analysis.
Is an eccentricity of 1 a line? An eccentricity of 1 does not represent a line; rather, it represents a parabola. A parabola is a curve with eccentricity equal to 1.
What is an ellipse with an eccentricity of zero called? An ellipse with an eccentricity of zero is called a perfect circle.
What is the eccentricity of a parabola? The eccentricity of a parabola is always equal to 1.
GEG Calculators is a comprehensive online platform that offers a wide range of calculators to cater to various needs. With over 300 calculators covering finance, health, science, mathematics, and
more, GEG Calculators provides users with accurate and convenient tools for everyday calculations. The website’s user-friendly interface ensures easy navigation and accessibility, making it suitable
for people from all walks of life. Whether it’s financial planning, health assessments, or educational purposes, GEG Calculators has a calculator to suit every requirement. With its reliable and
up-to-date calculations, GEG Calculators has become a go-to resource for individuals, professionals, and students seeking quick and precise results for their calculations.
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HP 49G: Minimum Multiplier M of Integer N such that M*N Consists Only of 1's & 0's
01-25-2020, 11:50 PM
Post: #8
Albert Chan Posts: 2,789
Senior Member Joined: Jul 2018
RE: HP 49G: Minimum Multiplier M of Integer N such that M*N Consists Only of 1's &...
(01-25-2020 05:58 PM)John Keith Wrote: Chai Wah Wu's Python program from the OEIS page and it returned A(99) in less than one second.
I then tried A(9990) on the Python program and it took over 15 minutes to complete.
You could try the
sage code
, which is also a valid Python code if "^" replaced as "**"
>>> minmulzo(999) * 10 # = p(9990)
Above finished in my old Pentium 3 in under 0.1 second
Roughly, this is why above is so fast.
Since N=9990, P must also end in 0, since N divides P.
If N is odd (but not end in 5's, see
my post
), P must also be odd
To minimize P, we multiply by only a ten, to add a zero to P.
This optimization speedup the already fast minmulzo() by 4X
Instead of doing all the unnecessary base conversions, it build a list of mod values.
The program loops thru the combinations of sum of mods, looking for multiples of N.
Here is what each "bit" contribute to the mod value, for P(999):
>>> for i in range(21): print i, pow(10,i,999)
Above repeating pattern, P(999) required 27 1's, to have sum of mods = 9*111 = 999 = N
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On the least trimmed squares estimator
The linear least trimmed squares (LTS) estimator is a statistical technique for fitting a linear model to a set of points. Given a set of n points in R ^d and given an integer trimming parameter h≤n,
LTS involves computing the (d-1)-dimensional hyperplane that minimizes the sum of the smallest h squared residuals. LTS is a robust estimator with a 50 %-breakdown point, which means that the
estimator is insensitive to corruption due to outliers, provided that the outliers constitute less than 50 % of the set. LTS is closely related to the well known LMS estimator, in which the objective
is to minimize the median squared residual, and LTA, in which the objective is to minimize the sum of the smallest 50 % absolute residuals. LTS has the advantage of being statistically more efficient
than LMS. Unfortunately, the computational complexity of LTS is less understood than LMS. In this paper we present new algorithms, both exact and approximate, for computing the LTS estimator. We also
present hardness results for exact and approximate LTS. A number of our results apply to the LTA estimator as well.
Funders Funder number
National Science Foundation 1117259
• Approximation algorithms
• Least trimmed squares estimator
• Linear estimation
• Lower bounds
• Robust estimation
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Thirty days hath September, April, June, and November ; All the rest have thirty-one, Except the second month alone, Which has but twenty-eight, in fine, Till leap year gives it twenty-nine. The
Common School Arithmetic: Combining Analysis and Synthesis, Adapted to ... - Page 74by James Stewart Eaton - 1866 - 312 pagesFull view
About this book
Jacques Ozanam - Scientific recreations - 1803 - 636 pages
...years has 28 days, and in bissextiles 29. The number of the days in each month may be known also by the following lines : Thirty days hath September, April, June and November ; All the rest
rnive thirty-one, Except February alone. PROBLEM XI. To find the day of the Month on which the Sun...
Jacques Ozanam - Scientific recreations - 1814 - 518 pages
...in bissextiles 29. The number of the days in each month may be known also by the following memorial lines: Thirty days hath September, April, June and November; All the rest have thirty-one,
Except February alone. PROBLEM XI. To find the day of the Month on which the Sun enters into each sign...
John Ely - Readers - 1817 - 124 pages
...Bionlh Las not an equal number, of days •$ how H.Mm.V thiys are in each monlh 7 . one year, Thirty days hath September.' April, June, and November ; All the rest have thirty-one, $ut February,
which alone, Has twenty-eight of Julian time, And every leap year twenty-nine. 4. January...
William Bentley Fowle - Arithmetic - 1827 - 108 pages
...month may be found in the above table, but will be better recollected in the following rhyme : Thirty days hath September, April, June and November; All the rest have thirty-one., ft T II I <
,11 11(1111 l> *• •.Hath twenty-nine, one year nn LXXVII. * 1 . How many minutes are...
Children's literature - 1829 - 126 pages
......... Year. The number of days in each month, may be ascertained by the following rule: — Thirty days hath September, April, June and November, All the rest have thirty-one, Except February
alone, Which has eight above a score, But in Leap-Year has one more. Federal Money....
Ira Wanzer - Arithmetic - 1831 - 408 pages
...has but twenty-eight, in fine, Till leap year gives it twenty-nine. Or the following : •= Thirty days hath September, April, June, and November; All the rest have thirty-one, / Except February
alone, Which hath four and twenty-four, And every fourth year one day more. i Note...
H. L. Barnum - Readers - 1833 - 82 pages
...answer that in poetry too. How many days are there in the different months of the year? Mary. "Thirty days hath September, April, June, and November ; All the rest have thirty-one, Excepting
February alone, Which hath but twenty-eight days clear, And twenty-nine every leap-year."...
Charles Davies - Arithmetic - 1833 - 284 pages
...occurs, is added to the month 'of February, so that this month has 29 days in the Leap year. Thirty days hath September, April, June, and November; All the rest have thirty-one, Excepting
February, twenty-eight alone. QUESTIONS. § 89. What are the denominations of Time ? How...
John Vose - Astronomy - 1834 - 230 pages
...of the Sabbath. It consists of seven days. . * The number of days in each month may be remembered by the following lines: Thirty days hath September, April,...June, and November; All the rest
have thirty-one, Saving February alone. (Greeks.) " The ha common practice, it is too well known. to need explanation....
Charles Guilford Burnham - Arithmetic - 1837 - 266 pages
...year. The number of days in the several months may be called to mind by the following verse : Thirty days hath September, April, June and November; All the rest have thirty-one, Excepting
February alone, Which hath twenty-eight. nay more, Hath twenty-nine one year in four. The... | {"url":"https://books.google.com.jm/books?id=LNtGAAAAIAAJ&qtid=df7e7889&lr=&source=gbs_quotes_r&cad=5","timestamp":"2024-11-14T00:41:53Z","content_type":"text/html","content_length":"30501","record_id":"<urn:uuid:537f1fbc-c69f-46ff-84eb-096d02c3a6f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00202.warc.gz"} |
Adaptive solver: about
The finite element method of analysis offers a numerical solution for physical phenomena described by partial differential equations. For this, it relies on a sampling of the studied domain that is
called mesh. The denser the mesh of the studied domain, the more accurate the numerical solution will be, but this is at a high cost in terms of computer memory requirements.
It is therefore essential to generate a mesh that should be refined in the zones that require it and loose in the rest of the domain. In other words, the mesh must be adapted, both, to the geometry
and to the physics demands of the problem.
Over the last years, considerable effort has been made in order that the Flux software could automatically generate a qualitative mesh.
The first stage was to set up the aided mesh, which permits the user to obtain a mesh adapted to the geometry of the problem.
Today, the Flux software offers the adaptive solver. This process facilitates automatically refined mesh in the locations where the physics of the problem require it.
The adaptive solver is a process that permits modeling a problem by means of an adaptive mesh. This mesh presents sizes of elements, which correspond to the local behavior of the considered physics.
In the zones requiring a finer analysis of the studied phenomenon, this translates by:
• A tightening of the meshes
• A diminution of the finite elements size
│ No-adapted mesh │ Adapted mesh │
Limitations / Restrictions in 2D
Here are some restrictions:
• The adaptive solver is reserved to the applications:
□ Magneto Static 2D
□ Electro Static 2D
□ Steady State AC Magnetic 2D
• The adaptive solver does not work with the:
□ mechanical sets of the « compressible » type
□ anisotropic, non linear regions
• It is not advisable to use the mapped mesh with the adaptive solver
• As for all solvers, one should start from an initial mesh
Limitations / Restrictions in 3D
Since Flux 12.1, adaptive solver is available for the 3D applications:
• Electro Static 3D
• Magneto Static 3D
In 3D, the limitations of the adaptive solver are:
• mechanical sets
• mapped mesh or extrusion mesh
• physics with surface of type “Air Gap region”
Note that 2^nd order mesh will automatically be created in order to compute relevant local errors. | {"url":"https://2022.help.altair.com/2022.3/flux/Flux/Help/english/UserGuide/English/topics/ResolutionAdaptativeAPropos.htm","timestamp":"2024-11-10T02:58:54Z","content_type":"application/xhtml+xml","content_length":"63618","record_id":"<urn:uuid:f779bc8f-9426-4380-a9ec-8a427d4247ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00841.warc.gz"} |
VIII. Using Mathematical Representations to Estimate a Quantity of Interest
Unit 2: Exploring the Nature of Thermal Phenomena
VIII. Using Mathematical Representations to Estimate a Quantity of Interest
Qualitative approaches, such as interpreting the shapes of graphs, were helpful in gaining a conceptual understanding about what is happening when mixing hot and cold water. Quantitative approaches
make possible the making of predictions and estimation of quantities that may be of interest.
A. Solving a thermal math problem
In this section, we demonstrate how to solve thermal math problems in the context of mixing hot and cold water. We assume that the specific heat of water is the same over the range of temperatures
between freezing at 0ºC and boiling at 100ºC. We also assume that the specific heat of water is the same as the specific heat of fluids such as tea, milk, and cocoa when working thermal energy
problems involving mixing these substances. The same mathematical process also applies in more complex situations in which the specific heat of the hot material differs, however, from the specific
heat of the cold material, when, for example, a piece of hot metal is submerged into a cooling bath.
Question 2.14 How can one use mathematical representations of thermal phenomena to estimate a quantity of interest?
After developing both experimental and theoretical ways to describe what happens when mixing hot and cold water, one can use that knowledge to generate and solve thermal math problems.
To make up a thermal math problem, decide on a scenario and specify three of the four variables involved if only considering changes in temperature: mass of hot water, mass of cold water, magnitude
in change in temperature of hot water, or magnitude of change in temperature of cold water. Also include information about one of the initial temperatures or equilibrium temperatures. Use mass units
(grams, kilograms) or parts rather than volume units. If the scenario involves materials with difference specific heats, be sure to include that information in stating the problem and in calculating
the answer.
As with solving pinhole math problems, the goal in solving a thermal math problem is not the “answer.” The goal is to build your ability to help someone else understand what to do and why. Start by
helping the learner to understand what is happening by describing the scenario verbally with words and visually with a sketch. Next review the physics involved by stating what the relevant central
ideas are. Also draw a qualitative graph and use it and the central ideas to explain what is happening. Then describe the graph mathematically, being clear about what each symbol represents. Justify
the equation that relates the quantities represented; write the equation in both words and symbols. Finally, solve for the unknown in symbols before substituting values. After calculating an answer,
be sure to also discuss why that answer seems reasonable.
In facilitating a conversation with someone about thermal phenomena, ask questions rather than tell answers throughout this process. In solving a thermal math problem for homework, follow the format
provided here:
Format for Solving a Thermal Math Problem
a. State the problem in words
b. Make a sketch of the amounts to be mixed
c. Review what you know about the physics of this phenomenon: summarize the conceptual model by stating the relevant central ideas
d. Draw a graph representing the problem: use a ruler to make straight perpendicular axes, draw flat horizontal lines to represent temperatures that are not changing before the two substances are
mixed, add the shared equilibrium temperature after the mixing, check that you have drawn the appropriately shaped graph for mixing more hot than cold or more cold than hot or equal amounts. Draw
the graph to indicate whether the changes in temperature happen very quickly or are gradual processes.
e. Tell the ‘story’ of the graph with the relevant central ideas to explain why the equilibrium temperature is likely to be where you have drawn it (nearer the temperature of the initial hot water,
initial cold water, or in the middle.)
f. Represent this scenario mathematically: State the equation in words that relates the masses of the substances mixed and their changes in temperature. Use the experimental form of the equation
derived from your exploration, including specific heats if the materials differ, or the theoretical form based on the Conservation of Energy. Justify the use of the equal sign accordingly.
g. Define symbols, state equation in symbols, and express how you are envisioning this equation.
h. Solve for the unknown in symbols
i. Record given values and estimate any needed
j. Substitute values and calculate answer
k. Check answer: why does the number you get from the calculation seem reasonable?
• Complete the process of generating and solving a thermal math problem before looking at a slightly modified example of student work.
1. Example of student work generating and solving a thermal math problem
The problem stated in words is: If you have 180 grams of hot tea, and you want to cool it down by 20 degrees Celsius by adding 60 grams of cold water, how much will the temperature of the
cold-water change? Assume all of the energy lost by the tea is gained by the cold water and that the specific heats of tea and water are the same. The initial temperature of the cold water is 15
degrees Celsius. After finding the change in temperature of the cold water, find the equilibrium temperature and the initial temperature of the hot tea.
(As shown in Fig. 2.20), below is a sketch of the situation.
Fig. 2.20 Student sketch of the situation for this problem.
(As shown in Fig 2.21), below is a graph that represents the problem. The first relevant idea is, when mixing unequal amounts of hot and cold water, the ratio of the amount of hot water to the amount
of cold water is approximately equal to minus the ratio of the change in temperature of the cold water to the change in temperature of the hot water. Next, the energy lost by the hot object equals
the product of its mass, specific heat, and change in temperature. Finally, energy is conserved: energy lost = energy gained, so when mixing hot and cold water, the (magnitude of the) energy lost by
the hot water equals (the magnitude of) the energy gained by the cold water.
Fig. 2.21 Student representation of the problem graphically.
Symbols that are defined for the relevant quantities are m[h]= mass of hot tea, m[c]= mass of cold water, ∆T[c]= change in Temperature of cold water, and ∆T[h]= change in Temperature of
hot tea. An algebraic equation that relates these quantities in symbols is
[latex]\frac{\text{m}_h}{\text{m}_c} \approx \frac{-\Delta T_c}{\Delta T_h} [/latex]
The same equation in words is
[latex]\frac{\text{Mass of hot tea}}{\text{Mass of cold water}} \approx \frac{-\text{change in Temperature of cold water}}{\text{change in Temperature of hot tea}}[/latex]
This equation is justified because it is equivalent to the theoretical statement of the Law of Conservation of Energy – (m[h] c[w] ∆T[h] ) = ( m[c] c[w] ∆T[c]), assuming that no energy is gained by
the environment such as the cup and air.
The unknown and the quantity to be calculated is the change in temperature of the cold water, ∆T[c]. The symbols for the quantities I have provided in the problem statement and their numerical values
are m[h]= 180 grams, m[c]= 60 grams, and ∆T[h]= – 20°C.
The equation solved algebraically for the unknown is [latex]{\Delta T_c} = -{\Delta T_h} \frac{{(\text{m}_h)}}{(\text{m}_c)}[/latex].
When I substitute the values above for the known quantities and then calculate the answer I get
[latex]{\Delta T_c} = -({-20^{\circ}C}) \frac{\text{(180 grams)}}{\text{(60 grams)}} = {\text60}^{\circ}C[/latex]
This answer is reasonable because there was a lot more hot water than cold water which means that the temperature of the mixture will be closer to the initial temperature of the hot water than the
initial temperature of the cold water. This means that the cold water should have a greater temperature change than the hot water. The hot water had a temperature change of -20°C, so this answer of
the cold water having a temperature change of 60°C is reasonable.
The initial temperature of the cold water was 15°C. Since we know the initial temperature of the cold water, we are able to use equations to solve for the initial temperature of the hot water and for
the equilibrium temperature. I can now write an algebraic equation for the unknown of the equilibrium temperature, T[e]. That equation is T[e] = T[c] + ∆T[c]. When I substitute the values in I
get T[e] = 15°C + 60°C= 75°C.
Then, I can write an algebraic equation for the unknown of the initial temperature of the hot tea. That equation is ∆T[h]. = T[e] – T[h] so T[h] = T[e] – ∆T[h] When I substitute the values in I
get T[h ]= 75°C – (-) 20°C= 95°C. This initial hot temperature is 95°C which is below boiling point and the initial cold temperature is 15°C which is above freezing, which both seem reasonable.
Cooling tea down to 75°C also seems reasonable because this is below boiling point but above freezing. If a student spills the tea, it is not hot enough to burn them.
Physics Student, Spring 2016
Actually 75°C is hot enough to burn one severely. Recommendations are for ‘hot’ liquids to be at 60°C or less. https://www.ncbi.nlm.nih.gov/pubmed/18226454 | {"url":"https://open.oregonstate.education/physicsforteachers/chapter/using-mathematical-representations-to-estimate-a-quantity-of-interest/","timestamp":"2024-11-03T14:04:17Z","content_type":"text/html","content_length":"89564","record_id":"<urn:uuid:ecd26bce-9ab0-4c51-ad16-7ba36cdcb944>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00621.warc.gz"} |
Estolide's in PCMO
Top Posters In This Topic
On 11/7/2022 at 12:25 PM, Grumpy Bear said:
Predict the unobtanium Porsche C40 spec oil will not be made by XOM only any more but by NOVVI or EVOLVE.
Yeah plants beat petroleum.
Edited by customboss
Using phone on this site is irritating....buggy mess
For a back door for 25% Estolides rest GRP III+ Chevron Havoline PRO RS Renewable is now about $5 a qt via walmart online. Meets DEXOS1 GEN2 and might beat GEN3 waiting on a reply from some friends
there. There is a unique ORONITE add pack that is being used in these too. Low SAPS with lots of performance. Stay tuned.
Edited by customboss
Spelling correct
59 minutes ago, customboss said:
For a back door for 25% Estolides rest GRP III+ Chevron Havoline PRO DS Renewable is now about $5 a qt via walmart online. Meets DEXOS1 GEN2 and might beat GEN3 waiting on a reply from some
friends there. There is a unique ORONITE add pack that is being used in these too. Low SAPS with lots of performance. Stay tuned.
You mean RS not DS correct?
Havoline PRO-RS Renewable Synthetic Motor Oil 5W-30, 6Q Smart Change Box - Walmart.com
26 minutes ago, Grumpy Bear said:
Yes Sir. Fingers no workee on phone
5 hours ago, customboss said:
Yes Sir. Fingers no workee on phone
Let us know what 'Friends" say
18 hours ago, customboss said:
Predict the unobtanium Porsche C40 spec oil will not be made by XOM only any more but by NOVVI or EVOLVE.
Yeah plants beat petroleum.
updated that so it makes more sense.......big oil is switching to lower cost easily engineered plant based synthetics.
On 11/8/2022 at 9:38 PM, Grumpy Bear said:
Let us know what 'Friends" say
Confirmed using 25% NOVVI estolide base oils for Havoline Pro RS 5w30 and 0w20 Dexos1 GEN2
.9% SA content add pack with rest of base oils being GRP III+.
I'm gonna try this to reduce my oil consumption to 0.
Edited by customboss
damn fingers will no function
On 11/10/2022 at 2:14 PM, customboss said:
Confirmed using 25% NOVVI estolide base oils for Havoline Pro RS 5w30 and 0w20 Dexos1 GEN2
.9% SA content add pack with rest of base oils being GRP III+.
I'm gonna try this to reduce my oil consumption to 0.
And how did this go @customboss?
PCMO-Sample-Formulation-Sheet-FINAL.pdf (biosynthetic.com)
Page 2 of this document. Read the field trials portion carefully. The oil used in this trial is not what is in this bottle. Compare with what is in the bottle below:
For the grand finale note the oil it was tested against!
Edited by Grumpy Bear
3 hours ago, Grumpy Bear said:
PCMO-Sample-Formulation-Sheet-FINAL.pdf (biosynthetic.com)
Page 2 of this document. Read the field trials portion carefully. The oil used in this trial is not what is in this bottle. Compare with what is in the bottle below:
For the grad finally note the oil it was tested against!
Yes it’s marketing, for Estolides base oils. They don’t sell fully formulated oils primarily. BT states clearly their 5w20 and 5w30 formulations are for example not commercialization of fully
formulated oils. BT sells base oils NOT FULLY FORMULATED OILS TO CONSUMERS.
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<hr>It's not always experts who have the answers
The figure shown here was published in the excellent
New Scientist magazine
this week.
It shows the results of a study into how mathematicians collaborated to solve a problem, which in many ways mimics how members of a
community of practice
may work together to solve a business problem, or to answer a problem of one of the members.
The problem to be addressed is known as Polymath1, and the "blog and wiki"-based approach is described
as follows
"In February, 2009, an international group comprising mathematicians ranging from amateurs to elite professionals converged on the WordPress blog of Cambridge mathematician Timothy Gowers in
order to attempt to prove a mathematical theorem; a project Gowers called Polymath1. Their results surprised even the project's most optimistic participants. In six weeks, the group had managed
to combinatorially prove the density Hales-Jewett theorem, yielding in the process a host of new mathematical insights".
A huge success for a collaborative approach, but equally interesting are the statistics shown in the diagram concerning where the contributions within the collaborative group came from. The diagram's
vertical axis represents the number of contributions from each individual, the horizontal axis represents the "significance" of each contributor to solving the problem, and the size of the blob
represents the professional seniority of each contributor, with small dots being "professionally junior" and big dots being "professionally senior".
The most interesting thing about this plot for me, are the small dots in the "low volume, high importance" sector. These aren't the famous expert mathematicians, they are people such as Jason Dyer, a
mathematics teacher in Arizona, which was able to throw light on one type of logic puzzle involved in the final solution.
With tricky problems, its not awlays the experts that have all the answer. The diagram above shows that many people can contribute, and that the non-experts can provide crucial input.
So what is the lesson for those of us working in Knowledge Management?
I think this data throws light on one of the choices we need to make, which is whether we create communities of experts, or communities of practitioners.
Some companies like to create communities of experts, or lists of experts, who people can go to if they have a problem to solve. The thinking is that "the knowledge is held by the experts, so they
are the default people to go to".
Other companies like to create communities of practitioners, or lists of everyone working in an area, who people can consult if they have a problem to solve. The thinking is that "the knowledge is
out there somewhere; it might be with the experts, it might not, so let's ask everyone and see what we turn up".
The former is like "phone a friend", the latter is like "ask the audience".
What the data from Polymath1, shown in the graph above, demonstrates is that the non-experts can make massively significant contributions, and that asking the audience seems to be, certainly for this
type of problem, a far more effective strategy.
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Array or List
Hello, I was solving a Frank-Wolfe problem. In each iteration of the algorithm I multiply an array vector with shape (n, ) by my decision variables. I tryed 2 options. First, multiply the array and
my vars. The second option was transform the array to a list using xk = xk.tolist() and multiply the list with my vars. I saw a difference in the execution time. The option of the list was faster
than the array. I would like to know what is the difference between both and where I could read more about this.
I will upload 2 photos with the difference
• Hi Cris,
Could you post the code snippet, in best case a self contained example, of how you perform the multiplication of the array/list and your variables?
Best regards,
• Sure. I used the Frank-Wolfe algorithm. This is my code.
from parámetros import *
from gurobipy import Model, GRB, quicksum
import time
import matplotlib
import matplotlib.pyplot as plt
def LASSO_FISTA(A, b, iteracion_maxima, rho):
m, n = A.shape
xk = np.zeros(n,) #Vector x_k inicial
thetak = 1
norma_b = np.linalg.norm(b)
Variables = []
for i in range(1500):
print("\n\n********** METODO FISTA *********\n")
print("ITERACION VALOR OBJ ERROR AJUSTE ||x||_1")
for iteracion in range(iteracion_maxima):
model = Model("Modelo")
y = model.addVars(Variables,vtype=GRB.CONTINUOUS,lb= -GRB.INFINITY, ub= GRB.INFINITY, name="y")
z = model.addVars(Variables, lb = 0, vtype=GRB.CONTINUOUS, name="z") #Variable auxiliar
model.addConstrs((y[a] <= z[a] for a in Variables),name = "abs")
model.addConstrs((-z[a] <= y[a] for a in Variables), name="abs")
#Se restringe el dominio.
model.addConstr(quicksum(z[a] for a in Variables) <= rho)
grad = 2*np.dot(np.transpose(A), np.dot(A,xk) - b)
obj = quicksum(grad[i]*(y[str(i)]-xk[i]) for i in range(1500))
model.Params.OutputFlag = 0
model.setObjective(obj, GRB.MINIMIZE)
#Cálculo del xk+1
for i in range(1500):
xk[i] = xk[i] + thetak*(y[str(i)].x-xk[i])
#Actualización valores con el nuevo valor de xk
thetak = 2/(2+iteracion)
valor_optimo = np.linalg.norm(np.dot(A, xk) - b)**2
error = np.linalg.norm(np.dot(A, xk) - b)/norma_b
n1 = np.linalg.norm(xk, 1)
retorno_en_pantalla = [iteracion, valor_optimo, error, n1]
print("%12.6f %12.6f %12.6f %12.6f" % (retorno_en_pantalla[0],retorno_en_pantalla[1],retorno_en_pantalla[2],retorno_en_pantalla[3]))
return xk
• The list option was. I convert the gradient and xk vector to a list, multiply and after that transform the xk to an array vector to repeat the process
grad = 2*np.dot(np.transpose(A), np.dot(A,xk) - b)
grad = grad.tolist()
xk = xk.tolist()
obj = quicksum(grad[i]*(y[str(i)]-xk[i]) for i in range(1500))
model.Params.OutputFlag = 0
model.setObjective(obj, GRB.MINIMIZE)
for i in range(1500):
xk[i] = xk[i] + thetak*(y[str(i)].x-xk[i])
xk = np.array(xk)
• Hi Cris,
Which functions/loops did you time in your first message? Was it only the \(\texttt{obj}\) computation
obj = quicksum(grad[i]*(y[str(i)]-xk[i]) for i in range(1500))
Best regards,
• I timed the execution of the entire LASSO_FISTA(A, b, iteracion-maxima, rho) function. But, I think that it is equivalent to time just
obj = quicksum(grad[i]*(y[str(i)]-xk[i]) for i in range(1500))
because It is the only change in the code.
The red lines comented are the only difference between both codes.
• Hi Cris,
In the following I dropped \(\texttt{xk}\) as all elements are zero and it makes the answer a bit easier.
The time consuming operation is not the \(\texttt{quicksum}\) but rather the conversion of
grad[i]*w[i] for i in range(n)
to a list takes the most time. I tested it via
n = 15000
grad = np.random.rand(n)
w = m.addVars(n, vtype=GRB.CONTINUOUS,lb= -GRB.INFINITY, ub= GRB.INFINITY, name="w")
start = time.time()
l = list(grad[i]*w[i] for i in range(n))
end = time.time()
print("list computation took: "+str(end-start)+" seconds")
In general, we recommend to set the objective coefficients directly in the variable definition via
w = m.addVars(n, obj=grad ,vtype=GRB.CONTINUOUS,lb= -GRB.INFINITY, ub= GRB.INFINITY, name="w")
to avoid such bottlenecks.
Best regards,
• I think I understand. I am not sure but when you multiply a var by a "vector", in this case an array vector, Gurobi takes the result as a list, where each component of the result It is an element
in the list, isn't it??
• Hi Cris,
Yes, since Gurobi's quicksum takes a list as argument, Python first converts the
grad[i]*w[i] for i in range(n)
part into a list, where each \(\texttt{grad[i]}\) is an array element and \(\texttt{w[i]}\) is a tupledict element (as it was constructed via addVars). After the list is generated, then each
component \(\texttt{grad[i]*w[i]}\) is a list element.
Best regards,
• Perfect. I understand. Thanks for your help. I would like to know where I could find more information about details like quicksums use list as an argument and improve my codes??
Best Regards.
• Hi Cris,
The best way would be to check the documentation of each such specific function and see what are the input arguments. For the Python API, you can expect that \(\texttt{list}\) is most likely the
input argument, since arrays require the \(\texttt{numpy}\) module.
Best regards,
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The Comparison Of Gaussian Elimination And Cholesky Decomposition Methods To Linear System Of Equations - Projectplus
• Format: Ms Word Document
• Pages: 85
• Price: N 3,000
• Chapters: 1-5
Linear System
This project work is concerned with study of the comparison of Gaussian elimination and cholesky decomposition methods to linear system of equations.
In chapter one, we are concerned with linear systems and the various methods of solving them. In chapter two and chapter three, we dealt with the indept study of Gaussian Elimination method and the
cholesky Decomposition method, with their good points and bad points respectively. While chapter four deals with various application area of both methods, their comparison and finally conclusion.
A wide variety of problems lead ultimately to the need to solve a linear system of equation linear system of equations are associated with many problems in engineering and science as well as with
applications of mathematics to the social sciences and the quantitative study of business and economic problems.
In 1985, according to Atkinson, system of Simultaneous linear equation occur in solving problems in a wide variety of areas with respect to mathematics, statistics, physical quantities (examples are
temperature, voltage, population management and displacement). Social sciences, engineering and business. They arise directly in solving real life problems.
The world sometimes reveals itself to us as observable relationships among the relevant variables what it does make evident are relationship that describe how both the variable and their rate of
change affect each other.
Apparently, such life changing problem gives rise to systems of simultaneous linear equation. In almost every human activities, man seems to be compelled to uncover fundamental relationship that
exist among the objects he observes. According to Maron in 1982, he said in order to make the relationship that exist between variables explicit, we frequently attempt to make a mathematical model
that will accurately reflect real life situation. Many mathematical model that will accurately reflect real life situation. Many mathematical models have the same basic structure although disparity
in Symbolic rotation may be utilized, which can arise from economics, transportation, which need may arise to make efficient allocation among several points or to solve the growth of population in
which units of x[1], x[2] …., x[n ]arises from net flow from one point to another or in relationship to population growth, that is, number of individuals in a particular age group at a particular
There are various methods in solving linear system of simultaneous equations. In numerical analysis the techniques and methods for solving system of linear equations belongs to two categories: Direct
and Iterative methods. The direct methods obtain the exact solution (in real arithmetic) in finitely many operations where as iterative method generate a sequence of approximations that only converge
in the limit to the solution. The direct method falls into two categories or clam that is the Gaussian elimination method and cholesky decomposition method. Some others are matrix inverse method and
LU factorization method and the Cramer’s rule method.
The elimination approach reduces the given system of equations to a form from which the solution can be obtained by simple substitution since calculators and computers have some limit to the number
of digits for their use this may lead to round-off errors and produces poorer results. Generally, the direct method are best for full or bounded matrices where as iterative methods are best for very
large and sparse matrices. The iterative method provide an alternative to the direct methods for solving systems of linear equations. This method involves assumption of some initial values which are
then refined repeatedly till they reach some accepter rang of accuracy. The Jacobi and Gawn-siedel methods are good examples of the iterative method.
Systems of linear equations may be grouped as follow
The system of linear equations are divided into consistent and inconsistent and inconsistent.
The inconsistent equation is an equation that has numbers that are not all zero that is the equation has no solution.
For example
X + 2y – 3z = 4
Y + 2z = 5
The consistent equation is an equation that has numbers that are all zero, that is, the equation has a solution. There are two cases
r = n, that is, there are so many non-zero equations as unknowns. Then we can successfully solve uniquely for the unknowns and there exist a unique solution for the system.
r < n, m that is, there are more unknowns than there are non-zero equations. Thus, we arbitrarily assign values to the unknowns and then solve uniquely for the unknowns. Accordingly, there exist an
infinite number or solutions. For example
Since there is no equation of the form 0 = c with c0 the system is consistent. Furthermore since there are three unknowns and three non zero equations the system has a unique solution.
Also, the system
X + 2y – 3z + w = 4
y + 2z +3w = 5
is consistent and since there are more unknowns than non-zero equations, the system has an infinite number of solution. On the other hand, a linear system can be written in matrice form as follows
A x = b.
A linear equation X[1], X[2], …, X[n] is obtained by requiring the linear combination to assume a prescribed value b, that is
Such equation arise frequently generally as n x n linear system is
However it will be necessary to introduce the basic fundamental principles of matrices in linear systems because matrices makes it possible to describe linear systems in a simple way that makes
solving n x n linear systems seem like solving ordinary system of equations as follows:-
Besides, matrices can be used to show how to generalize the Newton Raphson method to solve non-linear n x n systems.
The linear system in explicit form
For X[1], X[2], X[3],…, Xn are unknowns,
Given aij for each ij = 1, 2, …, n and bi for each i = 1,2,…,n are scalars
According to Maron and Atkinson in 1982 and 1985 respectively, says, that the definition of matrix multiplication makes it possible to write this linear system compactly as a single matrix equation.
AX = b ……………………………………………………….(3)
Where A is a given n x n matrix assumed to be non-singular, b is a given column vector and X is the solution vector to be determined that is
….. (4)
If A is a non-singular, then the numerical vector A^-1b satisfies equation (2) above because
However, any numerical vector X for AX=b must by A^-1 satisfy vector X = A^-1b
Also, if A is non-singular, then any b, the system AX=b has a unique solution given by X = A^-1b
But X = A^-1b makes it easy to solve the system AX=b where A^-1 is known.
However, the easiest way to solve a 2 x 2 linear system is to use
It is invertible if the only if ad – bc such that
The above equation can be called formula for A^-1 when n= 2
It can easily be seen or verified that if n > 2 and all that is required is the solution of a single n x n system AX = b, then finding A^-1 and then multiplying b is generally not the most efficient
way to obtain the solution. However, we shall consistently denote the solution of AX = b as A^-1b even if is not actually obtained as the product A^-1b.
• METHODS OF SOLVING LINEAR SYSTEM OF EQUATION.
We are interest at this point in discussing methods for solving linear systems and we shall restrict ourselves to system of the form AX = b where a unique solution exist.
In trying to find solutions to system for linear equation of the type AX = b, many methods can be adopted. In reality, there are many methods of solution but choice of methods varies in computation
and also depends on efficiency and accuracy.
A large part of numerical analysis is concerned with why and how errors exist in mathematical problems. So there is the search for methods which minimizes the totality of these errors, such methods
1) Cramer’s Rule
2) Triangular Decomposition method
3) Iterative method
4) Direct method
5) Repeated direct method
We consider the linear system (3) Supposed that A is non-singular, the equation (3) can be re-written as X = A^-1b
If det A 0, then the unique solution of AX = b is
And Aj is the matrix obtained by replacing the jth column of A by b.
Finding X by Cramer’s rule requires evaluating the determinant of A and of n additional n x n matrices A[1], A[2], …, A[n]. The arrow rules makes crammer’s rule convenient when n = 2 and reasonably
easy to use when n = 3. However, for n the efficient evaluation of det A alone is det A =
(-1)^p (product of the pivots of L/U) where P is the number of row interchanges used to get L/U is even or odd.
Be a linear system of n equations in n unknown and let A = [aij] be the coefficient matrix so that we can write the given system as AX = b where
If , the system has a unique solution
Where A[1] is the matrix obtained from A by replacing the ith column of A by b. If /A/0, then A is non-singular, hence
This means that
Now let
If we evaluate /Ai/ by expanding about the ith column, we find that /A[i]/ = A[1i] b[1] + a[2i]b[2 ]+ … + A[ni]b[n]
For i = 1, 2, …, n. In the expression for X[i], the determinant /A[i]/ of A[1] can be calculated by another method.
The method of triangular decomposition is set out to decompose a square matrix into a product of lower triangular matrix and the upper triangular matrix where A =
But for this project, we shall be discussing the method used by cholesky. There are two ways:-That which have unit lower triangular matrix and that which has unit upper triangular matrix .
In the case of a 4 x 4 matrix we have
Multiplying the rows of L by the first column of U we get L[11]=a[11], L[21] =a[21], L[31]= a[31], L[41] = a[41] the first column of L is the same as the first of a. we now multiply the first row of
L by the columns of U.
Thus the first row of U is determined. In this method, we alternate between getting a column of L and a row of U, so we get column of L equation for the second column of L by multiplying the rows of
Which gives
Proceeding in the same fashion, the equation we need are
[ ]The general formula for getting elements of L and U corresponding to the coefficient matrix for n simultaneously can be written as
For j = 1, the rule for L reduces to L[ij] = a[ij]
For i =1, the rule for U reduces to
In iterative method, we start from an approximation to the true solution and if successful, obtain better approximations from a computational cycle repeated as often as may be necessary for achieving
a required accuracy so that the amount of arithmetic depends upon the accuracy required. Iterative methods are used mainly in those problems for which convergence is known to be rapid and for systems
of large order but with many zero coefficients.
Generally, these methods have more modest storage requirements than direct methods and may also be faster depending on the iterative method and the problem. They usually also have better
vectorization and parallelization properties.
The jacobi and Gauss-siedel methods are examples of the iterative methods.
The jacobi iteration is also known as the method of simultaneous displacements.
By solving the first equation X[1] in terms of the remaining unknowns, solving the second equation for X[2] in terms of the remaining unknowns, solving the third equation for X[3] in terms of the
remaining unknowns e.t.c.
…………… (2)
The system 20X[1] + X[2] – X[3] = 17
X[1] – 10X[2] + X[3] = 13
-X[1] + X[2] + 10X[3] = 18
Would be re-written as
………………… (3)
……………………… (4)
To solve the system by Jacobi iteration, make an initial approximation to the solution. When no better choice is available use X[1] (0), i=1(1)n substitute the initial approximation into the right
hand side of equation (2) above and use the values of X[1], X[2] …, X[n] that result on the left hand side as a new approximation to the new solution. Thus, as new values are generated they are not
immediately used for rather are retained for the next iteration.
To solve system (3) by Jacobi method we would substitute the initial approximation X[1]=0, X[2]= 0, X[3]=0 into right hand side of system (4) and calculate the new approximation.
To improve the approximation we would repeat the subsequent process.
History had it that this iteration was used by Gauss in his calculations and was independently discovered by siedel. The striking factor about this method is that whenever each component of X^(r+1)
is computed, for this reason Gauss-siedel method is sometimes called the method of successive displacements recalls the system of linear equations
Gauss-siedel decided to write this equation
Provided the diagonal elements
Direct methods are those which in the absence of round-off or other errors, will yield the exact solution in a finite number of elementary arithmetic operations. In practise, because a computer works
with a finite word length, direct methods do not lead to exact solutions the methods used for exact solutions are Gaussians elimination method and cholesky decomposition method. We shall discuss
these methods in chapter two and three of this project work respectively.
If the system of equation is ill-conditioned, that is if the determinant of matrix coefficients is infinitesimal in comparison with the size of the element of the matrix, we use the repeated direct
method, that is we interprete the equation of the system as hyper-planes in an n-dimensional co-ordinate system, then the elements (a[11], a[12], …, a[1n]) of row represent the direction numbers of
the “normal” to the ith hyper-plan. In this context, we can interprete an ill-conditional system of linear equation as a set n hyper-planes whose “normal” are very nearly “parallel”. If the systems
of equation are thus ill-conditional, then the solution obtained from direct method may be a poor representation of the true solution because of the accumulating effect of round-off error. This
round-off error can be limited to some degree by using pivotal condensation. | {"url":"https://www.projectplus.com.ng/2017/04/10/linear-system-equations/","timestamp":"2024-11-03T22:25:16Z","content_type":"text/html","content_length":"124978","record_id":"<urn:uuid:2d03de35-bd82-4ca3-8126-17d9576c9ca5>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00815.warc.gz"} |
Herwig Hauser Classic
Herwig Hauser’s classic algebraic surfaces are compiled for the original IMAGINARY exhibition. Herwig Hauser’s forms and formulas are chosen in such a way that equations are simple and beautiful. The
figures are plain and natural and show interesting geometrical facts. Herwig Hauser is professor of mathematics at the University of Vienna and works in algebraic geometry and singularity theory.
Zitrus (Citric)
The equation x2+z2 = y3(1−y)3 of Citric appears as simple as the figure itself. Two cusps mirror-symmetrically arranged rotate around the traversing axis. The equation x2+z2 = y3 simplified by
omitting (1−y)3 provides for exactly one cusp, and x2+z2 = (1−y)3 yields the mirror image. Both are infinitely extending surfaces. The product on the right side of the initial equation ensures that
Citric remains bounded. You may consider the following: If the absolute value of y is getting larger than 1 the right side becomes negative and the equation does not admit real solutions of x and z.
This surface with the equation x2 = y3z3 is difficult to visualize in an appealing manner. Both the lighting and the shaping near the origin produce problems, since the image shall present the figure
as authentically as possible. This depends on the geometry and above all on the viewing angle of the camera. Artifacts occur, now and again, such as frayings or mottled colourings, which do not
correspond to mathematical reality. Also, reflections can have a disturbing effect. Alongside the edges, mist is coming up near the intercept point, despite the quality of the POVRay program. The
problem rather lies in the complexity to solve equations near singular points. The so-called resolution of singularities, which provides a parametrization of the surface, can help in many cases.
• x^2-x^3+y^2+y^4+z^3-z^4=0
Vis á Vis
Vis à Vis means opposite – and, here, two essential phenomena of algebraic geometry stand opposite each other. The singular tip on the left looks at a curved but smooth hill on the right. This
singularity is more exciting, because various changes to the equation can result in unpredictable changes to the figure, which does not happen at smooth points. By using the SURFER program, such
surfaces can be generated and modified quite easily and intuitively. The comparison of form and formula, i.e. of equation and corresponding surface, becomes an interactive experience which is
intriguing to understand.
The surface Calypso with equation x2+y2z = z2 contains three straight lines. The horizontal straight line is clearly visible, it passes through the origin (zero) where the upper and lower part meet.
The two other straight lines lie in a vertical plane, they also pass through 0 and intersect each other at that point. The section of the surface with this plane shows the two straight lines.
If you shift this plane a bit forward the section curve turns into a hyperbola. This can be easily checked by calculation. You set either y=0 or y=1. In the first case the result is x2 = z2 or (x−z)
(x + z) = 0, the equation of two straight lines in the plane. In the second case you get x2+z = z2. This can be rewritten in −x2+(z−1/2)2 = 1/4, the equation of a hyperbola with centre at (0, 1/2).
The surface Calyx with equation x2+y2z3 = z4 has a straight line as its singular locus. The lower part of the surface has cusp-shaped singularities alongside the straight line, whereas the upper part
tangentially touches the straight line at a point, the origin. The real image is misleading insofar as the defining polynomial is irreducible and the surface, as a result, consists just of one
algebraic component (not of two components as the figure suggests).
It can be shown that Calyx is an appropriate projection of Calypso. In a three-dimensional space a cylinder surface is contracted to the singular straight line of Calyx. Algebraically, the mapping is
defined by the requirement (x, y, z) → (xz, y, z). It is very simple. The respective substitution in x2+y2z3 = z4 and subsequent reducing of z2 yields the Calypso equation x2+y2z = z2.
• (x^2-y^3)^2-(z^2-y^2)^3=0
The equation (x2−y3)2 = (z2−y2)3 of Daisy implies by differentiation that the singular locus consists of two (plane) curves which transversally meet at their common singular point. In order to better
understand singularities the geometrician constructs their resolution by means of blowups. In finitely many steps they provide a surface without singularities (a manifold) together with a projection
map onto the original surface which interprets it as a shade of manifold.
• (x+b\,y)^2-(y^2+z^2)^2-a=0
The Diabolo equation x2 = (y2+z2)2 factorizes into the product (x−y2−z2)(x+y2+z2) = 0. Hence, the respective surface is the unification of the two single-shell rotating hyperboloids x = ±(y2+z2).
They touch each other tangentially at the origin. The contact can algebraically be gathered from the concurrent linear term x in the two factors. The tangent plane is the vertical plane x = 0.
The stripes in the images are shades due to lighting. If you modify the Diabolo equation by adding a constant term such as in x2 = (y2+z2)2+1/1000, then the two halves are separated. The substitution
of x by x+y, however, yields the variation (x+y)2 = (y2+z2)2 of the equation. The two shells are moved at an angle to each other.
Ding Dong
This surface described by the equation x²+y²+z³ = z² was one of the very first visualizations we tried. Equation and shape are simple: A vertical alpha-loop rotates around the z-axis. But there was
the problem with the colouring. Green is generally rather tricky in three-dimensional visualization of surfaces and, in addition, tends to be matt or yellowish. The lights and reflexions must be well
tested. Note the light blue hard shadow intensifying the spatial effect.
• x^2+y^2+z^2+1500\cdot(x^2+y^2)\cdot(x^2+z^2)\cdot(y^2+z^2)-1=0
Distel (Thistle)
The surface Thistle with the equation x2+y2+z2+c(x2+y2)(x2+z2)(y2+z2) = 1 excels through its extraordinary symmetry. The real image was provided with a very big coefficient c. The six spikes are
located on the three coordinate axes of the Euclidian space. Each twist permuting the three axes leaves Thistle unchanged. The symmetry group, as a result, is that of the cube and the octahedron
which is dual to it, two of the five Platonic solids.
Surprisingly, it is not possible to construct totally regular stars such as Thistle having any number of spikes. This results from considerations of group theory. There can be only four, six, eight,
twelve or twenty spikes according to the sides of the Platonic solids. We leave it to the curious viewer to find the appropriate equations for all these stars.
• (x^2+y^2+z^2)^2-(x^2+y^2)=0
If the audience in an oval stadium scream about a score (typically of the favoured team), the sound spreads like a quickly inflated floating tyre. After some split seconds the tyre meets itself at
its centre – the opening has closed – and that is what exactly happens at the kick-off spot. At that point sound waves from all directions meet simultaneously and are boosted accordingly. This is why
referees are advised to always stay level with the ball. Thus, when a goal is scored they do not stand in the middle circle and get a buzzing in their ears.
• (x^2+y^2)^3-x^2y^2\cdot(z^2+1)=0
Eistüte (Cone)
The horizontal section through Cone is a so-called rosette curve: A small wheel rolls round the interior of an annular body while a pencil fixed to the wheel is drawing a curve (like a Spirograph).
Different curves are formed depending on the ratio of the two radii. The curve closes if the ratio is a rational number.
In our case it is the four- leaf clover. Our cone has the advantage that four scoops of ice-cream will fit in!
The Lemniscate is the plane curve with the equation y4+z2 = y2. It results from the circle y2+z2 =1 by substituting z by z/y. Geometrically, the substitution corresponds to a distortion of the circle
to a figure of eight loop. If y4+z2 = y2 is conceived as equation with three variables x, y and z (where x is hidden) then the solution set is a surface in the three-dimensional space, i.e. the
cylinder above the lemniscate.
The Helix equation y4+x2z2 = y2 is yielded by substituting z by xz. From the geometrical view, this construction is a kind of folding. The symmetry with respect to x and z is clearly visible. For the
final formula we added the factors 2 and 6 to slightly stretch Helix. The singular locus is a crossing of straight lines. The sections of Helix with the planes x = c or z = c are lemniscates for c
≠ 0, whereas the sections y = c are hyperbola pairs.
Herz (Heart)
Despite the simple equation y2+z3 = z4+x2z2 the surface Heart possesses a subtle local and global structure. The singular set is a straight line on which the surface intersects itself. The origin 0
is the interesting point as we intersect Heart with planes x = c orthogonal to the singular straight line. The result is a loop which contracts like a knot if c tends to 0. A funnel is formed.
Viewed from the distance we see a circle shaped opening in the surface. The section with the vertical xy-plane is in fact a circle. The simplicity of the surface allows us to close our eyes, yet
still recall the figure in detail. However, it is much more difficult to describe verbally the geometric pattern because there are no common words to do so.
Himmel & Hölle (Heaven and Hell)
A piece of paper is folded and is held from beneath such that you can put your four fingers in the four corners so formed. By spreading your fingers the figure opens in two different ways so that two
of the four inner sides can be seen at a time.
This figure reminds us of the popular children’s fortune teller game where you have to predict which colour will show up – blue for heaven and red for hell (hence the name!)
By adding up the squares at y and z you get the highest exponent 4. This is called an equation of the 4th degree. The higher the degree the more complicated it is to calculate the surface.
Kolibri (Hummingbird)
You may notice that Hummingbird is a close-up, upside down view of Heart near the origin. The equation is x3+x2z2 = y2, compared with the Heart formula y2+z3 = z4+x2z2. It is possible to change the
coordinates so that one transforms into the other.
The Hummingbird is one of the smallest birds and is equipped with impressive abilities. Its wings are able to beat up to 200 times in a second so that it can hover in the air. This activity needs a
lot of energy, and is why the hummingbird must eat twice its body weight in food per day. Without constant food supply it would starve within a few hours. It lowers its body temperature substantially
at night so as to save energy.
• (60-x^2-y^2-z^2)^3-60 (x^2+y^2)z^4=0
Kreisel (Spinning Top)
To illustrate the shape of a spinning top as a solution set of a simple algebraic equation needs some inventive mathematics! It is, indeed, not clear mathematically, at first sight, what this
equation must look like because different equations can provide similar looking forms.
The image shows the equation 60(x2+y2)z4 = (60−x2−y2−z2)3. The rotation symmetry round the vertical z axis is identified by x and y occurring in the quadratic polynomial x2+y2. The third power on the
right hand side is necessary to produce the two points. And as z occurs only in even powers, the surface is symmetrical with respect to the reflection in the horizontal xy plane. The parameter 60 in
the equation is used for aesthetic reasons.
• x^2yz+x^2z^2+2y^3z+3y^3=0
This surface was developed by chance on a dull train journey (working with algebraic visualisation makes time pass more quickly on boring train rides!).
Deriving the algebraic equation for this surface would be very difficult. The challenge of Miau is the double opening with embedded singularity.
For mathematicians this is a treasure trove exploring the relationship between equation and form.
• (xy-z^3-1)^2-(1-x^2-y^2)^3=0
Let us look at the defining equation (xy−z3−1)2 = (1−x2−y2)3 of Nepali. The symmetry between x and y is enforced by the quadratic polynomial x2+y2, which is rotation symmetrical in contrast to the
monomial xy. Sections with horizontal planes z = c yield closed curves being almost circles. The simultaneous occurrence of squares and third powers produces the tapering at the top.
The lateral boundary curve of Nepali is not an exact circle but is arching up and down like the brim of a hat. Its projection to the horizontal xy plane, however, is a circle as can be seen from the
top view. The surface shown is bounded; hence there was no need to limit the view by sphere intersection. This fact can be directly derived from the formula by accurate analysis.
Seepferdchen (Seahorse)
If you want to find the equation of this surface it would take strong efforts. The soft tangential contact is not easy to achieve. It vanishes as you only slightly change the formula.
The elegance of the sea horse is an illusion: If you look at it from behind or from the side, it appears quite clumsy. Sea horses live worldwide in tropical and temperate climate zones. Its Latin
name is Hippocampus, you can find the equation next to it.
• x^2yz+xy^2+y^3+y^3z-x^2z^2=0
The Solitude equation x2yz+xy2+y3+y3z = x2z2 does not reveal its hidden geometrical diversity. Similarly, the image shows only part of the phenomena. What is informative, however, is the camera drive
round the surface as it can be experienced in the film “ZEROSET – I spy with my little eye”.
There are obviously two openings, a larger well visible one and a smaller one, which you would not suspect from the first perspective. The view from above shows the vertical singular straight line
alongside which horizontal sections force a sharp curve. The example of Solitude shows the complexity of the problem to deduce the visible real geometry from the equation. Similarly, one can question
the underlying complex or number-theoretical geometry.
Tanz (Dance)
By setting z = 0 the Dance equation 2x4 = x2 +y2z2 yields the equation 2x4 = x2. This is the section of Dance with the horizontal xy plane. If you rewrite the equation you will get x2(√2x+1) (√2x−1)
= 0 of three parallel straight lines. We, here, observe a typically real phenomenon: The straight line x = z = 0 belongs to the solution set of the Dance equation 2x4 = x2+y2z2, but is an isolated
one-dimensional component, because if x is near 0 there are only the solutions x = y = 0 or x = z = 0, that means the cross consisting of the y and z axis.
As straight lines are infinitely thin they are missed by the visualization program or are not indicated. Therefore, the existence of one-dimensional components in the solution set must be cleared up
by computation beforehand and then they must be added to the image as thin cylinders, if necessary.
Taube (Dove)
Dove possesses the amazing formula
256z3 − 128x2z2+16x4z+144xy2z−4x3y2−27y4 = 0.
The coefficients are not incidental. On the contrary: The equation results from another more general formula, the so-called discriminant. It describes the shade of a surface or variety which emerges
along with the projection on a surface or a linear space of higher dimension. The contour line is clearly defined by the surface and the projection and, from the algebraic point of view, also the
form of the equation.
Tülle (Nozzle)
The Nozzle surface is constituted by three smooth components, which intersect each other pairwise in a (likewise) smooth, plane curve. That way, three section curves are obtained, i.e. one straight
line and two parabolas. Note that these curves touch each other tangentially at the origin. We here have the simplest example of a surface with three pairwise transversal components so that the
section curves of every two of them do not intersect transversally. The surface, thus, is no Mikado Variety.
The transversal intersection of two smooth components of a surface – a fundamental concept of geometry - can, by way of the so-called ideal theory, be put into formulas in an algebraically precise
manner. It can perfectly be used for calculation and proofs. In the case of singular components a correct definition of transversality is still to await.
Zeck (Tick)
The simple Tick equation x2+y2 = z3(1−z) fully dictates the geometry just as with the other surfaces; that means, both the singular points and the outer shape, the curvature and the extension are
clearly defined by the four monomials x2, y2, −z3 und z4. As a result, the formula is a very efficient way to codify forms which appear complicated. However, the geometric information cannot always
be read from the formula. The local shape of the surface near a given point can be explicitly defined, in most cases; the techniques of local analytic geometry have a good effect. Defining the global
structure requires much more efforts and cannot always be satisfactorily accomplished. | {"url":"https://www.imaginary.org/gallery/herwig-hauser-classic","timestamp":"2024-11-11T06:09:50Z","content_type":"text/html","content_length":"130505","record_id":"<urn:uuid:6652a354-45d6-4b7e-950f-e22199b20f86>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00106.warc.gz"} |
What Is Supervised Learning? :A Brief Guide - Engineer's Planet
Supervised learning is a fundamental paradigm in machine learning where algorithms are trained on labeled data to make predictions or decisions. This approach is guided by the explicit supervision of
a labeled dataset, allowing the algorithm to learn the mapping between input features and the corresponding output. Understanding the core concepts of supervised learning is essential for building
and deploying effective machine learning models.
1.1 Definition of Supervised Learning
Supervised learning involves a predictive modeling task where the algorithm is provided with a labeled training dataset, consisting of input-output pairs. The goal is to learn a mapping function that
can accurately predict the output (target variable) for new, unseen inputs.
The term “supervised” refers to the training process where the algorithm is guided by the known outcomes in the training data. In this context, the algorithm learns from the labeled examples,
adjusting its parameters to minimize the difference between its predictions and the actual outcomes.
This iterative process continues until the model achieves a satisfactory level of accuracy. Consequently, supervised learning is particularly useful in scenarios where the goal is to predict or
classify outcomes based on historical data.
Supervised learning plays a crucial role in machine learning by enabling algorithms to make informed decisions based on historical data. Moreover, it is widely used in various applications such as
image and speech recognition, natural language processing, and predictive analytics.
1.3 Historical Evolution
The roots of supervised learning can be traced back to the mid-20th century. Early developments, such as the perceptron by Frank Rosenblatt in 1957, laid the foundation for the concept of learning
from labeled data. The field evolved over the years with advancements in algorithms, computing power, and the availability of large datasets, leading to breakthroughs in deep learning and neural
Fundamental Concepts
2.1 Target Variable and Features
In supervised learning, the target variable is the outcome or prediction that the model aims to generate. Features are the input variables or attributes that influence the target variable. For
instance, in a housing price prediction model, the target variable may be the price, and features could include factors like square footage, number of bedrooms, and location.
2.2 Training and Testing Data
The labeled dataset is typically divided into two subsets: the training set and the testing set. The model is trained on the training set to learn patterns and relationships. The testing set is then
used to evaluate the model’s performance on unseen data, providing an indication of its ability to generalize.
Table 1: Comparison of Training and Testing Data
Dataset Purpose Characteristics
Training Set Model Training Used to train the algorithm; labeled data with known outcomes.
Testing Set Model Evaluation Unseen data used to assess the model’s generalization performance.
2.3 Labels and Predictors
In supervised learning, the labeled data consists of pairs of inputs (predictors) and corresponding outputs (labels). The algorithm learns to map predictors to labels during the training process.
Once trained, the model can make predictions on new, unseen data based on the learned patterns.
2.4 Types of Supervised Learning
There are two main types of supervised learning: classification and regression. In classification, the goal is to predict a categorical outcome, such as whether an email is spam or not. Regression,
on the other hand, deals with predicting a continuous numerical outcome, such as predicting house prices based on features.
1. Classification: This type involves predicting a categorical outcome. For example, it can be used to determine whether an email is spam or not. The algorithm learns to classify input data into
distinct categories.
2. Regression: In regression, the goal is to predict a continuous numerical outcome. An example is predicting house prices based on features like square footage, number of bedrooms, etc.
Supervised Learning Algorithms
Supervised learning is a category of machine learning where algorithms are trained on labeled datasets, learning the mapping between input features and corresponding output labels. Here, we delve
into some commonly used supervised learning algorithms and explore how they function.
3.1 Linear Regression
Linear regression is a foundational algorithm used for predicting a continuous output variable based on one or more input features. The relationship between the inputs and output is assumed to be
linear. The formula for a simple linear regression with one feature is:
Table 1: Linear Regression Example
Input (x) Output (y)
Code 1: Linear Regression in Python using scikit-learn
from sklearn.linear_model import LinearRegression
import numpy as np
# Sample data
X = np.array([1, 2, 3, 4]).reshape(-1, 1)
y = np.array([3, 5, 7, 9])
# Create and fit the model
model = LinearRegression()
model.fit(X, y)
# Make predictions
predictions = model.predict([[5]])
print("Prediction for input 5:", predictions[0])
3.2 Decision Trees
Decision trees are versatile, They make decisions by recursively splitting the dataset based on features, creating a tree-like structure. Each leaf node represents a predicted outcome.
3.3 Support Vector Machines (SVM)
Support Vector Machines are powerful algorithms used for classification and regression. SVM seeks to find the hyperplane that best separates different classes or fits the regression data. It is
particularly effective in high-dimensional spaces.
Code 2: Support Vector Machines in Python using scikit-learn
from sklearn import svm
# Sample data
X = [[0, 0], [1, 1]]
y = [0, 1]
# Create and fit the model
model = svm.SVC()
model.fit(X, y)
# Make predictions
predictions = model.predict([[2, 2]])
print("Prediction for input [2, 2]:", predictions[0])
3.4 Neural Networks
Neural networks, inspired by the human brain, consist of layers of interconnected nodes. They excel at learning complex patterns and relationships in data, making them suitable for a wide range of
tasks, from image recognition to natural language processing.
How Supervised Learning Works
4.1 Training Phase
In the training phase, the algorithm learns from the labeled dataset, adjusting its parameters to minimize the difference between predicted and actual outputs.
4.2 Testing and Validation
After training, the model is tested on new, unseen data to assess its generalization performance. Validation sets help fine-tune hyperparameters to improve the model’s accuracy.
4.3 Evaluation Metrics
Evaluation metrics, such as accuracy, precision, recall, and F1 score, quantify the model’s performance on the test set.
Table 2: Example Evaluation Metrics
Metric Value
Accuracy 0.85
Precision 0.78
Recall 0.92
F1 Score 0.84
4.4 Overfitting and Underfitting
Overfitting occurs when a model learns the training data too well but performs poorly on new data. Underfitting happens when the model is too simple to capture the underlying patterns.
4.5 Model Interpretability
The interpretability of a model is crucial for understanding its decision-making process. Linear models are often more interpretable than complex models like neural networks.
In conclusion, supervised learning algorithms play a central role in machine learning, ranging from linear regression for simple relationships to neural networks for complex patterns. Understanding
how these algorithms work and the considerations in their application is essential for building effective and interpretable models.
Advantages and Limitations of Supervised Learning
5.1 Advantages of Supervised Learning
Supervised learning, a cornerstone of machine learning, offers numerous advantages that contribute to its widespread adoption.
Table 1: Advantages of Supervised Learning
Advantages Description
Clear Objective Supervised learning has a well-defined objective: to predict or classify based on labeled data.
Well-established Framework With labeled training data, supervised learning follows a structured framework for model training.
Generalization Capability Trained models generalize well to new, unseen data, making them applicable in various scenarios.
Versatility Suited for both regression and classification tasks, addressing a wide range of real-world problems.
Code 1: Example of Supervised Learning in Python using Scikit-learn
# Example: Regression task
X_train, X_test, y_train, y_test = train_test_split(features, target, test_size=0.2, random_state=42)
model = LinearRegression()
model.fit(X_train, y_train)
predictions = model.predict(X_test)
mse = mean_squared_error(y_test, predictions)
print("Mean Squared Error:", mse)
In this example, we use the scikit-learn library to perform a simple linear regression task. The model is trained on labeled data (X_train, y_train) and evaluated on a separate test set.
5.2 Limitations and Challenges
While supervised learning is powerful, it is not without limitations and challenges.
Table 2: Limitations and Challenges of Supervised Learning
Limitations and Challenges Description
Need for Labeled Data Supervised learning requires labeled training data, which can be time-consuming and costly to obtain.
Limited to Available Labels The model’s performance is constrained by the quality and diversity of the labeled data.
Lack of Explanation Some complex models, like deep neural networks, may lack interpretability, making it challenging to understand their decision-making process.
5.3 Mitigating Overfitting and Bias
To enhance the effectiveness of supervised learning, addressing overfitting and bias is crucial.
• Hyperparameter Tuning: Adjusting hyperparameters like learning rate or regularization helps find a balance between underfitting and overfitting.
• Cross-Validation: Implementing cross-validation techniques, such as k-Fold Cross-Validation, aids in evaluating model performance across different subsets of data.
• Data Augmentation: Increasing the diversity of labeled data through techniques like data augmentation mitigates bias and improves model generalization.
Comparison with Unsupervised Learning
6.1 Key Differences
While supervised learning relies on labeled data for training, unsupervised learning operates on unlabeled data, emphasizing patterns and relationships without predefined targets.
Table 3: Key Differences between Supervised and Unsupervised Learning
Key Differences Description
Labeled vs. Unlabeled Data Supervised learning requires labeled data, while unsupervised learning works with unlabeled data.
Objective Supervised learning predicts or classifies, while unsupervised learning identifies patterns and structures without predefined goals.
Common Algorithms Supervised learning includes algorithms like linear regression and decision trees, whereas unsupervised learning employs clustering and dimensionality reduction algorithms.
6.2 Use Cases and Applications
Both supervised and unsupervised learning find applications in diverse fields.
Table 4: Use Cases and Applications
Use Cases Applications
Supervised Learning Image classification, spam detection, speech recognition.
Unsupervised Learning Clustering customer segments, anomaly detection, topic modeling.
Real-World Applications
1.Health Care
In the realm of healthcare, supervised learning proves invaluable for predicting diseases. Imagine a scenario where doctors collect data on patients with labeled indicators, such as symptoms and test
results. With this information, a machine learning model is trained to predict the likelihood of various diseases. Once deployed, the model can assist healthcare professionals in early diagnosis,
optimizing patient care.
2. Financial Sector
In the financial sector, supervised learning takes center stage in credit scoring. Consider a bank gathering labeled data on individuals’ credit histories and whether they default on loans. A
supervised learning model can be trained to assess new loan applications, predicting the risk of default based on historical patterns. This enhances the efficiency of credit evaluation processes,
aiding in responsible lending.
3.Social Media
Natural Language Processing (NLP) leverages supervised learning for sentiment analysis. Imagine social media platforms collecting labeled data on user sentiments expressed in posts or comments.
A supervised learning model can then learn to classify text as positive, negative, or neutral. Deployed in real-time, this model can help companies gauge public opinion and respond to customer
feedback effectively.
Challenges and Future Directions
8.1 Data Quality and Bias
As machine learning and artificial intelligence (AI) systems become increasingly integral to decision-making processes, the quality and biases within training data pose significant challenges. Data
quality issues, such as missing or noisy data, can impact the performance and reliability of models. Additionally, biases present in the data can lead to unfair or discriminatory outcomes. Addressing
these challenges requires vigilant data preprocessing, bias detection, and mitigation strategies to ensure the ethical deployment of AI systems.
8.2 Interpretable AI
Interpretable AI remains a critical challenge as complex models, such as deep neural networks, often act as “black boxes,” making it challenging to understand their decision-making processes.
Achieving interpretability is essential for building trust in AI systems, especially in fields like healthcare and finance where transparent decision-making is crucial. Future directions involve
developing model-agnostic interpretability techniques and integrating them seamlessly into the machine learning pipeline.
8.3 Integration with Emerging Technologies
The future of machine learning involves integrating with emerging technologies, such as edge computing, blockchain, and quantum computing. These integrations bring new challenges, including
optimizing models for resource-constrained environments, ensuring security and privacy in decentralized systems, and adapting algorithms for the unique capabilities of quantum computing. Addressing
these challenges will be crucial for harnessing the full potential of machine learning in a rapidly evolving technological landscape.
Supervised Learning in Python: A Practical Guide
9.1 Setting up the Environment
To implement supervised learning in Python, start by setting up the environment. Use popular data science platforms like Jupyter Notebooks or Google Colab and install essential libraries such as
scikit-learn, pandas, and matplotlib.
9.2 Libraries for Supervised Learning
Python offers powerful libraries for implementing supervised learning algorithms. Scikit-learn is a comprehensive library that includes various algorithms for classification, regression, and more.
Pandas is useful for data manipulation, while matplotlib helps with data visualization.
Table 2: Essential Libraries for Supervised Learning
Library Purpose
scikit-learn Machine learning algorithms and tools
pandas Data manipulation and analysis
matplotlib Data visualization
10.1 Recap of Key Concepts
In this exploration of challenges and future directions in machine learning, we discussed the importance of addressing data quality and bias, achieving interpretability in AI systems, and integrating
with emerging technologies. Tables, flowcharts, and code snippets were used to illustrate key concepts, challenges, and potential solutions. | {"url":"https://engineersplanet.com/what-is-supervised-learning-a-brief-guide/","timestamp":"2024-11-07T18:38:33Z","content_type":"text/html","content_length":"610751","record_id":"<urn:uuid:9ebc963e-bc39-4233-b863-17a50193c496>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00334.warc.gz"} |
Incremental Packing Problems: Algorithms and Polyhedra
2022 Theses Doctoral
Incremental Packing Problems: Algorithms and Polyhedra
In this thesis, we propose and study discrete, multi-period extensions of classical packing problems, a fundamental class of models in combinatorial optimization. Those extensions fall under the
general name of incremental packing problems. In such models, we are given an added time component and different capacity constraints for each time. Over time, capacities are weakly increasing as
resources increase, allowing more items to be selected. Once an item is selected, it cannot be removed in future times. The goal is to maximize some (possibly also time-dependent) objective function
under such packing constraints.
In Chapter 2, we study the generalized incremental knapsack problem, a multi-period extension to the classical knapsack problem. We present a policy that reduces the generalized incremental knapsack
problem to sequentially solving multiple classical knapsack problems, for which many efficient algorithms are known. We call such an algorithm a single-time algorithm. We prove that this algorithm
gives a (0.17 - ⋲)-approximation for the generalized incremental knapsack problem. Moreover, we show that the algorithm is very efficient in practice. On randomly generated instances of the
generalized incremental knapsack problem, it returns near optimal solutions and runs much faster compared to Gurobi solving the problem using the standard integer programming formulation.
In Chapter 3, we present additional approximation algorithms for the generalized incremental knapsack problem. We first give a polynomial-time (½-⋲)-approximation, improving upon the approximation
ratio given in Chapter 2. This result is based on a new reformulation of the generalized incremental knapsack problem as a single-machine sequencing problem, which is addressed by blending dynamic
programming techniques and the classical Shmoys-Tardos algorithm for the generalized assignment problem. Using the same sequencing reformulation, combined with further enumeration-based
self-reinforcing ideas and new structural properties of nearly-optimal solutions, we give a quasi-polynomial time approximation scheme for the problem, thus ruling out the possibility that the
generalized incremental knapsack problem is APX-hard under widely-believed complexity assumptions.
In Chapter 4, we first turn our attention to the submodular monotone all-or-nothing incremental knapsack problem (IK-AoN), a special case of the submodular monotone function subject to a knapsack
constraint extended to a multi-period setting. We show that each instance of IK-AoN can be reduced to a linear version of the problem. In particular, using a known PTAS for the linear version from
literature as a subroutine, this implies that IK-AoN admits a PTAS. Next, we study special cases of the generalized incremental knapsack problem and provide improved approximation schemes for these
special cases.
In Chapter 5, we give a polynomial-time (¼-⋲)-approximation in expectation for the incremental generalized assignment problem, a multi-period extension of the generalized assignment problem. To
develop this result, similar to the reformulation from Chapter 3, we reformulate the incremental generalized assignment problem as a multi-machine sequencing problem. Following the reformulation, we
show that the (½-⋲)-approximation for the generalized incremental knapsack problem, combined with further randomized rounding techniques, can be leveraged to give a constant factor approximation in
expectation for the incremental generalized assignment problem.
In Chapter 6, we turn our attention to the incremental knapsack polytope. First, we extend one direction of Balas's characterization of 0/1-facets of the knapsack polytope to the incremental knapsack
polytope. Starting from extended cover inequalities valid for the knapsack polytope, we show how to strengthen them to define facets for the incremental knapsack polytope. In particular, we prove
that under the same conditions for which these inequalities define facets for the knapsack polytope, following our strengthening procedure, the resulting inequalities define facets for the
incremental knapsack polytope. Then, as there are up to exponentially many such inequalities, we give separation algorithms for this class of inequalities.
• Zhang_columbia_0054D_17542.pdf application/pdf 1.3 MB Download File
More About This Work
Academic Units
Thesis Advisors
Faenza, Yuri
Ph.D., Columbia University
Published Here
October 12, 2022 | {"url":"https://academiccommons.columbia.edu/doi/10.7916/52qd-2y13","timestamp":"2024-11-07T17:29:54Z","content_type":"text/html","content_length":"28235","record_id":"<urn:uuid:2b2ba016-c47d-4032-aa81-21ec4e85a9cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00880.warc.gz"} |
NEET Previous Year Question Paper, Solutions, Paper Analysis
NEET Previous Year Papers are known to be the best source for NEET aspirants to analyze their performance and preparation level. Cause Previous Year Question Papers provide a crystal clear
understanding of all the aspects of NEET Question Paper.
Being a NEET aspirant, it is very important to understand the importance of Previous Year NEET Question Papers. As they serve as one of the most important tool to crack the NEET Examination.
Question Paper Analysis, Answer Key and Video Solutions
Question Paper Analysis and Solutions by Resonance gives a student a detailed analysis of the NEET question paper in the form of Overall Analysis and Paper Analysis. Students can also download NEET
previous year question paper with answer key for future reference and revision. We also provide subject-wise Detailed Solutions and Video Solutions for Physics, Chemistry, Botany and Zoology, so that
a student can work on his/ her focus subject.
Another important benefit of going through the detailed analysis of NEET previous year papers is that it will help students to identify the questions which are often repeated and also help them find
the best way to solve the questions.
The Key to Success - ‘Practice’
Solving the Previous Year Question Papers helps a student in more ways than one. First, it helps a student to understand the various sections of the question paper so that he/she can manage his/her
time properly. Second, it allows a student to analyse from his/ her mistakes and figure out which areas need revision. And also allows them to solve their doubts at the earliest instance possible.
Students will get habituated to solve questions in a fixed time frame while solving the previous year papers.
While practising student can develop a thorough understanding of the pattern of the examination and the questions asked. NEET previous year papers also help in determining the weightage of marks
allotted to different topics thereby allowing aspirants to plan an effective preparation strategy centred around those topics.
Begin Now!
A Vital element of preparation for NEET Examination is to understand the solution of the Questions, which a student might not be able to solve. So, for the better understanding of the solution
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We hope students can grasp all the important details through these detailed analysis and solutions and become competent enough to tackle the real challenge. | {"url":"https://www.resonance.ac.in/answer-key-solutions/neet.aspx","timestamp":"2024-11-11T13:10:47Z","content_type":"application/xhtml+xml","content_length":"14617","record_id":"<urn:uuid:81b99d7b-d4dc-4d08-9699-7c0faf9f7a01>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00045.warc.gz"} |
If you flip a coin 3 times what is the probability of getting 3 heads? [Solved]
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
If you flip a coin 3 times what is the probability of getting 3 heads?
For answering this question we need to consider all the possibilities when you flip a coin 3 times.
Answer: If you flip a coin 3 times the probability of getting 3 heads is 0.125
When you flip a coin 3 times, then all the possibe 8 outcomes are HHH, THH, HTH, HHT, TTH, THT, HTT, TTT.
Possible outcomes are HHH, THH, HTH, HHT, TTH, THT, HTT, TTT.
The number of cases in which you get exactly 3 heads is just 1.
Thus, the probability of getting 3 heads = 1/8 = 0.125
Hence, if you flip a coin 3 times the probability of getting 1 heads is 0.125
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/questions/if-you-flip-a-coin-3-times-what-is-the-probability-of-getting-3-heads/","timestamp":"2024-11-02T21:54:14Z","content_type":"text/html","content_length":"187654","record_id":"<urn:uuid:f0514a83-cd34-4fb7-a76b-18dfbc515af5>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00530.warc.gz"} |
An asymptotically optimal algorithm for online stacking
Consider a storage area where arriving items are stored temporarily in bounded capacity stacks until their departure. We look into the problem of deciding where to put an arriving item with the
objective of minimizing the maximum number of stacks used over time. The decision has to be made as soon as an item arrives, and we assume that we only have information on the departure times for the
arriving item and the items currently at the storage area. We are only allowed to put an item on top of another item if the item below departs at a later time. We refer to this problem as online
stacking. We assume that the storage time intervals are picked i.i.d. from [0 , 1] × [0 , 1] using an unknown distribution with a bounded probability density function. Under this mild condition, we
present a simple polynomial time online algorithm and show that the competitive ratio converges to 1 in probability. The result holds if the stack capacity is o(n), where n is the number of items,
including the realistic case where the capacity is a constant. Our experiments show that our results also have practical relevance.
Dyk ned i forskningsemnerne om 'An asymptotically optimal algorithm for online stacking'. Sammen danner de et unikt fingeraftryk. | {"url":"https://pure.au.dk/portal/da/publications/an-asymptotically-optimal-algorithm-for-online-stacking","timestamp":"2024-11-07T03:43:49Z","content_type":"text/html","content_length":"59800","record_id":"<urn:uuid:e6fcabca-f7d5-41c7-96dc-21b7c3064d73>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00742.warc.gz"} |
Thin-plate smoothing spline
st = tpaps(x,y) is the stform of a thin-plate smoothing spline f for the given data sites x(:,j) and the given data values y(:,j). The x(:,j) must be distinct points in the plane, the values can be
scalars, vectors, matrices, even ND-arrays, and there must be exactly as many values as there are sites.
The thin-plate smoothing spline f is the unique minimizer of the weighted sum
with E(f) the error measure
$E\left(f\right)=\sum _{j}{|y\left(:,j\right)-f\left(x\left(:,j\right)\right)|}^{2}$
and R(f) the roughness measure
$R\left(f\right)={\int \left(|{D}_{1}{D}_{1}f|}^{2}+2{|{D}_{1}{D}_{2}f|}^{2}+{|{D}_{2}{D}_{2}f|}^{2}\right)$
Here, the integral is taken over all of R^2, |z|^2 denotes the sum of squares of all the entries of z, and D[i]f denotes the partial derivative of f with respect to its i-th argument, hence the
integrand involves second partial derivatives of f. The function chooses the smoothing parameter p so that (1-p)/p equals the average of the diagonal entries of the matrix A, with A + (1-p)/p*eye(n)
the coefficient matrix of the linear system for the n coefficients of the smoothing spline to be determined. This ensures staying in between the two extremes of interpolation (when p is close to 1
and the coefficient matrix is essentially A) and complete smoothing (when p is close to 0 and the coefficient matrix is essentially a multiple of the identity matrix). This serves as a good first
guess for p.
st = tpaps(x,y,p) also inputs the smoothing parameter, p, a number between 0 and 1. As the smoothing parameter varies from 0 to 1, the smoothing spline varies, from the least-squares approximation to
the data by a linear polynomial when p is 0, to the thin-plate spline interpolant to the data when p is 1.
[...,P] = tpaps(...) also returns the value of the smoothing parameter used in the final spline result whether or not you specify p. This syntax is useful for experimentation in which you can start
with [pp,P] = tpaps(x,y) and obtain a reasonable first guess for p.
Recover the Underlying Exact Smooth Values from Noised Data
The following code obtains values of a smooth function at 31 randomly chosen sites, adds some random noise to these values, and then uses tpaps to recover the underlying exact smooth values. To
illustrate how well tpaps does in this case, the code plots, in addition to the smoothing spline, the exact values (as black balls) as well as each arrow leading from a smoothed value to the
corresponding noisy value.
rng(23); nxy = 31;
xy = 2*(rand(2,nxy)-.5); vals = sum(xy.^2);
noisyvals = vals + (rand(size(vals))-.5)/5;
st = tpaps(xy,noisyvals); fnplt(st), hold on
avals = fnval(st,xy);
quiver3(xy(1,:),xy(2,:),avals,zeros(1,nxy),zeros(1,nxy), ...
noisyvals-avals,'r'), hold off
Use an Interpolating Thin-Plate Spline to Construct a Map
The following code uses an interpolating thin-plate spline to vector-valued data values to construct a map, from the plane to the plane, that carries the unit square $\left\{\mathit{x}:|\mathit{x}\
left(\mathit{j}\right)|\le 1,\mathit{j}=\mathrm{1}:2\right\}$ approximately onto the unit disk $\left\{\mathit{x}:{\mathit{x}\left(1\right)}^{2}+{\mathit{x}\left(2\right)}^{2}\le 1\right\}$.
n = 64; t = linspace(0,2*pi,n+1); t(end) = [];
values = [cos(t); sin(t)];
centers = values./repmat(max(abs(values)),2,1);
st = tpaps(centers, values, 1);
fnplt(st), axis equal
Note the choice of 1 for the smoothing parameter here, to obtain interpolation.
Input Arguments
x — Data sites
vector | cell array
Data sites of data values y to be fit, specified as a vector or as a cell array for multivariate data. Spline f is created with knots at each data site x such that f(x(j)) = y(:,j) for all values of
For multivariate, gridded data, you can specify x as a cell array that specifies the data site in each variable dimension: f(x1(i),x2(j),...xn(k)) = y(:,i,j,...,k).
y — Data values to fit
vector | matrix | array
Data values to fit during creation of the spline, specified as a vector, matrix, or array. Data values y(:,j) can be scalars, matrices, or n-dimensional arrays. Data values given at the same data
site x are averaged.
Data Types: single | double
p — Smoothing parameter
scalar in the range [0,1] | vector | cell array | empty array
Smoothing parameter, specified as a scalar value between 0 and 1 or as a cell array of values for multivariate data. You can also specify values for the roughness measure weights λ by providing p as
a vector. To provide roughness measure weights for multivariate data, use a cell array of vectors. If you provide an empty array, the function chooses a default value for p based on the data sites x
and the default value of 1 for the roughness measure weight λ.
The smoothing parameter determines the relative weight to place on the contradictory demands of having f be smooth or having f be close to the data. For p = 0, f is the least-squares straight-line
fit to the data. For p = 1, f is the variational, or natural, cubic spline interpolant. As p moves from 0 to 1, the smoothing spline changes from one extreme to the other.
The favorable range for p is often near 1/(1 + h^3/6), where h is the average spacing of the data sites. The function chooses a default value for p within this range. For uniformly spaced data, you
can expect a close fit with p = 1(1 + h^3/60) and some satisfactory smoothing with p = 1/(1 + h^3/0.6). You can input p > 1, but this choice leads to a smoothing spline even rougher than the
variational cubic spline interpolant.
If the input p is negative or empty, then the function uses the default value for p.
You can specify the roughness measure weights λ alongside the smoothing parameter by providing p as a vector. This vector must be the same size as x, with the ith entry the value of λ on the interval
(x(i-1)...x(i)), for i = 2:length(x). The first entry of the input vector p is the desired value of the smoothness parameter p. By providing roughness measure weights, you can make the resulting
smoothing spline smoother (with larger weight values) or closer to the data (with smaller weight values) in different parts of the interval. Roughness measure weights must be nonnegative.
If you have difficulty choosing p but have some feeling for the size of the noise in y, consider using spaps(x,y,tol) instead. This function chooses p such that the roughness measure is as small as
possible, subject to the condition that the error measure does not exceed tol. In this case, the error measure usually equals the specified value for tol.
Data Types: single | double
Output Arguments
st — Spline structure
spline structure
Spline, returned as a structure with these fields.
Form — Form of spline
st-tp00 | st-tp10 | st-tp01 | st-tp
Form of the spline, returned as st-tp00, st-tp10, st-tp01, or st-tp.
Centers — Sequence of sites
matrix | array
Sequence of sites, returned as a matrix or as an array for multivariate data.
Coefs — Coefficients of polynomials
matrix | array
Coefficients of polynomials for each piece, returned as a matrix or as an array for multivariate data.
Ncenters — Number of centers
scalar | vector
Number of sequence of sites.
Number — Number of polynomial pieces
scalar | vector
Number of polynomial pieces describing the spline, returned as a scalar or as a vector of numbers of pieces in each variable for multivariate data.
Dim — Dimensionality
Dimensionality of the target function, returned as a scalar.
Interv — Basic interval
cell array
Basic interval for the stform that contains all the given centers, returned as an array.
P — Smoothing parameter
scalar | cell array
Smoothing parameter used to calculate the spline, returned as a scalar or as a cell array of scalar values for multivariate data. P is between 0 and 1.
The determination of the smoothing spline involves the solution of a linear system with as many unknowns as there are data points. Since the matrix of this linear system is full, the solving can take
a long time even if, as is the case here, an iterative scheme is used when there are more than 728 data points. The convergence speed of that iteration is strongly influenced by p, and is slower the
larger p is. So, for large problems, use interpolation, i.e., p equal to 1, only if you can afford the time.
Version History
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Adding and Subtracting Ten Mentally - The Math Spot
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For some first grade students, adding and subtracting ten to a given number comes naturally. They know that 23 has 2 tens and 3 ones so, naturally adding another ten gives the answer of 33! For other
students, this skill isn’t so intuitive. You may have taught the skill using a place value chart and, while 80% of the class picked it up using that representation, there are still a few students who
will tell you, with confidence, that 23 + 10 = 24.
A place value chart model for adding ten.
I have written at length about the C-R-A framework of teaching math. When I think about the skill and understanding required to add or subtract ten and I think about using a place value chart to
promote understanding I see that the reason why some kids are falling out behind the rest of the class is that a place value chart alone is a representative (verging on abstract) model. What about
the students who needed to take this skill back to the concrete level?
A strong method for teaching addition and subtraction of 10 is to start with the concrete, link it to a representation and then move to a more abstract model. Read on to see how each of these steps
play out in the classroom:
I have a few preferred concrete models for teaching this skill. First, I love using base ten blocks . They are an accessible tool in the classroom, students likely have already gained familiarity
with the tool and they adequately demonstrate the difference in magnitude between a one and a ten. When teaching this skill I ask students to build a number with ten sticks and ones. I then ask them
to tell me about what block they would need to have to show ten more. We may have some discussion about It is so important that students can articulate that when they added a ten the ones remained
the same but the tens changed. If they aren’t noticing this pattern, help them to notice and then test the pattern. Without this understanding, they will have a very hard time generalizing the rule!
how we could use either a ten stick or ten ones but ultimately, push them to use a ten stick as it is the more efficient way of counting. Students then add a ten stick to their original number and
write an equation to demonstrate what happened. For example: 23 + 10 = 33. I then ask students to tell me what happened to the tens and what happened to the ones using place value language.
You can also use $10 bills and $1 bills to model this relationship. If your students are very strong in coin names and values you can also use dimes an pennies to demonstrate this relationship. If
your students aren’t super solid in their understanding of coins, I would save that representation for another day. I am a huge proponent of using dimes and pennies to teach and reinforce place value
concepts, however, I am a bigger proponent of staying focused and I don’t want to muddy the waters in terms of the focus of this lesson.
As mentioned above, a strong model for representing the number of tens and ones before and after
ten is added or subtracted would be a place value chart. If your students are demonstrating that they are able to use concrete materials but that the place value chart is a bit of a reach for them
at this time, consider having your students draw a representation of ten sticks and ones. They can do this parallel to their use of base ten blocks and then drop the manipulatives once they are able
to perform the skill AND talk about the problem using base ten language using their drawing alone.
Just because you are teaching using the CRA framework does not mean that you are teaching each of these three skills in isolation. As I mentioned in the “Representative” paragraph, the best way into
the representative model is to ask students to do so parallel to the “Concrete” model so that students can make explicit connections. The abstract model of this work being, for example, a number
sentence, can be linked to both the representative and the abstract. I ask students on day 1 of this skill when they are building with blocks to write a number sentence that matches what they have
built with blocks. As students are drawing place value drawings I also ask them to write an equation that represents what they have done.
And always, always, always, promote math talk and math language when your students are working. Their ability to articulate their thinking and the math processes will help to solidify their
understanding and build connections to future learning!
If you are looking for a math resource that will better guide you and your students through the development of this skill (1.NBT.5) from concrete to mental math please take a look at my 5 Day Focus:
10 More and 10 Less. This resource includes a pre and post assessment, 5 days of detailed lesson plans, independent, hands on activities for each day and tickets out the door for each lesson. | {"url":"http://k5mathspot.com/adding-and-subtracting-ten-mentally/","timestamp":"2024-11-07T02:44:08Z","content_type":"text/html","content_length":"97702","record_id":"<urn:uuid:6c7c4345-76e0-46e8-a821-006872225970>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00883.warc.gz"} |
NeurIPS 2019
Sun Dec 8th through Sat the 14th, 2019 at Vancouver Convention Center
Reviewer 1
This work addresses the task of designing experiments on bipartite graphs in the presence of SUTVA violation. Specifically, the authors propose an extension to the graph cluster randomization
framework to the bipartite setting by casting the problem as correlation clustering. Propositions are provided that tie the variance of the causal estimate to the quality of the discovered solution.
A heuristic is provided to solve the correlation clustering problem. Experiments are provided comparing the proposed method to balanced partitioning and random assignment which show the proposed
method performing favorably. Overall, I think this is a nice, simple, solution to a problem that occurs a fair amount for practitioners. The paper is clearly written and motivated. The connection to
correlation clustering is sensible and provides a decent interpretation of the results. I have two is that there are details missing from the experiment section that would improve reproducibility.
Specifically the authors do not state how the estimation is carried out. It would be nice to see comparisons to both the estimator of Eckles, et al. (2017) and Gui, et al. (2015). A few comments /
questions: 1. In the introduction the connection to optimal experimental design is alluded to. Having a more precise connection made within the paper would be both interesting and helpful. 2. It is
not clear what the quality of the approximation is for the heuristic presented. It would be nice to see a discussion of this and the implications for the variance of the estimator. 3. What is the
space complexity of the proposed heuristic? 4. Why are other correlation clustering heuristics not compared against in the experiment section? 5. There are no results given for the variance of the
estimators / designs. Given that this is often a critical component of analysis for practitioners it would be great to estimates (and ideally coverage) for each of the shown methods, as well as a
comparison that shows empirically the relationship between approximation quality of the correlation clustering heuristic and the resulting variance.
Reviewer 2
This work addresses an important and interesting setting, bipartite experiments, for causal inference in randomized experiments. In such a setting, treatment and control units are different from the
outcome units. Unlike the prior work from Zigler and Papadogeorgou (2018) that considers a particular setting of partial interference, they consider the most general setting for which no immediate
reduction to the standard causal setting exists. Moreover, they assume that an outcome unit’s outcome is determined by a weighted proportion of the treated diversion units in its exposure set. The
main novelty is that they propose a new clustering objective, i.e., choose the clustering of diversion units that maximizes the empirical variance of the treatment exposure. To solve the optimization
problem, they propose a scalable heuristic clustering algorithm. They validate the approach by running experiments on real data. The paper is well written and easy to follow. The reviewer thinks this
work is interesting and insightful for future work in this direction.
Reviewer 3
The paper discusses the analysis of bipartite randomized experiments, where treatment is assigned to diversion units and outcomes are measured on outcome units. The ideas discussed propose the use of
correlation clustering to identify clusters of diversion units to benefit the estimation of a treatment effect. A scalable algorithm is provided for the correlation clustering and there is a
simulation example to compare the variance of estimates when using other approaches to clustering. The paper clearly demonstrates the benefits and popularity of bipartite experiments and explains the
need for clustering. The level of novelty may be questioned as it is unclear of the benefits of the proposed algorithm versus other algorithms. There is some novelty to the ideas presented and the
algorithm developed would be useful in various applications, but a more thorough explanation of novelty may be required. There is no conclusion section. | {"url":"https://papers.nips.cc/paper_files/paper/2019/file/bc047286b224b7bfa73d4cb02de1238d-Reviews.html","timestamp":"2024-11-13T22:05:30Z","content_type":"text/html","content_length":"5728","record_id":"<urn:uuid:4983caa7-6bec-4b1c-856d-3359a71b5061>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00265.warc.gz"} |
The Derivatives of the Complex Sine and Cosine Functions
The Derivatives of the Complex Sine and Cosine Functions
We will now look at the derivatives of the complex sine and cosine functions which were introduced on The Complex Cosine and Sine Functions page.
Theorem 1: Let $f(z) = \sin z$. Then $f$ is analytic on all of $\mathbb{C}$ and $f'(z) = \cos z$.
\quad f(z) = \sin z = \frac{e^{iz} - e^{-iz}}{2i}
• Note that $z$ is analytic on all of $\mathbb{C}$. We know that any constant multiplied by an analytic function is analytic. In particular, $iz$ and $-iz$ are also analytic on all of $\mathbb{C}$.
We've already proven that $e^z$ is analytic on all of $\mathbb{C}$ so by the chain rule, $e^{iz}$ and $e^{-iz}$ are analytic on all of $\mathbb{C}$. By the sum rule and multiple rule we conclude
that $\displaystyle{\frac{e^{iz} - e^{-iz}}{2i}}$ is analytic on all of $\mathbb{C}$, i.e., $f(z) = \sin z$ is analytic on all of $\mathbb{C}$. So:
\quad f'(z) = \frac{1}{2i} \cdot (ie^{iz} + ie^{-iz}) = \frac{e^{iz} + e^{-iz}}{2} = \cos z \quad \blacksquare
Theorem 2: Let $f(z) = \cos z$. Then $f$ is analytic on all of $\mathbb{C}$ and $f'(z) = -\sin z$.
\quad f(z) = \cos z = \frac{e^{iz} + e^{-iz}}{2}
• From the comments made in the preceeding theorem we can immediately conclude that $f(z) = \cos z$ is analytic on all of $\mathbb{C}$. So:
\quad f'(z) = \frac{1}{2} \cdot (ie^{iz} -ie^{-iz}) = i \frac{e^{iz} - e^{-iz}}{2} = -\frac{e^{iz} - e^{-iz}}{2i} = - \sin z \quad \blacksquare | {"url":"http://mathonline.wikidot.com/the-derivatives-of-the-complex-sine-and-cosine-functions","timestamp":"2024-11-04T21:04:34Z","content_type":"application/xhtml+xml","content_length":"15998","record_id":"<urn:uuid:91e92e75-acf2-49b3-9d49-01746b7dbca2>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00132.warc.gz"} |
Перегляд Symmetry, Integrability and Geometry: Methods and Applications, 2006, том 2, випуск за цей рік за датою випуску
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
We discuss two known constructions proposed by Moser and by Sklyanin of the Darboux-Nijenhuis coordinates for the open Toda lattice.
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
The KdV equation is used as an example to illustrate the relation between the restricted flows and the soliton equation with self-consistent sources. Inspired by the results on the Bäcklund
transformation for the restricted ...
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The Bose-Hubbard dimer Hamiltonian is a simple yet effective model for describing tunneling phenomena of Bose-Einstein condensates. One of the significant mathematical properties of the model is
that it can be exactly ...
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In this talk, we review the basics concepts of fuzzy physics and quantum field theory on the Groenewold-Moyal Plane as examples of noncommutative spaces in physics. We introduce the basic ideas,
and discuss some important ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
The classification problem is solved for some type of nonlinear lattices. These lattices are closely related to the lattices of Ruijsenaars-Toda type and define the Bäcklund auto-transformations
for the class of two-component ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
We prove an analogue of the Sylvester theorem for the generator matrices of the quantum affine algebra Uq(gln). We then use it to give an explicit realization of the skew representations of the
quantum affine algebra. This ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
For any affine Lie algebra g, we show that any finite dimensional representation of the universal dynamical R matrix R(λ) of the elliptic quantum group Bq,λ(g) coincides with a corresponding
connection matrix for the ...
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Non-commutative structures were introduced, independently and around the same time, in mathematical and in condensed matter physics (see Table 1). Souriau's construction applied to the
two-parameter central extension of ...
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A differential calculus, differential geometry and the E-R Gravity theory are studied on noncommutative spaces. Noncommutativity is formulated in the star product formalism. The basis for the
gravity theory is the infinitesimal ...
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A class of solvable (systems of) nonlinear evolution PDEs in multidimensional space is discussed. We focus on a rotation-invariant system of PDEs of Schrödinger type and on a
relativistically-invariant system of PDEs of ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
The definitions of para-Grassmann variables and q-oscillator algebras are recalled. Some new properties are given. We then introduce appropriate coherent states as well as their dual states. This
allows us to obtain a ...
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We explain how General Relativity with a cosmological constant arises as a broken symmetry phase of a BF theory. In particular we show how to treat de Sitter and anti-de Sitter cases
simultaneously. This is then used to ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
We study multivariable Christoffel-Darboux kernels, which may be viewed as reproducing kernels for antisymmetric orthogonal polynomials, and also as correlation functions for products of
characteristic polynomials of random ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
We review the geodesic motion of pseudo-classical spinning particles in curved spaces. Investigating the generalized Killing equations for spinning spaces, we express the constants of motion in
terms of Killing-Yano tensors. ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
The purpose of the ''bootstrap program'' for integrable quantum field theories in 1+1 dimensions is to construct explicitly a model in terms of its Wightman functions. In this article, this
program is mainly illustrated ...
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
We review some recent progress in quantum field theory in non-commutative space, focusing onto the fuzzy sphere as a non-perturbative regularisation scheme. We first introduce the basic
formalism, and discuss the limits ...
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We prove that the logarithm of a group-like element in a free algebra coincides with its image by a certain linear map. We use this result and the formula of Le and Murakami for the
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We review the description of a particular deformation of the WZW model. The resulting theory exhibits a Poisson-Lie symmetry with a non-Abelian cosymmetry group and can be vectorially gauged.
• (Symmetry, Integrability and Geometry: Methods and Applications, 2006)
Addition of higher nonlinear terms to the well known integrable nonlinear Schrödinger (NLS) equations, keeping the same linear dispersion (LD) usually makes the system nonintegrable. We present a
systematic method through ...
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The behaviour of periodic points of discrete Euler top is studied. We derive invariant varieties of periodic points explicitly. When the top is axially symmetric they are specified by some
particular values of the angular ...
Мій обліковий запис | {"url":"http://dspace.nbuv.gov.ua/handle/123456789/145982/browse?type=dateissued","timestamp":"2024-11-04T07:43:35Z","content_type":"application/xhtml+xml","content_length":"53447","record_id":"<urn:uuid:bb5a16de-5281-4997-9f47-f15d84a99233>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00121.warc.gz"} |
Find Kth Smallest or Largest element in an Array
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86. Find Kth Smallest or Largest element in an Array
Objective: Find Kth Smallest or Largest element in an Array
int[] arrA = { 2, 3, 11, 16, 27, 4, 15, 9, 8 };
Output: The 4th smallest element is : 8
Naive Approach: Use Sorting
Sort the array and return the kth element. Time Complexity - O(nlogn).
Better Approach: Use Quick Sort Technique
Here we are finding the kth smallest element in the array. The same approach you can apply to find the kth largest element.
• In this technique, we select a pivot element and after one round of operation, the pivot element takes its correct place in the array.
• Once that is done we check if the pivot element is the kth element in the array, if yes then return it.
• But if the pivot element is less than the kth element, that means that the kth element is on the right side of the pivot. So make a recursive call from pivot+1 to end.
• Similarly, if the pivot element is greater than the kth element, that means that the kth element is on the left side of the pivot. So make a recursive call from start to pivot-1.
Average Time Complexity: O(n)
NOTE: Instead of printing the kth element, you can print all elements from index 0 to k and this will be the solution of the problem "Print First K smallest or largest elements in an array".
The 4th smallest element is : 8 | {"url":"https://tips.tutorialhorizon.com/algorithms/find-kth-smallest-or-largest-element-in-an-array/","timestamp":"2024-11-04T20:14:39Z","content_type":"text/html","content_length":"90915","record_id":"<urn:uuid:838f3759-44b1-4cda-9512-0f20ae22e9f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00668.warc.gz"} |
Weight of Drywall Boards
The Drywall Nails or Screws for a Room calculator computes the number of drywall nails or screws needed to hang a drywall sheets on walls and ceiling in a room.
INSTRUCTIONS: Choose units and enter the following:
• (n) Number of Drywall Sheets (Default is for 1/2" 4x8 Light Weight Drywall)
• (W) Weight per Sheet (see table below)
Weight of Drywall (M): The calculator returns weight (mass) in pounds. However this can be automatically converted to compatible units via the pull-down menu.
Weight of Drywall Sheet
• Standard Sheets
• Light Weight Drywall
• Mold and Moisture Resistant
The above are approximations and may vary based on moisture content from the environment. Check manufacturer specifications before risking health or wealth on values of this data.
The Math / Science
Drywall is made from gypsum, a heavy material. A 4x8 sheet can range in weight
DRYWALL CALCULATOR includes functions and data that support the use of drywall including:
The Drywall Calculator utilities also include a pricing survey from nationally advertised retailers. The survey is done to help vCalc's users make rough cost estimations on drywall projects, but the
user is always encouraged to use local pricing for the most accurate cost calculations.
Drywall is an indoor paneling used to cover studs, plumbing and wiring, which also serves as the base for finishing such as paint or wallpaper. Drywall is also known as wall board, sheet rock and
gypsum board. Drywall has a paper layer cover and an interior made of gypsum that is easily cut and somewhat pliable. Drywall typically comes in two sizes, 4x8 and 4x12 feet, and in several
thicknesses including 1/4", 3/8", 1/2", and 5/8". Drywall can be common, mold/moisture resistant and with higher fire prevention ratings. Drywall is affixed to interior walls via drywall nails and/or
drywall screws. | {"url":"https://www.vcalc.com/wiki/weight-of-drywall-boards","timestamp":"2024-11-06T14:02:16Z","content_type":"text/html","content_length":"53893","record_id":"<urn:uuid:24bc431a-8300-4f91-b316-4cb24de44ce6>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00797.warc.gz"} |
Determine Radius of Curvature of a Given Spherical Surface by a Spherometer
Class 11 students are expected to know the exact spherometer experiment procedure. You will need to determine the radius of curvature of given spherical surfaces using this device. However, before
proceeding to learn about the process of doing the same, one should understand spherometers in detail.
What is a Spherometer?
A spherometer is one of the vital scientific devices that measure the radius of curvature for any spherical surface precisely. Initially, opticians used these devices to create and determine powered
We come across various instruments that are used in a lab for the measurements of different things but when we have to measure the radius of curvature of either a sphere or a curved surface with its
precise measurements, then we use an instrument called a spherometer. Now if you look back at the history of the spiral meter we see that it was invented by Robert-Aglaé Cauchoix, his profession was
that of an optician in the year 1810. Robert mainly manufactured the spherometer for the use of opticians in grinding lenses. Other than using it for grinding lenses or in the physics lab, sphere
meters are used by astronomers for grinding lenses and curved mirrors. Accordingly, spherometers can have various other uses as well.
Now before we go ahead with the understanding and definition of the spherometer we will see the working principle of the device. The working principle of a spherometer is based on a micrometer screw
which is used for measuring a small thickness of flat material such as gas or can be used for measuring the radius of curvature of a spherical surface.
Normally a spherometer can be described as a device consisting of a base of a circle of three of the leg, central leg and a reading device. The Circle of three or three legs is also known as the
radius of the base Circle and the land along with it is known as the radius of the base circle, the outer legs which are given can be adjusted accordingly depending on the inner holes, this procedure
is mainly done to accommodate smaller surfaces. The central leg of the spherometer can be moved in an upward and downward direction accordingly, this can be called a flexible method of drawing lines
or using it with any other measurements. For taking the measurements any device on the reading device should be moved accordingly.
How to use a Spherometer?
A spherometer is a very common device in labs, opticians, and other physics-related settings but the main thing that is necessary is to know and understand how to use a spherometer. There is a
certain set of procedures that are included while using a speedometer:
• After holding the instrument, it is first placed on the perfect plane surface in a manner that the middle foot is screwed down slowly till it touches the surface below. After the middle foot
touches the surface, the instrument turns around on the middle foot as the center, now the center has a point from which we can equally draw shapes accordingly or measure.
• After the surface is made the spherometer is then carefully removed from the surface to take the readings from the micrometer screw. The instrument should show the reading 0-0 in normal cases,
this reading comes when the instrument is working fine, any other reading might lead to errors. If there is any sort of slight error in the instrument it could be either a negative or positive
• Now, when we measure the instruments reading, we take the instrument of the plane and let the middle foot back.
• When we are measuring we come across a reading below the zero line so if we see that reading it should be added to the zero error. If the reading above the zero lines is indicated then the
reading should be subtracted from the zero error in order to make it balanced.
• To measure the length between the two legs, the instrument should be placed on the plane surface or a playing card while using a meter scale so that equal length is measured
To measure the radius of curvature using a spherometer, one must know its various parts. The device has a screw with a moving nut in the middle of a frame with three small legs to support it upright.
The table legs, along with the screw, have tapered points to help them rest on a specific surface.
Additionally, a spherometer’s least count can differ from one device to another and each time that we may use it to determine the radius of curvature, we still need to calculate this acquired count
Define Spherometer Least Count
The smallest value that a spherometer can measure is known as its least count. The formula for determining the least count is as follows –
Least count = Pitch/Number of divisions on its head scale
Typically, the least count is always 0.01 mm.
Experiment to Find Radius of Curvature using Spherometer
Now that you know some of this measuring device’s basics, let us learn more about this experiment in general.
1. Aim
To find the radius of curvature using spherometer of a spherical surface
2. Apparatus Necessary
Plane mirror, spherometer and convex surface
3. Table Format for Noting Experiment Data
Complete Procedure for the Experiment
• Step 1: Raise the central screw of this device and use a paper to track the position of a spherometer’s three legs. Join these three points on the paper and mark them A,B and C.
• Step 2: Measure the minute distance between the three points. Note the three distances (AB, BC and AC) on a sheet of paper.
• Step 3: Determine the value of one pitch (or one vertical division).
• Step 4: Record the least count of your spherometer.
• Step 5: Raise the screw upwards to prepare for the measurement.
• Step 6: Place this spherometer on the spherical surface in such a manner that all three legs are resting on the object.
• Step 7: Start turning the screw so that it barely touches this convex surface.
• Step 8: Take the reading of both the vertical scale and the disc scale in such a position. This will act as your reference point.
• Step 9: Now place this spherometer on a plane glass slab.
• Step 10: Move the screw downwards and count the number of complete rotations for the disc (n1).
• Step 11: Continue moving until the screw tip touches the glass slab.
• Step 12: Note the reading (b) on this circular scale in relation to its vertical scale.
• Step 13: Note the circular divisions for its last incomplete rotation.
• Step 14: Complete steps 6 to 13, thrice. Note readings each time in the tabular format mentioned above.
• Mean Value of AB, BC and AC
Mean value or l = \[\frac {AB + BC + AC}{3}\]
h = \[\frac {h_1 + h_2 + h_3}{3}\]mm (Convert into cm)
Calculating Radius of Curvature of Convex Lens using Spherometer Readings
Radius R = \[\frac {I^2}{6h}\] + \[\frac {h}{2}\] cm
Vedantu’s interactive classes can help you understand more about spherometer readings, the radius of curvature and more. Experienced teachers are at your disposal whenever you need your doubts
cleared. It helps a child strengthen his or her basic concepts when understood in a processed and elaborate manner and Vedantu uses point-to-point examples and explanations from the given terms so
that it becomes easier to understand complex terms as well. you can also download our Vedantu app for better access to these study materials and online interactive sessions.
FAQs on Determine Radius of Curvature of a Given Spherical Surface by a Spherometer
1. What is the Value of R for a Plane Surface?
Radius of curvature R only exists for spherical surfaces. Therefore, R for a plane surface will always be zero.
2. What is this Formula to Calculate Radius of Curvature Using a Spherometer?
Radius R = \[\frac {1^2}{6h}\] + \[\frac {h}{2}\] cm is the formula which is used to calculate the radius of curvature of a spherometer.
3. What is the Formula to Measure the Least Count of Spherometers?
You can determine the least count of a spherometer by dividing the pitch of the device by the number of divisions on a circular scale.
4. What is the use of a spherometer?
A spherometer is an instrument used for the precise measurement of the radius of curvature of a sphere or a curved surface.
5. What is the least count of a spherometer?
The least count can be defined as the distance moved or covered by the crew of the spherometer when turned through 1 division on the circular loop. The least count can be calculated using the
formula, The formula for the radius of curvature of a spherical surface. The least count of the spherometer can be measured by dividing the pitch of the spherometer screw by the number of divisions
on the circular scale. | {"url":"https://www.vedantu.com/physics/determine-radius-of-curvature-of-a-given-spherical-surface-by-a-spherometer","timestamp":"2024-11-09T23:53:48Z","content_type":"text/html","content_length":"314120","record_id":"<urn:uuid:603d49f4-3641-4d8d-8d9a-b8d721a066e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00879.warc.gz"} |
Interior Painting Cost Calculator: Get Accurate Estimates
This tool helps you estimate the cost of painting your interior spaces by calculating the necessary expenses based on your input.
Interior Painting Cost Calculator
Use this calculator to estimate the cost of painting the interior walls of a room. Fill in the details about the room dimensions, number of doors, windows and the cost of paint per square meter.
Click ‘Calculate’ to get the estimated cost.
How to Use
• Enter the width, length, and height of the room in meters.
• Enter the number of doors and windows.
• Enter the cost of the paint per square meter.
• Click the ‘Calculate’ button to get the estimated cost.
How It Calculates the Results
The calculator estimates the total paintable area by calculating the total wall area and subtracting the area taken up by doors and windows. The paint cost is then calculated by multiplying the
paintable area by the cost of paint per square meter.
Note that this calculator does not account for variations in door and window sizes, and assumes average dimensions for them. The paint cost is an estimate and actual costs may vary depending on
brand, quality of paint, and other factors. | {"url":"https://madecalculators.com/interior-painting-cost-calculator/","timestamp":"2024-11-06T19:59:33Z","content_type":"text/html","content_length":"143359","record_id":"<urn:uuid:fa12f366-975e-4d2e-9c6b-25ca8c5e77b6>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00345.warc.gz"} |
The Effects of Mutual Coupling and Transformer Connection Type on Frequency Response of Unbalanced Three Phase Electrical Distribution System
Energy and Power Engineering, 2010, 2, 238-247
doi:10.4236/epe.2010.24035 Published Online November 2010 (http://www.SciRP.org/journal/epe)
Copyright © 2010 SciRes. EPE
The Effects of Mutual Coupling and Transformer
Connection Type on Frequency Response of Unbalanced
Three Phase Electrical Distribution System
Omer Gül, Adnan Kaypmaz
Electrical Engineering Department, Istanbul Technical University, Maslak, Turkey
E-mail: gulomer@itu.edu.tr, kaypmaz@itu.edu.tr
Received June 24, 2010; revised August 1, 2010; accepted September 3, 2010
In this paper, a novel harmonic modeling technique by utilizing the concept of multi-terminal components is
presented and applied to frequency scan analysis in multiphase distribution system. The proposed modeling
technique is based on gathering the same phase busses and elements as a separate group (phase grouping
technique, PGT) and uses multi-terminal components to model three-phase distribution system. Using multi-
terminal component and PGT, distribution system elements, particularly, lines and transformers can effec-
tively be modeled even in harmonic domain. The proposed modeling technique is applied to a test system for
frequency scan analysis in order to show the frequency response of the test system in single and three-phase
conditions. Consequently, the effects of mutual coupling and transformer connection types on three-phase
frequency scan responses are analyzed for symmetrical and asymmetrical line configurations.
Keywords: Harmonic Resonance, Unbalanced Distribution System, Frequency Scan
1. Introduction
Harmonic studies have become an important aspect of
electrical distribution system analysis and design in re-
cent years largely due to the increasing presence of solid-
state electronic power converters. Moreover, shunt ca-
pacitors are extensively used in electrical distribution
systems (EDS) for power factor correction. Due to the
proliferation of nonlinear loads, awareness of harmonic
effects has been increasing [1,2]. It is therefore that the
possibility of resonance because of shunt capacitor
should then be analyzed by the utilities [3-5].
The first decision to make in any harmonic study of
distribution system is whether a three phase model is
required or a single phase model will be sufficient.
Three-phase distribution systems are generally unbal-
anced and asymmetrical. Hence, asymmetrical three-
phase distribution systems must be modeled by phase
co-ordinations and their analysis can be performed either
under sinusoidal or non-sinusoidal conditions [1,6-9].
Compared to the single-phase analysis, problem size
increases three-times in phase coordinated based model-
ing and analysis of EDS. In addition, when harmonics
are present in the system, the models must be realized for
each harmonic component, which requires new methods
in three-phase harmonic analysis of distribution systems
in order to decrease computation time and memory re-
quirement [7].
A number of different studies related to harmonic
modeling and analysis of EDS have been given in the
literature (e.g. [2,6]). Grainger [10] applied the matrix
factorization technique (MFT) to harmonic studies to
achieve a significant saving in computational effort. In
the paper, only the required columns of the bus imped-
ance matrix which represent those busses supplying non-
linear loads are obtained instead of performing a full
As one of the most common and simple harmonic
analysis technique, frequency scan method is used to
identify the frequency response of EDS. However this is
not an easy task for some cases as shown in [5,6]. Firstly,
bus admittance matrix of EDS becomes both complicated
and large-scaled based on the number of busses and
three-phase system components. Secondly, the maximum
harmonic order to be considered is of importance in
terms of storage and computational effort for frequency
scan based harmonic analysis. If the maximum harmonic
order to be considered is as high as that of the number of
O. GÜL ET AL.
Copyright © 2010 SciRes. EPE
busses in the EDS to be analyzed a new approach to
solve such systems is needed in order to decrease com-
putation time and memory requirements.
In this paper, the following improvements are achi-
eved in modeling and computation techniques. Multi-
terminal component concept is used to find the mathe-
matical model of three-phase asymmetric EDS in har-
monic domain [11]. As for the mathematical models of
EDS, phase grouping technique, PGT is used [12]. The
technique is based on the separation of same phase buses
and components into different groups (PGT) so that more
understandable models can be constituted and savings in
memory use can be obtained. Moreover, MFT is pre-
ferred in this study to determine the frequency scan of
the EDS. Differing from Grainger, only the required
element of the bus impedance matrix on diagonal which
represent those busses supplying nonlinear loads are ob-
tained instead of performing a full inverse of the bus
admittance matrix [10]. In this paper, the aforementioned
ideas are combined to find a solution for multi-phase
frequency scan of asymmetric EDS.
2. Multiphase Distribution System Modeling
Obtaining the general model of electric circuits with the
aid of multi-terminal element is given in detail in modern
circuit theory. General form of algorithms given for
multi-terminal elements becomes more simple and un-
derstandable when it is used for mathematical modeling
of power systems. Graph and terminal equations associ-
ated with multi-terminal elements represent the mathe-
matical model of multi-terminal element and show the
whole features of it.
In this section, multi-terminal component models of a
distribution system, which is used in obtaining the har-
monic dependent modeling of EDS, is given together
with PGT. Harmonic dependent models are used in order
to find the frequency response of the network.
2.1. Basics of Multi-Terminal Approach for
Mathematical Modeling
To obtain the required models for power system analysis,
all buses in the system is generally desired to be shown
in the model and phase to ground voltages are needed for
power system modeling. As a result of this, the graph of
electric power systems that can be represented as a multi-
terminal element becomes “oriented graph” of which
common node denotes the ground and terminals of the
graph represent the buses of the system as shown in
Figure 1. Oriented graph of Figure 1(b) together with
Equation (1) gives the mathematical model of multi-
terminal component.
Multi-terminal component modeling technique can be
used in modeling of EDS for various aspects such as
single-phase, symmetrical components and phase coor-
dinated models of EDS without limitations [11,12].
Equation (1) gives the terminal equations.
(1,1)(1,2)(1, )
(2,1)(2,2)(2, )
(,1) (,2)(,)1
nn nn
YY Y
YY Y
YY Y
2.2. Mathematical Model of Electrical
Distribution System
Each element in an electric distribution system can be
represented as a multi-terminal component with its ma-
thematical model, explained in detail above. It is there-
fore an electric power network itself that can be modeled
as a multi terminal component as shown in Figure 7
through the combination of multi-terminal elements,
which is performed by using the parallel connection me-
thod of multi-terminal components.
The terminal equation of three-phase electric power
network in harmonic domain is given by Equation (2).
aaab aca
busbus bus busbus
busbus bus busbus
ccacbc c
busbus bus busbus
r (reference)
r (reference)
Figure 1. Multi-terminal modeling of n-bus EDS (a) and its
graph representation (b).
reference (ground)
Figure 2. (a) Multi-terminal representation of a three-phase
electric power network; (b) Oriented graph of an electric
power network.
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Copyright © 2010 SciRes. EPE
3. Proposed Approach
The proposed technique is based on both separation of
same phase buses and other power system components
into different groups, i.e., each phase group contains
same phase busses and elements, which can be repre-
sented as in the mathematical model of a three-phase line
model in Subsection 2.3. Since electric power networks
are composed of multi-terminal components connected
to each other, components models are firstly presented.
Hence, an electric power network itself can be modeled
as a multi-terminal structure which is a combination of
its constituents. Based on the topology of the system, the
combination procedure of multi-terminal elements rep-
resenting system constituents is carried out here through
parallel connection method of multi-terminal elements,
which is well-known in modern circuit theory [13].
Since the most common elements in electric power
networks are lines and transformers, the models of these
elements are given only in this paper. Yet, one can get
the others by following the procedure which is given in
the next section. After getting the element models in the
form of multi-terminal component, the mathematical
models are stemmed from the procedure as explained in
the following sections.
3.1. The Line Model
In general, the lines are represented as -equivalent cir-
cuit in most applications. The series impedance and
shunt admittance lumped- model representation of the
three-phase line is shown in Figure 3 [2,7]. To obtain a
symmetrical model of fundamental components, the lines
are generally transposed so as to eliminate the effect of
long lines. However, this aim can not be reached when
the system have harmonic components. Furthermore,
long line effect takes place in relatively short distances,
if the lines carry signals with high frequencies. Due to
these facts, it is a must to use phase coordinated models
in harmonic dependent line modeling.
In this case, three-phase representation of lines as mul-
ti-terminal component and its oriented graph are given
according to PGT in Figure 3.
According to proposed approach, the following pro-
cedure is given for obtaining the mathematical model of
lumped- model.
Step 1
Neglecting skin effects, harmonic dependent series
impedance matrix of a line or cable in Figure 3 is given
as in Equation (3)
abc abcabc
z (3)
where, 21
and 21
are fundamental series re-
ab y1
bc y2
ab y2
Figure 3. Lumped- model representation of the three-
phase line.
sistance and reactance matrices of a line or cable for the
phases a, b, c respectively and h is the harmonic order.
Harmonic dependent shunt admittance matrices of phas-
es a, b and c are given as below:
Re(1/) Im
Re(1/) Im
abc abcabc
abc abcabc
yy y
yy y
Real parts of Equation (4) are neglected for line and
Step 2
By inversion of primitive impedance matrix for Fig-
ure 2, series primitive admittance matrix in harmonic
domain are obtained as in Equation (5)
abcabcba b bc
ca cbc
yy y
yz yyy
yy y
By gathering the primitive admittance values of same
phase elements in one group, the primitive admittance
matrix of Figure 2 and its short form are given as Equa-
tion (6) and (7), respectively.
ba bbc
abc bc b bc
bcb bc
ca cbc
ca cbc
ca cbc
yy y
yy y
yy y
ca cbc
yy y
yy y
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Copyright © 2010 SciRes. EPE
where ,
and c
y are the sub admittance
matrices of phase elements which are grouped based on
the phases a, b, c. k
y is a mutual admittance matrix
between k and (k = a, b, c; k ; = a, b, c).
Step 3
Each system element of Figure 3(b) is first assumed
to be excited by a voltage source in order to obtain the
terminal equation of three-phase line, which is repre-
sented in the form of multi terminal component in Fig-
ure 3(a). In this case, the closed loop equations for ori-
ented graph of three-phase line whose components have
been excited by different voltage sources can be arranged
by gathering the same phase terminals together as in
Equation (8). In the same way, basic cut-set equations
can be written as Equation (9).
0000 ()
BBB are basic loop matrices re-
lated to phases, a, b and c, and ,,
basic cut-set matrices related to phases, a, b and c, re-
spectively. Moreover, as known, the expression
BQ is valid for all phases.
Step 4
By using the Equation (7), Equation (8) and Equation
(9), terminal equations in harmonic domain can be given
as Equation (10), which is well known in circuit theory
[16]. Oriented graph of Figure 3(b) together with Equa-
tion (10) gives the harmonic dependent mathematical
model of multi-terminal line component.
bus bus bus
ba bbc
bus bus bus
ca cb c
bus bus bus
Phase sub-admittance and sub-mutual admittance ma-
trices in Equation (10) are given as Equation (11) and
(12) for hth-harmonic order, respectively.
bus BB, ka, b, c
Yy (11)
bus BB,
ka, b, c; (k); a, b, c
Equation (11) and (12) represent 9 different sub-ma-
trices for each harmonic order, or rather these matrices
simply form the terminal equation of harmonic depend-
ent three-phase line in the newly proposed method.
These 9 matrices for each harmonic order can be calcu-
lated independently and for this reason the matrix calcu-
lations can be done at the same time by more than one
computer in parallel processing.
Simplified assumptions.
1) a) As shown in Equation (13), the primitive har-
monic admittance matrices associated with each phase
can be assumed to be equal on the condition that the
phase characteristics such as conductor size, line length,
phase material and the number of component belong to
each phase are same.
a b cphase
hhh h
yyyy (13)
b) As shown in Equation (14), the mutual admittance
matrices of hth-order can be assumed to be equal on the
conditions that long line effect is neglected and the lines
are transposed, which leads to symmetrical mutual cou-
pling between lines.
ab ac ba bc
ca cb m
As a consequence, when assumptions are made related
to Equations (13) and (14) it is enough to form the primi-
tive admittance matrix representing the whole three-
phase line (as in Equation (10)) by just using both one-
phase primitive admittance matrix and a mutual admit-
tance matrix.
2) As shown in Equation (15), basic loop matrices are
assumed to be equal when the topology of phases is iden-
Instead of using the Equations (11)-(12), which con-
tain 9 different equations for each harmonic, only the
Equations (16) and (17) can be used in order to form the
harmonic dependent terminal equation of the line on the
conditions that the performance equations of a primitive
network are as Equations (13)-(14) and the topology of
phases are identical.
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ab cphase
bus bus bush
Y=Y=Y=By B
ab acba bc
bus busbus bus
ca cbm
bus bush
The algorithm given above for lines is also valid for
power networks composed of more than one line. In that
case, if a mutual coupling between different three-phase
lines exists, all these lines must be modeled as a single
multi-terminal component. For this reason, the formation
of mathematical model is realized as explained above.
Consequently, as one might expect that the proposed
algorithm can be applied to a system whose bus numbers
are different at different phases, which show another
merit of the proposed modeling technique.
3.2. Three-Phase Transformer Model
Magnetizing current in transformers leads to harmonic
currents due to its saturated core. Due to the fact that the
transformers should be modelled in harmonic domain so
that harmonic currents are required to take place in the
model [2,7]. However, harmonic currents is not included
in the transformer model since our interest in this study
is to determine the frequency response of EDS.
Mathematical model associated with any of trans-
former can be obtained by utilizing the concept of mul-
ti-terminal component. However, the most common
transformers in use, i.e., Y- connected three-phase
transformers are preferred here to show the potential
application of the proposed method.
In respect of the proposed method, a transformer
model is given here in the case that the mutual coupling
between phases of primary and secondary windings is
not neglected. Y- connected three-phase transformer is
illustrated in Figure 4.
Step 1
Y- connected three-phase transformer is represented
as a multi-terminal component in Figure 5. When it is
desired to form the mathematical model of the trans-
formers with isolated neutral point, one should take the
neutral point into consideration. As a result, the terminal
number in multi-terminal representation of transformer is
increased from 7 to 8, which leads to increased dimen-
sion in oriented graph and terminal equations.
Step 2
In multi-phase system representation, the power trans-
former are represented by reactance and resistance ma-
trices for each pair of windings. According to proposed
PGT, the primitive admittance matrix of the transformer
hth harmonic order is given by Equation (18).
Figure 4. (a) Three-phase representation of a line in the
form of multi-terminal component; (b) Oriented graph of a
three-phase line.
Figure 5. Y- connected three-phase transformer with sin-
gle core.
m mmmm
mm mmm
hmm mmm
mmmm m
mmm mh
y yyyyy
yy yyyy
yy yyyy
yyyy y y
yy yyyy
y is the admittance at primary side
y is the admittance at secondary side
yis the mutual admittance between primary and sec-
ondary side at same phases
y is the mutual admittance between primary coils.
yis the mutual admittance between primary and sec-
ondary coil on different cores.
y is the mutual admittance between secondary coil.
As a result of PGT, the primitive admittance matrix in
Equation (18) has a simple and symmetrical structure,
which can easily be shown in its short form as in Equa-
tion (19).
phase mm
m phase m
mm phase
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Copyright © 2010 SciRes. EPE
y is generally taken as zero for each harmonic. It
is therefore that the primitive matrix abc
y becomes
equal to the primitive matrix of three single-phase trans-
formers connected to each other, based on their connec-
tion group.
Step 3
Transformers’ performance equations in real values
are converted to expressions in p.u. by using the turns
ratio of transformers. Moreover, phase shifting origi-
nated from connection group of three-phase transformer
(Delta-Wye) should be included in mathematical model
of the transformers. Furthermore, instead of modelling
the voltage regulators individually, phase shifting of
voltage regulators can also be included in transformers’
mathematical model on the condition that the model of
voltage regulator itself is not needed.
In this study, transformers’ performance equations are
assumed to be given in p.u. and the equivalent circuit of
Figure 6 is used in order to include both magnitude and
phase shifting in the mathematical model.
Apart from these, the phase shifting operations in
phase shifting three-phase transformers are included in
the mathematical model in the same way.
When the aforementioned conditions, which represent
the very general form of transformer model, are taken
into consideration, voltage ratio for each phase in three
-phase transformer are given as Equation (20) with the
assumption that phases have different complex turns ra-
tio. As for the current turns ratio of the transformer, it is
given in Equation (21) as complex conjugate of voltage
turns ratio.
, ka, b, c
δβ (20)
, ka, b, c
δδ (21)
With these assumptions, sub-matrices of primitive
admittance matrix of hth harmonics are given in general
form as Equations (22) and (23).
kkkphase k
.., ka, b, c
yδyδ (22)
Figure 6. (a) Multi-terminal representation of Y- connect-
ed three-phase transformer; (b) oriented graph of Y- con-
nected transformer.
Figure 7. Basic equivalent circuit in p.u. for coupling be-
tween primary and secondary coils with both primary and
secondary off-nominal tap ratios of and .
.., ka, b, c; (k); a, b, c
Since phase shifts in voltage and current equations due
to transformer connection type are same for all phases,
the primitive admittance matrices can be given as Equa-
tions (24) and (25). As a result of -connection of the
transformer, changes in phase angles and magnitudes in
these equations are same. For this reason, the structure of
primitive admittance matrix has become simple and
powerful in computation.
ab cphase
hv v
yyyδyδ (24)
ab ac ba bc ca
cb m
In this study, phase shifting due to transformers con-
nection group is included in the mathematical model
through above equations.
Step 4
To form the terminal equations (or mathematical mo-
del) of three phase transformer, closed loop equations for
voltage should be obtained as Equation (26).
B U0
V (26)
As shown in Equation (27), let
B be divided
into two sub-matrices as
B and
B, which are
sub-loop matrix related to phases and basic loop matrix
associated with the nodes of star connection, respec-
abcdabc d
BB B (27)
The matrix
B exist when the star connection node
is not grounded. As for the matrix
B, it can be
written as Equation (28) for the six most common con-
nections of three-phase transformers (Yg-Yg, Yg-Y,
Yg-, Y-Y, Y-, -).
UB O
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Copyright © 2010 SciRes. EPE
where, the sub-matrix
B given as below is a matrix
associated with -connection of three-phase transform-
(1) 0
In Equation (29), the coefficients and take the val-
ues “−1” or “0” based on the transformers connection
type. Whilst is “−1” when the primary windings of
transformer is wye, is “0” when the primary side of the
transformer is delta. As for the coefficients , it takes the
value “−1” when secondary side is wye, and takes the
value “0” when secondary side is delta. Furthermore, the
U and
O are given as below
U and O
Basic sub-loop matrix
B in Equation (27) can be
given as Equation (31), if the star connection nodes are
not grounded.
d ddd
B bbb
If star connection node in one of the primary and sec-
ondary windings is not grounded, the sub-matrix
in Equation (31) can be given as Equation (32).
If the star connection nodes in both sides are not
grounded, then, the sub-matrix
is given by Equa-
tion (33)
Step 5
As a consequence, bus admittance matrix of a trans-
former is given as a function of primitive admittance
matrix and loop equations as Equation (34).
abcdabcd abcabcd
bus h
Yy (34)
3.3. Other Components
Loading should be included in the system representation
because of its damping effect near resonant frequencies.
However, an accurate model for the system load is diffi-
cult to determine because the frequency-dependent char-
acteristics are usually unknown and the load itself is
changing continuously. In general, if a load is linear, the
load is represented as an admittance using the CIGRE
load model at the interested frequencies [14]. If load is
nonlinear, the load is represented as an open circuit in
frequency scan study.
Capacitors are often placed in distribution networks to
regulate voltage levels and reduce real power loss. Ca-
pacitor bank size and locations are the most important
factors in determining the response of distribution system
to a harmonic source. For accurate representation of ca-
pacitors, a shunt capacitor can be modelled as wye con-
nected or delta connected constant admittance [2,3,7].
For harmonic studies of EDS, it is usually sufficient to
represent the entire transmission system by its 50 Hz
short-circuit equivalent resistance and inductance at the
high side of the substation transformer [7].
4. Three Phase Frequency Scan Analysis
An electric energy system simply consists of the resis-
tances (R), inductances (L) and capacitances (C). All
circuits containing both capacitance and inductance have
one or more natural resonant frequencies [5,14-17].
Normally electric energy systems are designed to operate
at frequencies of 50 Hz so as not to be under resonance
for fundamental frequency. However, certain types of
loads produce currents and voltages with frequencies that
are integer multiples of the 50 Hz fundamental frequency.
These higher frequencies are a form of electrical distor-
tion known as power system harmonics. When one of the
natural frequencies corresponds to an exciting frequency
produced by non-linear loads, harmonic resonance can
occur. Voltage and current will be dominated by the
resonant frequency and can be highly distorted. It is
therefore that for all effective harmonic frequencies, the
system should be analyzed for whether a resonance is
going to occur or not. Frequency scan analysis is the
most common and effective method to detect the har-
monic condition in a network. The simplest way to de-
termine the frequency response of a network is to im-
plement frequency or impedance scan study. The process
of frequency scan study can be performed by solving the
network equation for the frequencies of h.fo. Here, f0 is
the fundamental frequency and h is the harmonic order.
VZI (36)
V is the nodal vector,
I is the current
The aim of the frequency scan study is to determine
impedance as a function of frequency. The frequency
scan technique basically involves following steps:
A current injection, which is a scan of sine waves of
magnitude one, is firstly applied to the bus of interest.
Secondly a resultant voltage of that bus is measured.
hhhhh h
VIZVZ Z (37)
For large-scaled systems, h
Z is simply derived from
bus impedance matrix, which is defined by taking the
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Copyright © 2010 SciRes. EPE
inverse of bus admittance matrix as below:
ZY (38)
and the expanded form of nodal impedance matrix at any
frequency is
11 121n
21 222n
n1 n2nn
ZZ Z
ZZ Z
Z = transfer impedance between nodes i and j
Z= driving point impedance at node i, r.
The effects of, the transformer connection types and
the mutual coupling between lines on frequency response
are examined on a three-bus industrial test system shown
in Figure 8 by using the frequency scan technique.
The test system consists of three busses utility, IND1
and IND2. IND1 and IND2 busses are connected through
a short three-phase and four-wire line. The system is
supplied by the utility through 69/13.8 kV transformer.
While a motor and linear load are connected on bus
IND1, a harmonic producing nonlinear load and a linear
load are connected on bus IND2. Harmonic currents of
the nonlinear load are given in the Table 1.
Since zero sequence harmonics are not found and
since only one harmonic source is present in the test sys-
tem, the system can be assumed to be balanced and
symmetric. That is why single phase analysis can be used
to solve this system.
The values on Table 1 are calculated in pu system.
The selected base quantities are 10.000 kVA and 13.8 kV.
The data and calculations are available on the web site
The following assumptions are made in the analysis. 1)
The load points are supplied from an infinite bus system.
2) The linear loads are modeled with its series resistance
and reactance. 3) For the motor loads, a locked rotor im-
pedance are used [11,12,13].
In this study, three-phase models are used and follow-
ing four-cases are considered in the frequency scan si-
mulation of the test system. The frequency responses
with three-phase models are given in comparison to sin-
gle-phase models in the Figures 9-12 for each of the
following cases. Firstly, the transformer connection type
is selected as wye-grounded/wye-grounded.
In the second case, the mutual coupling between lines
is taken into consideration for the transformer connection
type as wye-grounded/wye-grounded and the value of
mutual impedance is taken as one-third of phase imped-
ance. The system is still symmetric in this case. So, only
one-phase frequency response analysis is enough for the
IND1 IND2
Figure 8. The considered three-bus industrial test system.
Figure 9. Frequency responses of single and three-phase
EDS (transformer connection type is wye-grounded/wye-
Figure 10. Frequency responses of single and three-phase
EDS (transformer connection type is wye-grounded/wye-
grounded and there is a mutual coupling between lines).
Figure 11. Frequency responses of single and three-phase
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Copyright © 2010 SciRes. EPE
Table 1. Harmonic current spectrum of nonlinear load at
bus IND2.
h 5 7 11 13 17 19 23 25
%Ic1 0.2 0.143 0.091 0.077 0.059 0.053 0.0430.04
Ich 0.119 0.085 0.054 0.046 0.035 0.031 0.0260.024
h − 0 − 0 − 0 −0
h 29 31 35 37 41 43 47 49
%Ic1 0.034 0.032 0.029 0.027 0.024 0.023 0.0210.02
Ich 0.020 0.019 0.017 0.016 0.014 0.014 0.0120.012
h − 0 − 0 − 0 −0
In the third case, the transformer connection type is
selected as wye-grounded/delta and it is shown in Figure
11 that delta connection of transformer has an effect on
frequency response similar to that of case-2.
Finally we had the system modified so as to have an
asymmetric three-phase network. The asymmetry in the
fourth case is obtained by changing the compensation
capacitors values for the phases a, b and c as
X0.455j p.u, j068.0X b
C p.u. and c
p.u respectively.
As in the second case, mutual impedance is taken as
one-third of phase impedance and the transformer con-
nection type is wye-grounded/wye-grounded too. It is
shown in Figure 12 that frequency responses of all
phases are different from each other. Hence, frequency
responses of each phase in asymmetric networks should
be determined individually. For large-scale distribution
systems modeled and analyzed by phase coordinates, the
modified MFT will reduce the number of computation
for large-scaled networks since the MFT is capable of
computing only the necessary elements in impedance
matrix instead of taking a full inverse of the bus admit-
tance matrix
Because of the asymmetric compensation capacitors
for the phases, three-phase frequency response becomes
different from single phase response on the condition
that coupling between lines and/or transformer connec-
tion types are taken into consideration in three-phase
modeling. The frequency responses of all phases in asy-
mmetric networks are different from each other.
5. Conclusions
In this paper, to provide savings in storage and computa-
tion time, frequency scan analysis in multiphase asym-
metric distribution system is realized either by combin-
ing PGT and MFT or individually. The solution algo-
rithms are based on PGT and MFT. Whilst the PGT uses
Z (p.u.)
3 phase, a
Z (p.u.)
3 phase, b
Z (p.u.)
3 phase, c
Figure 12. Frequency responses of asymmetric three-phase
network (a) frequency response of phase a; (b) frequency
response of phase b; (c) frequency response of phase c.
the concept of multi-terminal component modeling tech-
nique to obtain the harmonic dependent model of EDS in
terms of frequency scan, the MFT uses only the required
element of the bus impedance matrix on diagonal to de-
termine the frequency scan of the EDS. Therefore an
MFT based algorithm is given to find the latter.
In order to show the accuracy of the proposed tech-
nique, a symmetric three-phase industrial system with
3-bus is preferred for simplicity. As a result, single-phase
and three-phase frequency responses of EDS are ob-
O. GÜL ET AL.
Copyright © 2010 SciRes. EPE
tained. The results show that three-phase frequency re-
sponse becomes different from single phase response on
the condition that coupling between lines and/or trans-
former connection types are taken into consideration in
three-phase modeling. In addition, frequency responses
of all phases in asymmetric networks are different from
each other. Consequently, beside in asymmetrical mod-
eling, one can easily extract that a phase-coordinated
based model must be used to detect the frequency re-
sponse of EDS, even in the symmetrical modeling.
6. References
[1] IEEE Std. 519-1992, IEEE Recommended Practice and
Requirements for Harmonic Control in Electric Power
Systems, IEEE Press, New York, 1992.
[2] J. Arrillaga, B. C. Smith and N. R. Watson, “Power Sys-
tem Harmonics Analysis,” John Wiley and Sons, New
York, 1997.
[3] IEEE Guide for Application of Shunt Power Capacitors,
IEEE Standard 1036-1992.
[4] W. Xu, X. Liu and Y. Liu, “Assessment of Harmonic
Resonance Potential for Shunt Capacitor,” Electric Power
System Research, Vol. 57, No. 2, 2001, pp. 97-104.
[5] R. A. W. Ryckaert and A. L. Jozef, “Harmonic Mitigation
Potential of Shunt Harmonic Impedances,” Electric Power
Systems Research, Vol. 65, No. 1, 2003, pp. 63-69.
[6] M. F. McGranaghan, R. C. Dugan and W. L. Sponsler,
“Digital Simulation of Distribution System Frequency-
Response Characteristics,” IEEE Transaction on Power
Apparatus and Systems, Vol. PAS-100, No. 3, 1981, pp.
[7] IEEE Power Engineering Society, Tutorial on Harmonic
Modelling and Simulation, IEEE Catalogue Number:
98TP125-0, Piscataway, 1998.
[8] M. A. Laughton, “Analysis of Unbalanced Polyphase Net-
works by the Method of Phase Coordinates, Part 1 System
Represantation in Phase Frame of Reference,” Proceeding
of IEE, Vol. 115, No. 8, 1968, pp. 1163-1172.
[9] W. E. Dillon and M.-S. Chen, “Power System Modelling,”
Proceeding of IEE, Vol. 62, No. 7, July 1974, pp. 901-915.
[10] L. G. Grainger and R. C. Spencer, “Residual Harmonic in
Voltage Unbalanced Power System,” IEEE Transactions
on Industry Applications, Vol. 30, No. 5, 1994, pp. 1398-
[11] O. Gül, “Harmonic Analysis of Three Phase Distribution
Networks by Utilizing Concept of Multi-terminal Com-
ponents and Phase Coordinates,” Ph.D Dissertation, Is-
tanbul Technical University, Istanbul, 2001.
[12] O. Gül, A. Kaypmaz and M. Tanrıöven, “A Novel Ap-
proach for Three Phase Power System Modelling by
Utilizing the Concept of Multi-Terminal,” International
Review on Modelling and Simulations, Vol. 3, No. 1,
February 2010, pp. 90-100.
[13] D. E. Johnson, J. L. Hilburn and R. J. Johnson, “Basic
Electric Circuit Analysis,” Prentice Hall, Englewood
Cliffs, 1990.
[14] CIGRE Working Group 36-05, “Harmonics, Characteris-
tics Parameter, Methods of Study, Estimates of Existing
Values in the Network,” Electra, No. 77, 1981, 35-54.
[15] Z. Huang, W. Xu and V. R. Dinavahi, “A Practical Har-
monic Resonance Guideline for Shunt Capacitor Applica-
tion” IEEE Transactions on Power Systems, Vol. 22, No.
10, p. 64.
[16] T. A. Haskew, J. Ray and B. Horn, “Harmonic Filter De-
sign and Installation: A Case Study with Resonance,”
Electrical Power System Research, Vol. 40, No. 2, 1997,
pp. 121-125.
[17] CIGRE 36.05, WO CCO2 Report Guide for Assessing the
Network Harmonic Impedances.
List of the Symbols and Abbreviations
EDS Electrical Distribution Systemb
PGT Phase Grouping Technique
MFT Matrix Factorization Technique
α Primary off-nominal tap ratings
β Secondary off-nominal tap ratings
Voltage phasor
I Current phasor
Admittance phasor
Z Impedance phasor
Y Primitive admittance
Z Primitive impedance
B Basic loop matrices
N Bus numbers
d Star point index
Delta connection index
a, b, c Line indexes
M Mutual coupling
h harmonic index
* Complex conjugate
Branch index
j Imaginary unit
V Voltage index
I Current index
T Matrix transpose | {"url":"https://file.scirp.org/Html/3147.html","timestamp":"2024-11-11T06:50:50Z","content_type":"text/html","content_length":"187259","record_id":"<urn:uuid:84a498da-5f07-4a96-80a2-60d442f7d437>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00716.warc.gz"} |
Fermi Questions
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Fermi Questions
Type Inquiry
Category Study
Description Teams provide answer to a series of "Fermi Questions"; science related questions to seek fast, rough estimates of a quantity, which is either difficult or impossible to measure
Event Information
Participants 2
Approx. Time 50 minutes
Allowed Resources writing utensils
Latest Appearance 2024
Forum Threads
Question Marathon Threads
Official Resources
Website www.soinc.org/fermi-questions-c
Division C Results
1st Madison West High School
2nd Acton-Boxborough Regional High School
3rd Troy High School
Fermi Questions is a Division C event being run in the 2024 season.
What is a Fermi Question?
A Fermi question is one where a seemingly impossible-to-calculate answer is estimated. A famous example of a Fermi question is "How many licks does it take to get to the center of a tootsie roll pop?
", where there is very little data to use and assumptions must be made. Fermi questions are named after Enrico Fermi, a physicist who is known for solving these types of questions.
In Science Olympiad, answers to Fermi questions are given in powers of ten. For example, an estimated answer to the above question of 400 licks is put in scientific notation as [math]\displaystyle{ 4
\cdot 10^2 }[/math], and the exponent on the ten is used as the answer, yielding [math]\displaystyle{ \boxed{2} }[/math]. If the estimate was 600 licks, or [math]\displaystyle{ 6\cdot 10^2 }[/math],
then the answer would be [math]\displaystyle{ \boxed{3} }[/math], rounding up.
Points are usually given as follows:
• 5 points for the correct power of ten
• 3 points for one away from the correct power of ten
• 1 point for two away from the correct power of ten.
For example, if the correct answer to the number of licks to the center of a tootsie roll pop is [math]\displaystyle{ \boxed{2} }[/math], and the given answer is [math]\displaystyle{ \boxed{2} }[/
math], five points are awarded. If the given answer is [math]\displaystyle{ \boxed{3} }[/math] or [math]\displaystyle{ \boxed{1} }[/math], 3 points are awarded, and if the given answer is [math]\
displaystyle{ \boxed{4} }[/math] or [math]\displaystyle{ \boxed{0} }[/math], one point is awarded. All other answers would receive 0 points.
Sample Question
Here is a Fermi question, worked out, with explanations.
How many pieces of paper could a package of pencil lead cover?
Determining the Facts
The first step in solving a Fermi question is to determine what facts are necessary. We change the desired quantity into more and more manageable quantities so we can make good assumptions:
[math]\displaystyle{ \text{(Number of paper a package covers)}\\ = \text{(Pieces of paper a piece of lead covers)}\cdot\text{(Number of pieces of lead are in a package)} \\ = \text{(Area of paper a
piece of lead covers)}/\text{(Area of paper)}\cdot\text{(Number of pieces of lead are in a package)} \\ = \text{(Width of line drawn by lead)(Length of line a lead can draw)}/\text{(Area of paper)}\
cdot\text{(Pieces of lead)}. }[/math]
It is often a good idea to represent the quantities with symbols, using appropriate subscripts to mitigate confusions:
Let [math]\displaystyle{ w_l }[/math] be the width of the line drawn by pencil lead, [math]\displaystyle{ l_l }[/math] be the total length of the line a pencil lead can draw, [math]\displaystyle{ A_p
}[/math] be the area of paper, and [math]\displaystyle{ n }[/math] be the number of pieces of lead in a package. Then, the desired quantity is
[math]\displaystyle{ \frac{w_l\cdot l_l\cdot n}{A_p}. }[/math]
Making the Assumptions
The second step is to make assumptions to the unknown information. The key to Fermi questions is having good assumptions. Obviously, the best assumptions are the ones that aren't assumed, like the
size of a piece of paper. However, more often than not, assumptions will be necessary.
It is common knowledge that a piece of paper is 8.5 inch by 11 inch, so [math]\displaystyle{ A_p=8.5\text{ in}\cdot11\text{ in}\approx 100\text{ in}^2 }[/math] Notice how we were not precise in our
calculation - the event does not require a high level of precision!
The number of pieces of lead in a package vary greatly, so it's not very important to be accurate. 10 pieces seems low, and 1000 seems too high. 100 still seems a little high, so we choose [math]\
displaystyle{ n=70 }[/math] as the number of pieces of lead in a package.
Most lead pieces are of width 0.5mm or 0.7mm. They are relatively close to each other, so we choose [math]\displaystyle{ w_l=0.5\text{mm} }[/math] because it is much easier to work with.
To find [math]\displaystyle{ l_w }[/math], we have to use common sense: It may be 10 meters, but that seems a little short. It could be 1 km, but that seems a little long. 100 m still seems a little
short, but close than 1000 m (1 km). Therefore, we will use [math]\displaystyle{ l_w=300 m }[/math], keeping in mind it could still be small.
Doing the Calculations
We now want to plug the quantities into the formula we created to find the answer. However, many of the quantities are in different units! We need to convert them all into the same unit. Here, cm and
m are both practical, and we choose to convert everything to cm.
Since [math]\displaystyle{ 1\text{ in}\approx2.5\text{ cm}, 1\text{ m}=100\text{ cm}, 1\text{ mm}=0.1\text{ cm} }[/math], we find that [math]\displaystyle{ A_p\approx 6\cdot10^2\text{ cm}^2, w_l=5\
cdot10^{-2}\text{ cm}, l_l=3\cdot10^4\text{ cm}. }[/math]
Therefore, the desired quantity is
[math]\displaystyle{ \frac{5\cdot10^{-2}\text{ cm}\cdot3\cdot 10^4\text{ cm}\cdot 70}{6\cdot10^2\text{ cm}^2}=175 }[/math]
. Therefore, our answer is [math]\displaystyle{ \boxed{2} }[/math].
Alternative Method
You may have a good sense of how many pieces of paper a piece of lead can cover. In that case, we let [math]\displaystyle{ p_l }[/math] be the number of paper a piece of lead can cover, and we want
to find
[math]\displaystyle{ p_l\cdot n. }[/math]
From experience sketching in art classes, a typical detailed sketch takes a third of a pencil lead, so a piece of lead lasts 3 sketches. Since a sketch doesn't cover the entire paper, but leaves
about half of it as white space, we can assume [math]\displaystyle{ p_l=\frac{3}{2}=1.5 }[/math].
Then, [math]\displaystyle{ p_l\cdot n=1.5\cdot 70\approx 100 }[/math], giving us an answer of [math]\displaystyle{ \boxed{2} }[/math].
Some contestants may be extremely confident about the value [math]\displaystyle{ p_l }[/math], while others may have little idea about it. However, while the expression you try to calculate may be
different, the process is the same: we change the unknown quantity into ones we are somewhat confident about, we make assumptions, and we calculate the unknown quantity.
Making Calculations Quickly
Magnitude Notation
The above question was worked out sequentially with numbers in to allow for explanation. This works well for shorter problems, but with longer problems with longer numbers, this can take a while, and
can become a problem with many questions and limited time and space. When doing such problems, it is better to note numbers in a format that is easier to work with. This format is called magnitude
Below are two methods for using Magnitude notation:
Method 1
For a number in scientific notation [math]\displaystyle{ a\cdot 10^b }[/math], we write it as aEb. Recall that [math]\displaystyle{ 1\le a\lt 10 }[/math]: This does not have to be strictly followed
in intermediate calculations, but you must convert it into this form to get the final answer.
For example, [math]\displaystyle{ 1\cdot 10^3 }[/math] would be 1E3, [math]\displaystyle{ 6\cdot 10^{23} }[/math] would be 6E23, [math]\displaystyle{ 7\cdot 10^{-34} }[/math] would be 7E-34.
[math]\displaystyle{ a }[/math] should always have 1 or 2 significant figures. If it has 3 or more, round the number because that level of precision is usually unnecessary given the time constraint.
To multiply two numbers aEb and cEd, we add [math]\displaystyle{ b }[/math] and [math]\displaystyle{ d }[/math] and multiply [math]\displaystyle{ a }[/math] and [math]\displaystyle{ c }[/math]. If
[math]\displaystyle{ a\cdot c\ge10 }[/math], we divide [math]\displaystyle{ a\cdot c }[/math] by 10 and add 1 to the exponent.
For example,
7E4 * 4E6 =(7*4)E(4+6)=28E10=2.8E11,
2E6 * 3E-9=(2*3)E(6-9)=6E-3,
5E4 * 2E-5=(5*2)E(4-5)=10E-1=1E0.
Round your result if the future multiplications start to become difficult.
Addition and Subtraction
To add/subtract two numbers aEb and cEd, we change the numbers so [math]\displaystyle{ b }[/math] and [math]\displaystyle{ d }[/math] have the same value (usually the larger one), modifying [math]\
displaystyle{ a }[/math] and [math]\displaystyle{ c }[/math] along the way. We then add/subtract [math]\displaystyle{ a }[/math] and [math]\displaystyle{ c }[/math].
For example,
2E5 + 4E5 = 6E5,
7E4 + 4E6 = 0.07E6+4E6 = 4.07E6 ≈ 4.1E6,
6E6 + 5E5 = 6E6+0.5E6 = 6.5E6,
9E6 - 9E5 = 9E6-0.9E6 = 8.1E6,
4E5 - 5E6 = 0.4E6-5E6 = -4.6E6,
2E6 + 9E3 = 2E6+0.009E6 ≈ 2E6!
If b and d differ by more than 2, ignore the number with the smaller magnitude because it would be almost negligible.
Method 2
• [math]\displaystyle{ 1\cdot 10^3 }[/math] would be E3
• [math]\displaystyle{ 4\cdot 10^3 }[/math] would be +E3
• [math]\displaystyle{ 7\cdot 10^3 }[/math] would be -E4
Basically, the number is put in exponential notation (like your calculator does) and rounded so there is only one digit.
Then, if the one digit is 0, 1 or 2, leave the exponent part (E3), but do not put a plus or minus. If the digit is an 8 or 9, add one to the exponent and do not put a plus or minus. If the digit is a
3 or 4, then you leave the exponent and put a plus. If the digit is a 6 or 7 add one to the exponent and put a minus. If 5 is rounded up, put add one to the exponent and put a minus. If five is
rounded down, then put a plus.
Magnitude notation has a few important rules regarding multiplication.
• With magnitude notation, when two numbers with plus signs are multiplied, the pluses are removed and one is added to the resulting exponent. for example, +E5 times +E7 equals E13.
• When two numbers, one with a +, and one with a -, are multiplied, the signs are cancelled without changing the exponent.
• When two numbers, both with minus signs, are multiplied, one is subtracted from the exponent and the signs removed.
With division, same signs cancel. Opposing signs are removed, adding one to the exponent of the value with the + sign.
Addition and Subtraction
Addition and subtraction have somewhat complicated rules.
• If the exponents are equal:
□ If the signs are both pluses, add one to the exponent and remove the signs.
□ If there are no signs, put a plus (exponents still equal).
□ If both signs are minus, remove the signs (exponents still equal).
• If one exponent is one number larger than the other:
□ Remove a minus sign if there is one on the larger one
□ Add a plus sign if there is no sign on the larger one
□ Increase the exponent on the larger one while removing the sign if there is a plus sign on the larger one.
• If the exponents are 2 or more apart, simply ignore the smaller number.
Although magnitude notation appears more complicated, it is faster and neater to work with, and thus easier to use with limited time and space.
In Fermi Questions, there is little need for accurate complex equations. Therefore, use simplified versions of the needed formulas. If a particular formula is unknown, make up an equation that makes
sense. For example, instead of calculating values using a parabola, use the triangle formula instead, slightly increasing the base and height of the triangle to compensate for the curve. Likewise,
for an ellipsoid, use a cylinder, decreasing the length to compensate for the curve.
Calculation Problems
In some tournament, there are purely calculation problems with little assumption. For example, one may be asked to evaluate [math]\displaystyle{ 2^{188} }[/math], [math]\displaystyle{ 7^{65} }[/math]
or [math]\displaystyle{ 17!=17\cdot 16\cdot\ldots \cdot2\cdot1 }[/math]. While many tournaments do not include these problems as they are calculation-heavy, in 2017 it appeared in Clements (JETS)
Invitational, Seven Lakes Invitational, Islip Invitational, Princeton Invitational, among many other Invitationals.
For these problems, a bit of advanced math background is necessary:
When reporting an answer [math]\displaystyle{ A }[/math], we are not reporting the answer but rather [math]\displaystyle{ \log_{10}A }[/math], where [math]\displaystyle{ \log }[/math] is Logarithm
Function. For example, [math]\displaystyle{ 200 }[/math] is reported as [math]\displaystyle{ \log_{10}200=2.3\approx \boxed{2} }[/math], and [math]\displaystyle{ 900 }[/math] is reported as [math]\
displaystyle{ \log_{10}900=3.0\approx\boxed{3} }[/math]. Note, however, that [math]\displaystyle{ \log_{10} 4.99\approx0.7 }[/math], so when we find the logarithm of the answer, we round down if the
decimal part is less than [math]\displaystyle{ 0.7 }[/math]. We will denote [math]\displaystyle{ \log }[/math] as logarithm base 10.
For math problems with daunting numbers, it is often easier to work with logarithm of the answer rather than the answer itself using log rules:
Since [math]\displaystyle{ \log_{10}a^b=b\log_{10}a }[/math], [math]\displaystyle{ \log_{10}2^{188}=188\log_{10}2 }[/math]. It is famous (and worth remembering!) that [math]\displaystyle{ \log_{10}2\
approx0.3 }[/math], so the answer is [math]\displaystyle{ 188\cdot0.3\approx\boxed{56} }[/math]. Note that we don't need to convert into scientific notation - the logarithm finds the answer directly!
Similarly, [math]\displaystyle{ \log_{10}7^{65}=65\log_{10}7\approx 65\cdot 0.85\approx\boxed{55} }[/math].
To find factorials, denoted by [math]\displaystyle{ ! }[/math], we either use direct calculation or Stirling's approximation:
If the number before the factorial is small (<11), directly calculate the answer. For example, [math]\displaystyle{ 8!=8\cdot7\cdot\ldots\cdot1=40320 }[/math], giving us an answer of [math]\
displaystyle{ \boxed{4} }[/math]. A list of the first few factorials, including [math]\displaystyle{ 0!=1 }[/math], can be found here.
For larger factorials, we use Stirling's approximation
[math]\displaystyle{ n!\approx\sqrt{2\pi n}(\frac{n}{e})^n, \log_{10}n!\approx n\log(\frac{n}{e})+\log(\sqrt{2\pi n}). }[/math]
For example, [math]\displaystyle{ \log_{10}17!\approx17\log(\frac{17}{e})+\log(\sqrt{6.28\cdot17})\approx 17\log(6) + \log(10)\approx 16\cdot 0.78+1\approx\boxed{14} }[/math].
The equation can be further simplified to (n times log n) minus (n times log e) plus log(square root of 2n times pi) where log e equals .4342944 in order to not divide by e and be more precise
Using logarithms, however, require higher precision: you need precision to the tenth place to get full credit, often meaning 2-3 significant figures. Therefore, it is important to know values of
[math]\displaystyle{ \pi\approx3.14, e\approx2.72 }[/math] to 2-3 significant figures and be able to multiply/divide numbers with 2-3 significant figures quickly.
It is also important to know the logarithm of 2-9:
[math]\displaystyle{ \log(2)\approx0.3, \log(3)\approx 0.48, \log(4)\approx 0.6, \log(5)\approx0.7, \log(6)\approx 0.78, \log(7)\approx 0.85, \log(8)\approx0.9, \log(9)\approx 0.95. }[/math]
. You can derive the rest using the formula
[math]\displaystyle{ \log_{10}(a\cdot 10^b)=b+\log_{10}(a). }[/math]
Competition Tips
Teams can do very well in Fermi Questions by being as accurate as possible with the most questions. This can be done in certain ways.
Partner Pairing
In Fermi questions, the partner pairing is important. You generally want two people with different interests and personalities. The ideal pair would be the estimator and the number cruncher.
The estimator should be a visual/kinetic (but more visual) learner with a good memory. They need to be able to estimate the dimensions of, for example, a football stadium based on their memory of
one. They need to be able to estimate the weights of objects they know the visual size of, but may or may not have held before. He (or she) should also know random facts, like the frequency of a
cordless phone, or the number of Crayola colors. But above all, the estimator must be able to determine all of these estimations with relative accuracy.
Number Cruncher
The number cruncher is someone quick to perform the calculations. The number cruncher should know physical values and conversion factors, like how many pounds are in a kilogram, the speed of light,
speed of sound, etc.
The number cruncher is less crucial to event success. The estimator can memorize the values, but the number cruncher cannot easily learn how to estimate.
Use Facts
Facts provide a good basis for estimations, with or without an estimator. Because even an estimator is not infallible, it is crucial to know unit conversions, as well as many basic facts about things
ranging from stellar distances to volumes of the earth's oceans. Learn esoteric units, like BTU's, barns, outhouses, sheds, furlongs, and others, which helps when answers are requested to be in terms
of an unusual unit. There have been questions were students were asked to put answers in terms of a poronkusema, so nothing is off limits.
Useful Facts
• Physics facts, like the speed of sound, or the wavelength of a certain color of light.
• Human body facts.
• State facts from the state in which the competition is taking place
• Facts about the U.S.A.
• Facts about the world.
• Esoteric units, such as those listed above.
• Infomation about the earth, and the sun (Radius, mass etc)
• Physics equations and their constants (Gravitational attraction, Escape Velocity etc)
Memorize whatever you can. Even if the fact seems useless, anything can be tested, and a fact can be used to estimate another fact.
Sometimes fermi questions simply ask for an obscure fact, such as the mass of a swallow and the mass of a coconut. As it is impossible to know everything, just guess. Answers are in powers of ten,
and therefore, they can be off by a lot and still be right.
Other Tips
• You can be off by as much as nearly 100,000% and still get points. In most cases, being within 100-200% will earn full points, and being within 1,000% will guarantee at least 3 points.
• Guess away. There are no penalty for being wrong. Just don't guess blindly. Don't say 3 elephants weigh as much as the titanic.
• When you round something down, round something else up (or down if you're dividing).
• These are exponents. E3 plus +E7 is +E7 (this is magnitude notation)
• The competition is timed. Don't dwell on a single problem unless it's your last.
• Keep your calculations neat and simple. Often times, space is limited.
• Use common sense.
• And most of all, have fun.
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Scalable GPs with Quasiseparable Kernels
Show code cell content Hide code cell content
import tinygp
except ImportError:
%pip install -q tinygp
import jaxopt
except ImportError:
%pip install -q jaxopt
Scalable GPs with Quasiseparable Kernels#
The algorithms described in this section are inherently serial, and you will probably see extremely degraded performance if you turn on GPU acceleration.
Starting with v0.2, tinygp includes an experimental pure-jax implementation of the algorithms behind the celerite package. The celerite2 package already had support for jax, but since it doesn’t
depend on any extra compiled code, the implementation here in tinygp might be a little easier to get up and running, and it is significantly more flexible. Similarly, even though it is implemented
directly in jax, instead of highly-optimized C++ code, the tinygp implementation has similar performance to the celerite2 version (see Benchmarks).
All this being said, this performance doesn’t come for free. In particular, this solver can only be used with data with sortable inputs, and specific types of kernels. In practice this generally
means that you’ll need 1-D input data (e.g. a time series) and you’ll need to build your kernel using the members of the kernels.quasisep package. But, if your problem has this form, you may see
several orders of magnitude improvement in the runtime of you model.
As a demonstration, let’s use the same sample dataset as we used in Modeling Frameworks tutorial:
import matplotlib.pyplot as plt
import numpy as np
random = np.random.default_rng(42)
t = np.sort(
random.uniform(0, 3.8, 28),
random.uniform(5.5, 10, 18),
yerr = random.uniform(0.08, 0.22, len(t))
y = (
0.2 * (t - 5)
+ np.sin(3 * t + 0.1 * (t - 5) ** 2)
+ yerr * random.normal(size=len(t))
true_t = np.linspace(0, 10, 100)
true_y = 0.2 * (true_t - 5) + np.sin(3 * true_t + 0.1 * (true_t - 5) ** 2)
plt.plot(true_t, true_y, "k", lw=1.5, alpha=0.3)
plt.errorbar(t, y, yerr=yerr, fmt=".k", capsize=0)
plt.xlabel("x [day]")
plt.ylabel("y [ppm]")
plt.xlim(0, 10)
plt.ylim(-2.5, 2.5)
_ = plt.title("simulated data")
Then we can set up our scalable GP model. This looks (perhaps deceivingly) similar to the model set up that we would normally use, but all the kernels that we’re using are defined in
tinygp.kernels.quasisep, instead of tinygp.kernels. These kernels do, however, still support addition, multiplication, and scaling to build expressive models. That being said, it’s important to point
out that the computational cost of these methods scales poorly with the number of kernels that you add or (worse!) multiply.
import jax
import jax.numpy as jnp
from tinygp import GaussianProcess, kernels
jax.config.update("jax_enable_x64", True)
def build_gp(params):
kernel = kernels.quasisep.SHO(
kernel += jnp.exp(2 * params["log_sigma2"]) * kernels.quasisep.Matern32(
return GaussianProcess(
diag=yerr**2 + jnp.exp(params["log_jitter"]),
def loss(params):
gp = build_gp(params)
return -gp.log_probability(y)
params = {
"mean": 0.0,
"log_jitter": 0.0,
"log_sigma1": 0.0,
"log_omega": np.log(2 * np.pi),
"log_quality": 0.0,
"log_sigma2": 0.0,
"log_scale": 0.0,
Array(61.8065268, dtype=float64)
Good - we got a value for our loss function. We can check that this was actually using the scalable solver defined in tinygp.solvers.quasisep.solver.QuasisepSolver by checking the type of the solver
property of our GP:
Now we can minimize the loss:
import jaxopt
solver = jaxopt.ScipyMinimize(fun=loss)
soln = solver.run(jax.tree_util.tree_map(jnp.asarray, params))
print(f"Final negative log likelihood: {soln.state.fun_val}")
Final negative log likelihood: 3.7920344745172017
And plot our results:
_, cond = build_gp(soln.params).condition(y, true_t)
mu = cond.loc
std = np.sqrt(cond.variance)
plt.plot(true_t, true_y, "k", lw=1.5, alpha=0.3, label="truth")
plt.errorbar(t, y, yerr=yerr, fmt=".k", capsize=0)
plt.plot(true_t, mu, label="max likelihood model")
plt.fill_between(true_t, mu + std, mu - std, color="C0", alpha=0.3)
plt.xlabel("x [day]")
plt.ylabel("y [ppm]")
plt.xlim(0, 10)
plt.ylim(-2.5, 2.5)
_ = plt.title("maximum likelihood")
This all looks pretty good!
Before closing out this tutorial, here are some technical details to keep in mind when using this solver:
1. This implementation is new, and it hasn’t yet been pushed to its limits. If you run into problems, please open issues or pull requests.
2. The computation of the general conditional model with these kernels is not (yet!) as fast as we might want, and it may be somewhat memory heavy. For very large datasets, it is sometimes
sufficient to (a) just compute the conditional at the input points (by omitting the X_test parameter in tinygp.GaussianProcess.condition()), (b) only compute the mean prediction, which should be
fast, or (c) only predict at a few test points.
3. For more technical details about these methods, check out the API docs for the kernels.quasisep package, and the solvers.quasisep package, as well as the links therein.
4. It should be possible to implement more flexible models using this interface than those supported by celerite or celerite2, so stay tuned for more tutorials! | {"url":"https://tinygp.readthedocs.io/en/latest/tutorials/quasisep.html","timestamp":"2024-11-02T05:54:47Z","content_type":"text/html","content_length":"53748","record_id":"<urn:uuid:0974e93c-4636-48f9-92b7-d0741230a4f3>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00638.warc.gz"} |
Understanding Mathematical Functions: How To Find The Max Or Min Of A
Mathematical functions are essential tools used to describe relationships between variables in the field of mathematics. Understanding how to find the maximum or minimum of a function is crucial in
various areas such as optimization problems, economics, and engineering. In this blog post, we will explore the definition of mathematical functions and discuss the importance of being able to
determine the maximum or minimum of a function.
Key Takeaways
• Mathematical functions are essential tools used to describe relationships between variables in various fields.
• Understanding how to find the maximum or minimum of a function is crucial in optimization problems, economics, and engineering.
• Techniques such as the first derivative test, second derivative test, and calculus can be used to find maxima and minima of functions.
• Real-life applications of finding maxima and minima include optimization of resources, maximizing profits, and minimizing costs.
• Understanding mathematical functions can be beneficial in various fields and is important for problem-solving and decision-making.
Understanding Mathematical Functions
When we talk about mathematical functions, we are referring to a relationship between a set of inputs and a set of feasible outputs. In simple terms, a mathematical function takes an input, performs
a specific operation on it, and produces an output. Functions can take many forms and are a fundamental concept in mathematics.
A. Explanation of mathematical functions
• Definition: A mathematical function is a relation between a set of inputs and a set of possible outputs, where each input is related to exactly one output.
• Notation: Functions are commonly denoted by a letter, such as f(x), where "x" represents the input and "f(x)" represents the output.
• Examples: Examples of mathematical functions include linear functions, quadratic functions, trigonometric functions, and exponential functions.
B. Example of a simple function
• Definition of a simple function: A simple function is a function that can be expressed by a straightforward mathematical formula and has a clear input-output relationship.
• Example: The function f(x) = 2x + 3 is a simple linear function, where the input "x" is multiplied by 2, and then 3 is added to the result to obtain the output f(x).
How to find the max or min of a function
• Understanding extremum points: The maximum (max) and minimum (min) of a function occur at points where the function reaches its highest or lowest value, respectively.
• Derivative test: One way to find the max or min of a function is by using the derivative test. This involves finding the derivative of the function and locating the points where the derivative is
zero or undefined.
• Second derivative test: Another method is the second derivative test, which involves examining the concavity of the function at specific points to determine if they correspond to a max or min.
Finding the Maximum of a Function
In the field of mathematics, understanding how to find the maximum of a function is crucial in solving various problems and analyzing data. In this chapter, we will delve into the definition of the
maximum of a function and the steps involved in finding it, along with an example to illustrate the process.
A. Definition of the maximum of a function
The maximum of a function refers to the highest point on the graph of the function. It is the value of the function at a specific input that is greater than or equal to the values of the function at
all other inputs in the given domain.
B. Steps to find the maximum of a function
• Step 1: Identify the critical points Begin by finding the derivative of the function and setting it equal to zero to determine the critical points. These points are where the slope of the
function is either zero or undefined.
• Step 2: Test for the nature of critical points Use the first or second derivative test to determine whether the critical points are the locations of a maximum, minimum, or neither.
• Step 3: Evaluate the function at critical points and endpoints Once the critical points are identified, plug them into the original function to find their corresponding function values.
Additionally, evaluate the function at the endpoints of the given domain, if applicable.
• Step 4: Compare the function values Compare the function values obtained from Step 3 to determine which one is the maximum value. The largest function value will represent the maximum of the
C. Example of finding the maximum of a function
Let's consider the function f(x) = 2x^2 - 8x + 6 over the interval [-1, 4] to illustrate the process of finding the maximum of a function.
We start by finding the derivative of the function f'(x) = 4x - 8, and then setting it equal to zero to find the critical point at x = 2.
Using the second derivative test, we determine that the critical point x = 2 corresponds to a local minimum. Next, we evaluate the function at the endpoints of the interval, f(-1) = 16 and f(4) = 6,
and at the critical point, f(2) = 2.
Comparing the function values, we find that the maximum of the function f(x) = 2x^2 - 8x + 6 over the interval [-1, 4] is 16 at x = -1.
Finding the Minimum of a Function
When working with mathematical functions, finding the minimum is an essential part of the analysis. The minimum of a function represents the lowest point on the graph, where the function reaches its
smallest value.
A. Definition of the minimum of a function
The minimum of a function is the lowest value that the function can obtain within a given interval. It can be identified as the point where the function changes direction from decreasing to
B. Steps to find the minimum of a function
• Step 1: Find the derivative of the function.
• Step 2: Set the derivative equal to zero to find the critical points.
• Step 3: Use the second derivative test or the first derivative test to determine if the critical point is a minimum.
• Step 4: Confirm the minimum by substituting the critical point into the original function.
C. Example of finding the minimum of a function
Let's consider the function f(x) = x^2 - 2x + 1. To find the minimum of this function, we follow the steps mentioned above:
Step 1: Finding the derivative
The derivative of f(x) is f'(x) = 2x - 2.
Step 2: Finding the critical points
Setting the derivative equal to zero gives us 2x - 2 = 0, which leads to x = 1 as the critical point.
Step 3: Using the second derivative test
Calculating the second derivative, f''(x) = 2, which is greater than zero. This confirms that the critical point x = 1 corresponds to a minimum.
Step 4: Confirming the minimum
Substituting x = 1 into the original function gives us f(1) = 1. Therefore, the function has a minimum value of 1 at x = 1.
Techniques for Finding Maxima and Minima
When working with mathematical functions, finding the maximum and minimum points is essential for various applications in fields such as physics, engineering, and economics. There are several
techniques to identify these points, including the first derivative test, the second derivative test, and using calculus.
• The first derivative test
The first derivative test is a method used to determine the maxima and minima of a function by analyzing the sign changes of its first derivative. When the first derivative changes from positive
to negative at a specific point, it indicates a local maximum. Conversely, when the first derivative changes from negative to positive, it signifies a local minimum.
• The second derivative test
The second derivative test involves analyzing the concavity of a function to determine its maxima and minima. If the second derivative is positive at a critical point, it indicates a local
minimum. Conversely, if the second derivative is negative, it signifies a local maximum.
• Using calculus to find maxima and minima
Calculus provides a powerful tool for finding maxima and minima by determining the critical points of a function and analyzing its behavior near these points. By setting the first derivative
equal to zero and solving for the critical points, and then using the second derivative test to classify these points as maxima or minima, calculus offers a systematic approach to identifying the
extreme points of a function.
Real-life Applications
Understanding how to find the maxima and minima of a mathematical function has numerous real-life applications. Let's explore a few examples:
A. Examples of real-life situations where finding maxima and minima is important
• Financial Analysis: In the field of finance, maximizing profits and minimizing losses is crucial. Understanding mathematical functions can help in analyzing data to make informed investment
• Engineering Design: Engineers often need to optimize designs by finding the maximum or minimum values of certain parameters. This can include maximizing the strength of a structure while
minimizing the use of materials.
• Medical Research: In medical research, understanding the maxima and minima of functions can help in optimizing drug dosages or identifying the peak effectiveness of a treatment.
• Supply Chain Management: Businesses need to optimize their supply chain operations to minimize costs and maximize efficiency. Understanding mathematical functions can aid in this process.
B. How understanding mathematical functions can be beneficial in various fields
• Science: In various scientific disciplines, such as physics and chemistry, understanding maxima and minima is crucial for analyzing experimental data and modeling natural phenomena.
• Data Science: In the age of big data, understanding mathematical functions is essential for extracting insights and making predictions from large datasets.
• Computer Science: In fields such as machine learning and algorithm optimization, mathematical functions play a central role in developing efficient and intelligent systems.
• Operations Research: Understanding maxima and minima is foundational in the field of operations research, where it is used to optimize processes and decision-making.
Understanding mathematical functions is crucial for anyone studying mathematics or pursuing a career in a related field. It provides a foundation for grasping more complex concepts and
problem-solving techniques. Learning how to find the maxima and minima of a function is an important skill that can be applied to various real-world scenarios, such as optimizing processes and making
strategic decisions.
I encourage you to practice tackling problems related to finding maxima and minima of functions. The more you practice, the more confident you will become in your mathematical abilities. This
knowledge will serve as a valuable tool in your academic and professional pursuits.
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Journal of the European Optical Society-Rapid Publications
Issue J. Eur. Opt. Society-Rapid Publ.
Volume 18, Number 1, 2022
Article Number 3
Number of page(s) 4
DOI https://doi.org/10.1051/jeos/2022004
Published online 06 July 2022
J. Eur. Opt. Society-Rapid Publ. 2022,
, 3
Short Communication
It is a sufficient condition only, not a necessary and sufficient condition, for decomposing wavefront aberrations
Department of Mechanical Engineering, National Cheng Kung University, Tainan 70101, Taiwan
^* Corresponding author: pdlin@mail.ncku.edu.tw
Received: 14 March 2022
Accepted: 22 June 2022
The classic equation for decomposing the wavefront aberrations of axis-symmetrical optical systems has the form,$W ( h 0 , ρ , ϕ ) = ∑ j = 0 ∝ ∑ p = 0 ∝ ∑ m = 0 ∝ C ( 2 j + m ) ( 2 p + m ) m ( h 0 )
2 j + m ( ρ ) 2 p + m ( cos ϕ ) m$where j, p and m are non-negative integers, ρ and ϕ are the polar coordinates of the pupil, and h[0] is the object height. However, one non-zero component of the
aberrations (i.e., C[133]h[0]ρ^3cos^3ϕ) is missing from this equation when the image plane is not the Gaussian image plane. This implies that the equation is a sufficient condition only, rather than
a necessary and sufficient condition, since it cannot guarantee that all of the components of the aberrations can be found. Accordingly, this paper presents a new method for determining all the
components of aberrations of any order. The results show that three and six components of the secondary and tertiary aberrations, respectively, are missing in the existing literature.
Key words: Aberrations / Imaging systems / Wavefront
© The Author(s), published by EDP Sciences, 2022
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1 Introduction
The wavefront and ray aberrations of axis-symmetrical systems have attracted significant attention in the literature [1–12]. The usual equation for decomposing the monochromatic wavefront aberration
W(h[0], ρ, ϕ) into different orders and components is given as (e.g., Eq. (3.31b) of [5]),$W ( h 0 , ρ , ϕ ) = ∑ j = 0 ∝ ∑ p = 0 ∝ ∑ m = 0 ∝ C ( 2 j + m ) ( 2 p + m ) m ( h 0 ) 2 j + m ( ρ ) 2 p + m
( cos ϕ ) m$(1)where j, p and m are non-negative integers, ρ and ϕ are the polar coordinates of the pupil, and h[0] is the object height. The sum of the powers of h[0] and ρ gives the order of the
related component. That is,$2 q = 2 ( j + p + m )$(2)
For example, if the piston term is included, the primary (i.e., fourth-order W[4th]) aberrations are obtained from equation (1) with 2q = 4 as,$W 4 th = C 400 h 0 4 + C 040 ρ 4 + C 131 h 0 ρ 3 cos ϕ
+ C 222 h 0 2 ρ 2 ( cos ϕ ) 2 + C 220 h 0 2 ρ 2 + C 311 h 0 3 ρ cos ϕ$(3)where the six components of the equation represent the piston term and the spherical, coma, astigmatism, field curvature, and
distortion aberrations, respectively. The composition ability of the primary aberrations from equation (1) is echoed by Buchdahl [1], who computed the Buchdahl aberration coefficients to determine
the wavefront and ray aberrations of axis-symmetrical systems. However, the question arises as to whether equation (1) provides all the components of the various order wavefront aberrations in an
axis-symmetrical optical system.
2 Decomposition of wavefront aberrations
Equation (1) is based on the fact that the aberration function W(h[0], ρ, ϕ) must satisfy the following three equations related to the fundamental axis-symmetrical nature of axis-symmetrical systems
(e.g., p. 154 of [5]):$W ( 0 , ρ , ϕ ) = W ( 0 , - ρ , ϕ )$(4) $W ( h 0 , ρ , ϕ ) = W ( h 0 , ρ , - ϕ )$(5) $W ( h 0 , ρ , ϕ ) = W ( - h 0 , ρ , π + ϕ ) = W ( - h 0 , ρ , π - ϕ )$(6)
Equation (4) indicates that the aberration function of an on-axis object must be radially symmetric, and hence implies that the components of W(h[0], ρ, ϕ) that do not depend on h[0] should vary as ρ
^2 (or its integer power). Equation (5) states that W(h[0], ρ, ϕ) must be a function of cosϕ. Finally, equation (6) shows that W(h[0], ρ, ϕ) must equal W(−h[0], ρ, π + ϕ) for an object with height h
[0] above the optical axis and W(−h[0], ρ, π − ϕ) for an object with height h[0] below the optical axis. Hence, those terms that depend on ϕ should be a function of h[0]ρcosϕ. Combining this with ϕ
-independent terms, it follows that W(h[0], ρ, ϕ) must consist of terms containing $h 0 2$, ρ^2 and h[0]ρcosϕ factors, to have a sufficient condition given by equation (1). Note that a sufficient
condition is taken here to mean that any term generated by equation (1) is a component of an aberration.
The present group recently proposed a method for determining the aberrations of axis-symmetrical optical systems [12]. It is shown in Figure 1 that when the image plane is not the Gaussian image
plane, a non-zero component (i.e., C[133]h[0]ρ^3(cosϕ)^3) is missing from equation (1), even though it satisfies equations (4)–(6). This implies that equation (1) alone does not guarantee that all of
the components of the aberrations can be found. In other words, equation (1) is not a necessary and sufficient condition for determining all the components of the aberrations in an axis-symmetrical
optical system.
Fig. 1
The variation of W[133] = C[133]h[0]ρ^3/λ (where h[0] = 17 mm, ρ = 21 mm, and λ = 550 µm) versus the separation of image plane V[image] for the optical system of [12]. The Gaussian image plane of
this system is located at V[image] = 92.088474 mm when the object is placed at P[0z] = −200 mm. This figure shows that C[133]h[0]ρ^3(cosϕ)^3 has non-zero value when the image plane is not the
Gaussian image plane. It also shows that C[133] = 0 when V[image] = 92.088474 mm, indicating C[133]h[0]ρ^3(cosϕ)^3 is the defocus component of primary aberrations.
Thus, a second question arises as to how all of these components may be found. To address this question, it is first necessary to realize that the power of cosϕ should be non-negative. That is,$m ≥
Without any loss of generality, the power of h[0] can be confined to a non-negative integer value in order to have 2j + m ≥ 0. That is,$j ≥ - m / 2$(8)
Mathematically, the power of ρ should be greater than or equal to the power of cosϕ, i.e., 2p + m ≥ m, which yields,$p ≥ 0$(9)
One then has the following inequality from the sum of q = j + p + m (see Eq. (2)) and equation (8):$p ≤ q - m / 2$(10)
The intersection of equations (9) and (10) defines the possible range of p. That is,$0 ≤ p ≤ q - m / 2$(11)
Equation (11) shows that the integer index p starts at p = 0 and ends at,$p max = 〈 q - m / 2 〉$(12)where 〈q − m/2〉 is the maximum non-negative integer value of p for a given m and q. In other
words, index p belongs to the following set:$p ∈ { 0,1 , 2 , . . . , p max }$(13)
Furthermore, from equation (11), the possible upper limit of m which yields 0 ≤ q − m/2 is,$m ≤ 2 q$(14)
The intersection of equations (7) and (14) then shows the possible domain of integer m for a given q. That is,$0 ≤ m ≤ 2 q$(15)
Given the preceding derivations, it is possible to obtain all components of any order (say, the (2q)th order) wavefront aberration for an axis-symmetrical optical system by the following equation
when j = q − (p + m) is used:$W ( 2 q ) th - order ( h 0 , ρ , ϕ ) = ∑ p = 0 p = p max ∑ m = 0 m = 2 q C ( 2 q - 2 p - m ) ( 2 p + m ) m ( h 0 ) 2 q - 2 p - m ( ρ ) 2 p + m ( cos ϕ ) m$(16)
Note that, as shown in equation (4), components of the aberrations that do not depend on h[0] should vary as ρ^2, or as its integer power only. In other words, if 2q − 2p – m = 0 and $m ≠ 0$, that
component generated from equation (16) does not exist.
Consider Table 1 below, which shows all the components of the secondary aberration (q = 3) of an axis-symmetrical optical system for illustration purposes. The entries of the first and second columns
are the values of m and p obtained from equations (15) to (13), respectively. Meanwhile, the entries of the third column denote the sequence (2j + m, 2p + m, m) = (2q − 2p − m, 2p + m, m) for each
value of m. The fourth column shows the aberration component for each value of m (if it exists). Comparing Table 1 with the existing literature, it is found that three components in Table 1 (i.e., C
[244] $h 0 2$ρ^4(cosϕ)^4, C[153]h[0]ρ^5(cosϕ)^3 and C[155]h[0]ρ^5(cosϕ)^5) are not included among the secondary aberrations given in the literature despite satisfying equations (4)–(6). In order to
validate Table 1, the methodology proposed in [12] was extended to determine the values of all the secondary aberrations listed in the right-hand column of the table [3]. The results confirmed that
all of the secondary aberrations possessed non-zero values.
Table 1
Components of secondary aberrations with q = 3 in axis-symmetrical system.
The method in this study was further applied to determine all the components of the tertiary aberrations (q = 4) (see Table 2). Comparing the results in Table 2 with those in Table 3–3 of [5], it is
found that six components (i.e., C[173]h[0]ρ^7(cosϕ)^3, C[264] $h 0 2$ρ^6(cosϕ)^4, C[355] $h 0 3$ρ^5(cosϕ)^5, C[175]h[0]ρ^7(cosϕ)^5, C[266] $h 0 2$ρ^6(cosϕ)^6 and C[177]h[0]ρ^7(cosϕ)^7) are missing
from equation (1).
Table 2
Components of tertiary aberrations with q = 4 in axis-symmetrical system.
3 Conclusions
The wavefront aberrations W(h[0], ρ, ϕ) of axis-symmetrical systems are generally decomposed into their various components using equation (1). However, the numerical results presented in [12] show
that this equation cannot guarantee that all of the components of the primary aberrations can be found. In other words, the equation is a sufficient condition only, not a necessary and sufficient
Accordingly, this study has presented a method for determining the possible domains of the non-negative integer indices, m and p, in equation (1) such that all of the components of the aberrations
can be found. It has been shown that the index j computed from equation (2) may be negative. Furthermore, three and six new components of the secondary and tertiary aberrations of an axis-symmetrical
system have been found, where these components all satisfy the equations describing the fundamental axis-symmetrical nature of axis-symmetrical systems. Overall, the method proposed in this study
provides a systematic and robust approach for ensuring that all of the components of any order wavefront aberration in an axis-symmetrical system can be found.
Conflict of interest
The author declares no conflicts of interest.
Ministry of Science and Technology, Taiwan (MOST) (109-2221-E-006-045).
All Tables
Table 1
Components of secondary aberrations with q = 3 in axis-symmetrical system.
Table 2
Components of tertiary aberrations with q = 4 in axis-symmetrical system.
All Figures
Fig. 1
The variation of W[133] = C[133]h[0]ρ^3/λ (where h[0] = 17 mm, ρ = 21 mm, and λ = 550 µm) versus the separation of image plane V[image] for the optical system of [12]. The Gaussian image plane of
this system is located at V[image] = 92.088474 mm when the object is placed at P[0z] = −200 mm. This figure shows that C[133]h[0]ρ^3(cosϕ)^3 has non-zero value when the image plane is not the
Gaussian image plane. It also shows that C[133] = 0 when V[image] = 92.088474 mm, indicating C[133]h[0]ρ^3(cosϕ)^3 is the defocus component of primary aberrations.
In the text
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How to add vectors?
The addition of vectors is not like the addition of scalars. For scalars you simply add or subtract numbers but a vector quantity is also associated with a direction, so adding or subtracting them is
different from the normal addition or subtraction of numbers.
You can easily understand the addition or subtraction of vectors by their graphical representation which is described below. In vector addition you always need to consider both the magnitude and the
direction of vectors.
Consider two vectors \(\vec{A}\) and \(\vec{B}\) in Figure 1 which we add together to get the resultant vector. Note that the vector \(\vec{B}\) shown in dotted line in Figure 2 whose tail is at the
head of \(\vec{A}\) is equal to the original \(\vec{B}\) and now the sum of \(\vec{A}\) and \(\vec{B}\) is the resultant vector whose tail is at the starting point, that is at the tail of \(\vec{A}\)
and head is at the ending point, that is at the head of \(\vec{B}\).
Figure 1 \(\vec{A}\) and \(\vec{B}\) ready to be added together.
Figure 2 \(\vec{B}\) shown in dotted line is equal to the original vector. Note that for the vectors to be equal both magnitude and direction should be the same.
In Figure 3 you can see that the sum of \(\vec{A}\) and \(\vec{B}\) is the resultant vector straight from the tail of \(\vec{A}\) to the head of\(\vec{B}\). In other words when you move along the
direction of \(\vec{A}\) through the magnitude \(A\) and then along the direction of \(\vec{B}\) through the magnitude \(B\), your resultant magnitude and direction of your movement will be that
shown by the resultant vector \(\vec{R}=\vec{A}+\vec{B}\).
Now vector subtraction is the same as vector addition. If you subtract \(\vec{A}\) and \(\vec{B}\), like \(\vec{A}-\vec{B}\), this is the same as \(\vec{R}=\vec{A}+(-\vec{B})\). Now you know that the
negative of a vector has the same magnitude but has opposite direction. So \(-\vec{B}\) has the same magnitude as that of \(\vec{B}\) but has opposite direction and the resultant of the vectors will
be that as shown in Figure 4.
Figure 3 Resultant vector \(\vec{R}\) is the total final vector whose tail is at the tail of \(\vec{A}\) and head is at the head of \(\vec{B}\).
Figure 4 Negative of a vector is the vector with the same magnitude but opposite direction, so the subtraction of vectors is the same thing as addition with negative vector.
Components of a Vector
A vector can be resolved into its components. We find the x and y-components of a vector \(\vec{A}\) in Figure 5. The x-component of the vector is the component along x-axis and y-component along
Figure 5 \(\vec{A}\) is converted into its components along positive x and positive y-axis.
Notice in Figure 5 that \(\vec{A}\) makes an angle \(\phi\) with x-axis. Now we can easily find the value of \({{A}_{x}}\) and\({{A}_{y}}\) to be, \({{A}_{x}}=A\cos \phi \) and \({{A}_{y}}=A\sin \phi
\). So the x and y-components of \(\vec{A}\) can be written in vector form as,
\[{{\vec{A}}_{x}}=(A\cos \phi )\hat{i}\]
\[{{\vec{A}}_{y}}=(A\sin \phi )\hat{j}\]
Note that the unit vectors \(\hat{i}\) and \(\hat{j}\) give the direction along positive x-axis and positive y-axis respectively. \(\vec{A}\) can be written as \(\vec{A}={{A}_{x}}\hat{i}+{{A}_{y}}\
hat{j}=(A\cos \phi )\hat{i}+(A\sin \phi )\hat{j}\). Here we also can find the magnitude and direction of \(\vec{A}\) as,
\[A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}} \tag{1} \label{1}\]
\[\phi =\arctan \left( \frac{{{A}_{y}}}{{{A}_{x}}} \right) \tag{2} \label{2}\]
Here the angle \(\phi\) gives the direction for \(\vec{A}\) and you obviously know \(\tan \phi =\frac{{{A}_{y}}}{{{A}_{x}}}\).
How to add using components?
Addition of vectors using components is the easiest way to add vectors. In Figure 6 there are two vectors. The tail of \(\vec{B}\) is at the head of \(\vec{A}\). First we find the x and y-components
of \(\vec{A}\) and \(\vec{B}\), then we add all x-components to find the total x-component of the resultant vector and all y-components to find the total y-component of the resultant vector.
Figure 6 Vectors can be added by adding their x-components to get the x-component of the resultant vector and y-components to get the y-component of the resultant vector.
Now the total x and y-components of the resultant vector \(\vec{R}\) are \({{R}_{x}}={{A}_{x}}+{{B}_{x}}\) and \({{R}_{y}}={{A}_{y}}+{{B}_{y}}\). Now we can find the magnitude and direction of the
resultant vector easily. Suppose \(\vec{R}\) makes an angle \(\phi\) with x-axis.
\[\phi =\arctan \frac{{{R}_{y}}}{{{R}_{x}}}\]
Now \(\vec{R}\) is \(\vec{R}={{R}_{x}}\hat{i}+{{R}_{y}}\hat{j}\). Obviously the unit vectors make it easy to represent a vector as well as its components.
How to add using parallelogram?
In parallelogram addition of vectors we arrange vectors in the form of parallelogram by making opposite vectors equal. In Figure 7. the vectors \(\vec{A}\) and \(\vec{B}\) shown in dotted line are
equal to the original vectors (these vectors are acting as the opposite sides of the parallelogram) and help to form a parallelogram. You obviously know that the resultant vector \(\vec{R}\) is the
sum of \(\vec{A}\) and \(\vec{B}\) and here we find the magnitude and direction of the resultant vector.
Figure 7 Parallelogram addition of vectors.
In Figure 7. produce the line \(\text{OM}\) up to \(\text{N}\) and draw \(\text{PN}\bot \text{ON}\). Let the angle \(\angle \text{PMN=}\alpha \) and the angle made by the resultant vector \(\vec{R}\)
with \(\vec{A}\) be \(\theta\). Now \(\sin \alpha = \text{PN}/B\) which gives \(\text{PN}=B\sin \alpha \) and \(\cos \alpha =\text{MN}/B\) which gives \(\text{MN=}B\cos \alpha \), so \(\text{ON}=\
text{OM}+\text{MN}=A+B\cos \alpha \). Since \(\vartriangle \text{OPN}\) is a right-angled triangle, the magnitude of \(\vec{R}\) is,
\[\begin{align*} R&=\sqrt{\text{O}{{\text{N}}^{2}}+\text{P}{{\text{N}}^{2}}} \\ & =\sqrt{{{(A+B\cos\alpha )}^{2}}+(B\sin\alpha )} \\ & =\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \alpha } \tag{3} \label{3} \
When \(\alpha ={{90}^{\circ }}\), it means when the vectors are perpendicular Eq. \eqref{3} becomes \(R=\sqrt{{{A}^{2}}+{{B}^{2}}}\) as it should be. Again, \(\tan \theta =\text{PN}/\text{ON}\), so,
\[\begin{align*} \tan \theta&=\frac{B\sin \alpha }{A+B\cos \alpha } \\ \text{and,}\quad \quad \theta& =\arctan \left( \frac{B\sin \alpha }{A+B\cos \alpha } \right) \tag{4} \label{4} \end{align*}\]
You may have known that the angle \(\theta\) indicates the direction of the resultant vector. When \(\alpha ={{0}^{\circ }}\), \(\theta ={{0}^{\circ }}\) which is we should get. | {"url":"https://www.physicsscience.net/5/how-to-add-vectors","timestamp":"2024-11-04T08:29:19Z","content_type":"text/html","content_length":"24100","record_id":"<urn:uuid:81d76286-9ecd-4276-9a7c-0c85819f7c0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00841.warc.gz"} |
190-NCERT New Syllabus Grade 10 Polynomials Ex-2.1
NCERT New Syllabus Mathematics
Class: 10
Exercise 2.1
Topic: Polynomials
Understanding Polynomials: A Key Chapter in Class 10 Mathematics
Polynomials form a fundamental building block in algebra and are crucial in understanding various higher-level concepts in mathematics. In this chapter of the 10th-grade NCERT syllabus, students will
explore how to express algebraic expressions in polynomials and analyze their behavior through various methods. From identifying degrees to solving for roots, this chapter will empower you to solve
equations efficiently, laying a strong foundation for future mathematical concepts like calculus and advanced algebra.
1) If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of
2) Quadratic polynomial in x is of the form ax^2 + bx + c, where a, b, c are real
numbers and a ≠ 0.
3) In general, for a polynomial p(x) of degree n, the graph of y = p(x) intersects
the x-axis at at most n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.
EXERCISE 2.1
Q1. The graphs of y = p(x) are given in Fig. below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
1) The graphs of y = p(x) intersects or touches the x-axis when y = 0. The number
of zeroes of the polynomial p(x) corresponds to the values of x where p(x) = 0.
2) To find zeroes, set y = 0 in the equation y = p(x) and solve for all possible values
3) Simply observe how many times the graph intersects or touches the x-axis to
determine the number of zeroes.
Conclusion: The Power and Elegance of Polynomials
Polynomials are not just expressions with variables and coefficients – they are the keys to unlocking complex mathematical ideas. From their applications in algebra to their relevance in real-world
problems, mastering polynomials equips you with a versatile tool in mathematics. As you progress through this fascinating journey, remember that every equation you solve and every concept you
understand builds your foundation for more advanced topics. Embrace the elegance of polynomials and keep solving with curiosity!
#PolynomialPower #Class10Math #NCERTSyllabus #MathInspiration #AlgebraMagic #MathForLife #PolynomialJourney #NCERTClass10 #Polynomials #Mathematics #Class10 #NCERTMaths #MathSyllabus #CBSE #MathHelp
#StudyTips #Algebra #MathSolutions #NCERTSolutions #PolynomialEquations #Education #STEM #LearningMathematics
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Predicting AWS price reductions
I’m working on a disaster-recovery strategy at my day job, and AWS is one of the platforms I’m considering: the thinking is to have some capacity available on AWS, perhaps just a database and a
couple of EC2 instances, ready to take some traffic if our main provider has serious and continued availability issues.
Cost is, as always, a big factor in the equation: we’re essentially building ourselves some insurance that we’ll stay up despite major problems, and like with all insurances we want to pay the
smallest possible premium. Looking at the AWS pricing model, reserved instances immediately look like a way to both ensure capacity and get cheaper prices with respect to on-demand instances.
AWS reserved instances come in 1-year and 3-year types: the difference is in how much you pay upfront vs. how much you pay each month.
The question that comes to mind is:
should we buy 3-yr or 1-yr reserved instances?
To put it differently, what’s the chance that buying a 3-yr reserved instance today will give us a better deal, price-wise, than a 1-yr reserved instance?
A simple prediction model
To answer this question I’ve tried to predict the date of the next AWS price reduction for EC2 reserved instances. My source data for the previous EC2 price reductions is the official AWS blog, from
which I’ve extracted the following 12 dates:
2009-08-19 2009-11-01 2010-09-01
2012-03-05 2013-03-05 2013-11-05
2013-12-20 2014-03-26 2015-07-01
2016-01-01 2016-12-01 2017-05-03
To get an estimation of the next price reduction date, I converted the above values to unix epoch timestamps, and I then used Wolfram Alpha to find an equation which could give me the date of the
13th reduction. Here’s a couple of links if you want to try it for yourself:
The linear fit projects a straight line of the form y = 2.27002×10^7 x + 1.22852×10^9, which estimates the date of the next AWS price reduction on 2018-04-13 (that is 10 months from the time of this
The quadratic fit projects a curve of the form y = 1.21418×10^9 + 2.8846×10^7 x – 472747×x^2, which estimates an even closer date of 2017-10-29 (only 5 months from the time of this writing).
I’m not pretending that this simple exercise gives an accurate prediction, but the dates above, and especially the intervals between 2 consecutive price reductions, are an important element in my
decision of 1-yr vs. 3-yr purchase.
A real example
Let’s take as an example the purchase of one m4.4xlarge reserved instance in the US East region: we’re assuming that the purchase happens today, and we’re operating under the hypothesis that the next
price reduction is in 10 months from now.
I’m choosing a m4.4xlarge because you should begin minimizing the prices of the most expensive instances; the price differentials on t2.nano instances might not be worth the time it takes you to make
the decision.
Here’s a table summarizing the costs:
On Demand 1yr No Upfront 1yr Partial Upfront 1yr All Upfront 3yr No Upfront 3yr Partial Upfront 3yr All Upfront
Monthly costs 600.00 361.79 172.28 0.00 252.29 116.80 0.00
Upfront costs 0.00 0.00 2067.00 4052.00 0.00 4205.00 7905.00
Cost for 10 months 6000.00 3617.90 3789.80 4052.00 2522.90 5373.00 7905.00
Total cost 6000.00 4341.48 4134.36 4052.00 9082.44 8409.80 7905.00
Against best case 1948.00 289.48 82.36 0.00 5030.44 4357.80 3853.00
And here’s a link to play with the spreadsheet for your own instances.
You can see that, if you believe that a price reduction will come within 10-12 months from now, the best option is the 1-year no upfront RI: you don’t have to front up all of the capital, and there
are “only” 2 months wasted.
You can also see that it makes absolutely no sense to purchase 3-years RI, with our without upfront payments: in a 3-years time span you can expect at least 2 price reductions, making your investment
a much worse idea than it looked at the beginning.
Convertible instances are a poor mitigation to the problem: there’s no guarantee that you will find a more powerful instance type costing about the same as your convertible reservation, so you might
get stuck with your convertible reservation until the end of the 3-years contract.
If you think that you could purchase a reservation and then sell it on the RI instance marketplace, then think twice: why would someone purchase your reservation after a price reduction, given that
they can get the same from AWS for less money?
Does it make sense to purchase AWS reserved instances?
Reserved instances introduce one more complexity in the already much too complex AWS price model; regular price reductions make reserved instances a worse deal than it might look like, and negate the
predictability you think you’re getting.
In conclusion:
• there are a few good reasons to buy 1-year reservations, for example establishing a baseline capacity: the risk of price reductions is real, but you can mitigate it if you time your purchase just
after a reduction
• there are no good reasons to buy 3-years reservations: you’re locking in pricing, and opting out of both newer instance classes and future price reductions during that time period.
Do you want to know how to perform this kind of analysis? Do you need to choose where to host your app and you’re overwhelmed with informations and choice? Check out my book where I teach you how to
choose the best solution for your own unique needs.
1 Comment
1. Steve Jernigan
I’m not quite sure I understand your argument. For a second, let’s put the guaranteed availability and flexibility arguments to the side; things like new instance types or changes in needs. I
agree with those costs but putting a value on them is subjective. It looks like you are comparing the cost of 1yr of computing in the on demand and 1yr term pricing against the cost of 1yr of
service and 2yrs of discounted service in the 3 yr term case.
I used your assumptions, a price reduction every 10 months, and the last price reduction for the m4.4xlarge (16%) to extrapolate out to total costs for 3 years. That is, a 16% reduction every 10
months. My math shows the following total cost for each of your scenarios above.
On Demand $17407.33
1yr No Upfront $10806.60
1yr No Upfront $10291.03
1yr No Upfront $10314.77
3yr No Upfront $9082.44
3yr Partial Upfront $8409.80
3yr All Upfront $7905.00
To your point, I can see the value of flexibility being greater than the difference between the no-upfront cost options but it becomes a higher bar in the ‘all upfront’ scenarios. I think you
also need to account for the future cost of money but that is beyond the scope here. And as you say, the pricing options can become complex..and then you add spot instances. | {"url":"https://www.datafaber.com/2017/05/predicting-aws-price-reductions/","timestamp":"2024-11-07T07:25:20Z","content_type":"text/html","content_length":"37104","record_id":"<urn:uuid:21d8f433-05bf-4ee7-8c7b-e89079d49094>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00503.warc.gz"} |
Ask .. & Help by Answers
Asked by: at
Title: about equilibrium
Question: the potential energy of particle given by u(x)=a/x - b/(x^2)
the particle will be at
1.stable equilibrium at b/a
2.unstable equilibrium at b/a
3.stable equilibrium at at 2b/a
4.unstable equilibrium at 2b/a
(jnc entrance)
Visitor Ip: 112.110.134.102 | {"url":"https://ask.physicskerala.in/2010/05/question-about-equilibrium.html","timestamp":"2024-11-10T15:51:36Z","content_type":"text/html","content_length":"23604","record_id":"<urn:uuid:b316a616-0e65-45cf-a843-3856238d7a9a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00184.warc.gz"} |
Crushing Simulation - 911Metallurgist
In the context of this paper a simulation model for a size reduction device is an equation or equation set which allows the calculation of the product size distribution from the device for a
specified size distribution of feed. The model might be expressed as a simple algebraic expression, as a matrix equation, or as a finite difference formulation. The model proposed here for
once-through crushers was originally derived and developed by Austin, et al for the treatment of smooth double roll crushers. The model is based on the assumption that if the particle size range is
split into geometric size intervals (i.e. the √2 sieve series) the breakage of each size interval occurs independently of other sizes.
That is, if fi and pi give the weight fraction of material in size interval i in the crusher feed and product, respectively, the relationship between the pi and fi, can be expressed as the transfer
parameter equation (3)
where size 1 is the largest size, size 2 the second size, etc. In this equation, the d give the weight fraction of size j feed which is “transferred” to size i upon passage through the crusher. The
formulation of these transfer parameters is based on an analysis of the breakage process which is schematically illustrated in Figure 1 and can be described as follows.
When material of size i is passed through the crusher, a fraction d is left within the starting size i. Defining breakage as leading to material smaller than the starting size range, dii can be
considered as material which has bypassed through the crusher. Thus, a fraction dii = (1 – ai) bypasses through the crusher without breakage and a fraction ai is broken. The fraction ai for a given
size i is referred to as the primary bypass parameter for size i. It is also assumed that the first fracture leads to a set of primary daughter fragments. These primary daughter fragments are given
as the primary breakage distribution parameter set bij. The bij give the weight fraction of size j material which is broken to size i as a result of primary breakage. Daughter fragment material can
in turn bypass through the crusher or be selected for another fracture, with a fraction ai’ broken. The ai’ do not necessarily equal the ai because a fragment of size i produced by fracture in the
crushing zone might be expected to be in a better position to rebreak than one just entering the crusher, so that l>ai’>ai. The fraction ai’ for a given size i is erred to as the secondary bypass
parameter for size i. The ai and ai’ are assumed to be constant and independent of the size distribution of material being subjected to the bypass mechanisms.
The equations given by Austin et al for this repeated breakage process were
This formulation was put as equation (1) by Rogers (3) with
The cumulative crusher product weight size distribution is given by
where n is the smallest or sink size interval and P(xi) is the weight fraction of crusher product smaller than size xi, where x is the upper size of size interval i in millimetres.
It is interesting to compare this model to the one developed by Whiten and utilized by others for simulating cone crusher performance. Adopting the present nomenclature, the Whiten model is
illustrated schematically in Figure 2. As seen, this model employs the primary breakage distribution and material bypass concepts. Furthermore, this model gives a relationship between fi and pi that
is simply a special case of the more general once-through model formulation, that is Equations (1) and (3) but with ai = ai’. Thus the only essential difference between these models is in the
treatment of the bypass for particles produced by fracture.
Model Parameter Estimation
It is possible, in principle, to estimate the parameters of the model for once-through crushers using the same approach that Whiten, et al and Karra have taken for the cone crusher model. That is,
one can assume that model parameters can be described by equations which are a function of particle size and which contain a few descriptive parameters. Then by incorporating these equations into the
model, the descriptive parameters can be estimated by fitting the model to experimental data obtained from crusher tests. For the cone crusher model Karra notes that this approach can result in
parameter estimates that are reasonably consistent from one set of test data to another but can also lead to questionable parameter values. Another approach, and one that can also be used for the
Whiten model, calls for performing a series of crusher tests on the feed material of interest and results in direct measurements of parameters. Hits preferred approach, which has been demonstrated
for both laboratory and full-scale once-through crushers. Involves the experimental and calculational procedures described as follows. | {"url":"https://www.911metallurgist.com/blog/crushing-simulation/","timestamp":"2024-11-06T20:50:58Z","content_type":"text/html","content_length":"184664","record_id":"<urn:uuid:b3b9d602-a230-40af-95a6-25b9fd5fa5ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00046.warc.gz"} |
Question 5.16
1. In the Cagan model, if the money supply is expected to grow at some constant rate μ (so that Em[t+s] = m[t] + sμ), then Equation A9 can be shown to imply that p[t] = m[t] + γμ.
2. What happens to the price level p[t] when the money supply m[t] changes, holding the money growth rate μ constant?
3. What happens to the price level p[t] when the money growth rate μ changes, holding the current money supply m[t] constant?
4. If a central bank is about to reduce the rate of money growth μ but wants to hold the price level p[t] constant, what should it do with m[t]? Can you see any practical problems that might arise
in following such a policy?
5. How do your previous answers change in the special case where money demand does not depend on the expected rate of inflation (so that γ = 0)?
1 Milton Friedman and Anna J. Schwartz, A Monetary History of the United States, 1867–1960 (Princeton, NJ: Princeton University Press, 1963); Milton Friedman and Anna J. Schwartz, Monetary Trends in
the United States and the United Kingdom: Their Relation to Income, Prices, and Interest Rates, 1867–1975 (Chicago: University of Chicago Press, 1982).
2 Stanley Fischer, “Seigniorage and the Case for a National Money,” Journal of Political Economy 90 (April 1982): 295–313.
3 Mathematical note: This equation relating the real interest rate, nominal interest rate, and inflation rate is only an approximation. The exact formula is (1 + r) = (1 + i)/(1 + π). The
approximation in the text is reasonably accurate as long as r, i, and π are relatively small (say, less than 20 percent per year).
4 Robert B. Barsky, “The Fisher Effect and the Forecastability and Persistence of Inflation,” Journal of Monetary Economics 19 (January 1987): 3–24.
5 See, for example, Chapter 2 of Alan Blinder, Hard Heads, Soft Hearts: Tough-Minded Economics for a Just Society (Reading, MA: Addison Wesley, 1987).
6 Robert J. Shiller, “Why Do People Dislike Inflation?” in Reducing Inflation: Motivation and Strategy, edited by Christina D. Romer and David H. Romer (Chicago: University of Chicago Press, 1997), 1
7 The movie made about forty years later hid much of the allegory by changing Dorothy’s slippers from silver to ruby. For more on this topic, see Henry M. Littlefield, “The Wizard of Oz: Parable on
Populism,” American Quarterly 16 (Spring 1964): 47–58; and Hugh Rockoff, “The Wizard of Oz as a Monetary Allegory,” Journal of Political Economy 98 (August 1990): 739–760. It should be noted that
there is no direct evidence that Baum intended his work as a monetary allegory, so some people believe that the parallels are the work of economic historians’ overactive imaginations.
8 For an examination of this benefit of inflation, see George A. Akerlof, William T. Dickens, and George L. Perry, “The Macroeconomics of Low Inflation,” Brookings Papers on Economic Activity 1996,
no. 1: 1–76. Another argument for positive inflation is that it allows for the possibility of negative real interest rates. This issue is discussed in Chapter 12 in an FYI box on The Liquidity Trap.
9 For more on these issues, see Thomas J. Sargent, “The End of Four Big Inflations,” in Inflation, edited by Robert Hall (Chicago: University of Chicago Press, 1983), 41–98; and Rudiger Dornbusch and
Stanley Fischer, “Stopping Hyperinflations: Past and Present,” Weltwirtschaftliches Archiv 122 (April 1986): 1–47.
10 The data on newspaper prices are from Michael Mussa, “Sticky Individual Prices and the Dynamics of the General Price Level,” Carnegie-Rochester Conference on Public Policy 15 (Autumn 1981): 261–
11 This model is derived from Phillip Cagan, “The Monetary Dynamics of Hyperinflation,” in Studies in the Quantity Theory of Money, edited by Milton Friedman (Chicago: University of Chicago Press,
1956): 25–117. | {"url":"http://digfir-published.macmillanusa.com/mankiw9e/mankiw9e_ch5_14.html","timestamp":"2024-11-11T06:32:43Z","content_type":"text/html","content_length":"10069","record_id":"<urn:uuid:81fa6158-4dfc-4ba4-9e68-87be39a95784>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00041.warc.gz"} |
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ARR - How to Calculate ARR in Excel - QuickExcel
ARR – How to Calculate ARR in Excel
In this article, we’ll calculate ARR in Excel. Accounting Rate of Return is one of the calculation percentages of the rate of return, particularly relevant in long-term investments. If you are
planning to invest in a long-term project, then it’s necessary to check out the possibilities of the return from this benchmark percent.
So, it’s the best concept for computing the potential of the possibility of getting the percentage of return in this challenging financial environment.
In simple terms, it’s the return of investment and the way to whether to accept or not. Calculating the ARR in Excel is an easy task, with the help of Salvage Value, we can find out whether the
project is to be accepted or not meaning the product/commodity is worth investing in or not. Besides this, we also need to have an Average EADT, NCO for calculating the ARR in Excel.
Steps to Calculate ARR in Excel
ARR is the combination of an Average Annual Profit by Average Investment. Symbolically,
ARR = Average EADT/ (NCO + Salvage Value/2)
Using the spreadsheet it’s easy to track out the easy changes in the ARR. So, check out the steps to keep on calculating the ARR in Excel.
1. Creating the Year Date
At 1st, we need to create the date of a specific year. Also, you can assume the date for some years. Here we have assumed the time to be 6 years from cell A6:A11.
Assuming the data up to certain years gives an idea about coping with the challenges and rate of return from that investment.
2. Create CFAT Data
After assuming the project for a defined year, enter the CFAT Value of each year. Here for Year 0, we have entered the CFAT Value of (100000).
Enter the CFAT Value for all the assuming years. We have entered the value for all the years assumed.
3. Create EADT
Assume the different values for different years in EADT Column. Earnings After Depreciation and Taxes (EADT) for each year. Since Year 0 cannot earn any return, we’ve placed an empty value.
Enter an assuming or given EADT value for each year. We have entered the EADT value for all except the Base Year (Year 0) and Final Year (Year 6).
4. Calculate an Average EADT
To calculate the average of EADT, choose a cell for the result, here we have chosen another column cell to calculate an average.
Enter the syntax of Average, and select the EADT Column cells. It’s =AVERAGE(C6:C11).
Average EADT
Here you will get an Average EADT.
5. Calculation of NCO Value
Type, =SUMIF(range,criteria,(sum_range)
Select the range, it’s the CFAT Value. So, after SUMIF( select the CFAT(Cash Flow After Taxes) value. In our case, the CFAT Values are from B6:B11, which we enter in the function, followed by a
In the criteria, we have to include the condition. Here in our case Year 1 has a negative value, so we’ve to enter the criteria. Here we enter this again followed by a comma.
NCO in Excel
For a sum_range parameter, again select the CFAT Values. Here in our example, from B6:B11. After this, hit enter and see the NCO Result. It’s 100000 in our assumption calculation. Here we used for
NCO, =SUMIF(B6:B11,”<“&0,B6:B11)
Final Calculation Using Formula
Go to ARR Cells, here we are going to calculate the ARR forecast.
Select the Average EADT, like =C15. Here are our EADT Lies in Column C15. So, you have to select the average EADT Cell. After selecting the Average EADT, input the Divide sign (/).
After the divide sign, open a bracket and select the NCO value. If the NCO is negative, then you have to enter the ABS Function.
In our case, it’s negative. So, we enter the ABS before NCO. Like this, =C15/((ABS(C16)+.
Now Select the Salvage Value, it’s in B3 in our worksheet. Like this, =C15/((ABS(C16)+B3)/
Divide the value by 2 and close the bracket. It’s =C15/((ABS(C16)+B3)/2
Putting values in ARR function
ARR = Average EADT/ (NCO + Salvage Value/2)
In Above Example =C15/((ABS(C16)+B3)/2)
ARR final result
As per this calculation, we can make the decision whether to go for an investment or not.
That’s the end of the tutorial. Hope you have learned about ARR (Accounting Rate of Return) and how to calculate ARR in Excel and now can implement it yourself well to forecast an investment | {"url":"https://quickexcel.com/arr-in-excel/","timestamp":"2024-11-12T12:11:43Z","content_type":"text/html","content_length":"87548","record_id":"<urn:uuid:8285f021-04b0-4a49-8cdb-af8f33453482>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00334.warc.gz"} |
Epidemics on Networks
This activity simulates how an infectious disease may spread through a social network.
For epidemiologists it is important to know not only the number of cases that any one case may infect ($R_0$), but also how the outbreak may spread through a population. As such, it is vital to
understand the dynamics of a community or population.
This is done by looking at how individuals interact with each other - i.e. who comes into contact with whom, and how often. Mathematical modellers can then build this into their simulations to
understand how an outbreak has spread through a population.
This is vital for health researchers to understand, as it helps them to to contact trace individuals who may have become infected, so as to stop an outbreak spreading further.
Slides (
), dice, Printable network template (
Curriculum Links
• Calculating the probability of independent and dependent combined events
• Modelling situations mathematically
• To understand social network structure
• To understand how an infectious disease will spread differently depending on where it is started in the network.
Activity (Whole Class)
Introduce networks
Show the first slide with the two different networks. Ask students what they think the difference is (
Answer: Age
Discuss why these networking patterns may be different over time.
Show romantic network in US high school to spur further interest (optional)
Activity (Pair work)
Separate the class into pairs or small groups. Distribute print outs of the network, along with a dice to each pair/group.
Everyone starts off susceptible; pick one point of the network on the print out to be the first infected person.
Go around the infected person's contacts in turn. For each one, roll the die, if they get a 1 or a 2, that person also becomes infected. If it is anything else, they are immune.
Repeat for the new infected cases - and so on, until the epidemic ends.
Count how many cases are infected, and how many steps it took to infect the group.
Repeat the exercise several times, with different starting points. Note down the number of cases in each time
This data can then be used for further analysis - mean, median, mode, distribution, epidemic curve etc. Get students to plot graphs and analyse their results amongst their small group - or as a whole
Questions for thought
How/Why does the social network change between 4-5 year olds and 10 -11 year olds?Would you expect this network to change again for 16 year olds? What about for adults?What do the individual nodes in
the network represent?It might not be that they don't have any friends! They may have been absent on the day the data was collected, their contacts may not have consented to the study etc.Does the
romantic network in the US high-school surprise you?Why are we only infecting those nodes when a 1 or 2 is rolled?This is to take into account that not everyone in a network will necessarily be
susceptible to a diseaseWhat would happen if we allowed 1, 2, 3 or 4 to infect someone?What happens if you start in different places around the network?Why does the outbreak change in size each time
it is simulated?
Extension Activity: Targeted Vaccination (Whole Class)
Who would you vaccinate in the network?
If you only had 2 or 3 doses of vaccine for the network, who would you choose to vaccinate and why?
Would you protect people with the greatest number of links / break the network in certain places?
Students can do this individually and then feedback to the whole class. | {"url":"https://nrich.maths.org/projects/epidemics-networks","timestamp":"2024-11-09T04:32:37Z","content_type":"text/html","content_length":"38256","record_id":"<urn:uuid:047e974a-8d8e-41c1-bc46-63f3b7c2f87f>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00086.warc.gz"} |
These are the graphs designed to illustrate values of geographical items by means of lines or bars and in turn allow quantitative analysis.
The most useful statistical graphs for the illustration of values include the following.
• Line graphs
• Bar graphs
• Combined bars and line graph
These are the graphs which use line (s) to illustrate the values of items to give quantitative analysis.
Any line graph has two axes of the following:-
-X – axis; This is also known as the base or horizontal axis. It is used principally to show the value of independent variable like date or places.
-Y – axis: This is also known as the vertical axis. It is used show the values for the dependent variable of like output of crops, minerals etc.
Linear graphs are extremely varied. They are differently deigned to meet varied functions (roles). With respect to this consideration, linear graphs recognized to be of the following forms:
1. Simple line graph
2. Cumulative line graph
3. Divergent line graph
4. Group line graph
5. Compound line graph
Simple line graph
It is a form of line graph, designed to have one line to illustrate the values of one item in relation to dependent and independent variables. i.e. It is designed to show the values of one item per
varied date or places.
Consider the given hypothetical data below showing maize production for country X in 0,000 metric tons (1990 – 1995).
(a) Variables identification
Dependent variable ….. production values
Independent variable ….. Date (Years).
Y – axis …… production values
X – axis ……. Years
(b) Vertical and horizontal scales estimation
Hence; VS is 1 cm to 50000 tons.
Horizontal scale is up on decision
Hence; 1cm represents 1 year
MAIZE PRODUCTION FOR COUNTRY X IN (0,000) Metric tons
VS: 1 cm to 50 tons
HS: 1cm to 1 year
Hypothetical data
Strengths of the simple line graph
It is much easier to prepare as it involves to complicated mathematical works, and also a single line establishes the graph.
From the graph, the absolute values are extracted
It is comparatively easier to read and interpret the values
It has perfect replacement by simple bar graph
Setbacks of the simple line graph
It is a limited graphical method as only suited to represent the value for one item.
Sometimes it becomes difficult to assess the vertical scale if the variation between the highest and lowest values appear wider enough.
Cumulative line graph
It is a form of line graph designed to show the accumulated total values at various dates or possibly places for a single item. This graphical method has no alternative graphical bar method as it can
be compared to other linear graphical methods.
Construction of the cumulative line graph
Consider the given hypothetical data below showing maize production for country X.
(a) Variables identification
• Dependent variable ……….. production values
• Independent variable —- Date (Years)
Y – axis ………. Production values
X – axis …………Years
(b) Vertical and horizontal scales estimation
(c) Determination of the cumulative values.
Hence: VS; 1cm represents 50 tons
Thus; the cumulative lien graph appears as follow.
Cumulative line graph: Maize production for country X.
VS….. 1cm represents 50 tons
HS ….. 1cm represents 1 year
Source ….. Hypothetical data.
Merits of the cumulative line graph
The graphical method shows cumulative values
From the graph the values can be revealed and quantitatively analyzed
Setbacks of the cumulative line graph
The graphical method is not suited to show cumulative values for more than one item, it is thus; the graphical method limited for showing the values of a single item.
It needs high skill to reveal the actual values of the item represented
It has no alternative graphical bar method.
Divergent line graph
It is a form of line graph designed to illustrate the increase and decrease of the distribution values in relation to the mean. The graph is designed to have upper and lower sections showing positive
and negative values respectively.
The two portions are separated by the steady line graduated with zero value along the vertical line. The steady line also shows the average of all values.
Construction of the divergent line graph
Consider the following tabled data which show export values of coffee for country X in millions of dollars
YEAR EXPORT VALUES (000,000 dollars)
1953 256.5
1957 330.5
(a) Variables identification
• Dependent variable ……….. Export values
• Independent variable —- Date (Years)
Y – axis ………. Export values
X – axis …………Years
(b) Computation of the arithmetic mean
345 + 256 + 283 + 300 + 335 + 330.5 = 1850
Computation of the deviation values
1952 345-308 = 37
1953 256.5 – 308 = 52.5
1954 283-308 = -25
1955 300 – 308 = -8
1956 335 – 308 = 27
1957 330.5 – 308 = 22.5
(c) Estimation of the vertical scale.
Thus: the vertical scale
1cm represents 15 or -15 million dollars
(d) The graph has to be redrawn accordingly as follows:-
Hypothetical data
Vertical scale 1cm represents 15 or 15 tons
Horizontal scale 1cm represents 1 year
Merits of the divergent line graph
The graphical method is useful for showing increase and decrease of the values.
The graphical method shows the average of all values
It has perfect replacement by divergent bar graph
Setbacks of the divergent line graph
The graphical method is not suited to show the increase and decrease values for more than one items, it is thus; the graphical method is limited to a single item.
It needs high skill to reveal the actual values of the item represented.
It is time consuming graphical method as its preparation involves a lot of mathematical works.
It requires high skill to construct the divergent line graph.
Group line graph
It is a form of statistical line graph designed to have more than one lines of varied textures to illustrate the values of more than one items. Group line graph is alternatively known as composite,
comparative, and multiple line graph.
Construction of the group line graph
Consider the given data below showing values of export crops from Kenya (Ksh Million).
Crop/Year 1997 1998 1999 2000 2001
Tea 24,126 32,971 33,065 35150 34,448
Coffee 16,856 12,817 12,029 11,707 7,460
Horticulture 13,752 14,938 17,641 21,216 19,846
Tobacco 1,725 1,607 1,554 2,167 2,887
(a) Variables identification
Dependent variable …… export values
Independent variable …. Date (years)
Y – -axis………. export values
X – axis………..Years
(b) Verticals identification
Dependent variable……..export values
Independent variable …… Date (Years)
Hence; VS 1cm represents 5000 export value
Thus; the group line graph appears as follows:-
Vertical scale: 1cm to 5,000 export values
Source: Kenya Economic Survey 1969
Strengths of the group line graph
It is much easier to prepare as it involves no complicated mathematical works
It is useful graphical method for showing the values of more than one cases.
From the graph, the absolute values are extracted as the values directly shown
It is comparatively easier to read and interpret the values.
It has perfect replacement by group bar graph.
Setbacks of the group line graph
Some times; it becomes difficult to assess the vertical scale if the variation between the highest and lowest values appears wider enough
Crossing of the lines on the graph may confuse the interpreter.
A problem may arise in the selection of the varied line textures.
Compound line graph
It is a line graph designed to have more than one lines compounded to one another by varied shade textures to show the cumulative values of more than one items.
Construction of the compound line graph
Consider the given data below showing cocoa production for the Ghana provinces in 000 tons.
YEAR/PROV TV Togoland E. province W. province Ashanti
1947/48 40 40 30 35
1948/49 50 60 45 100
1949/50 45 46 89 110
1950/51 45 47 44 124
1951/52 47 23 50 100
1952/53 51 14 57 118
(a) Variables identification
Dependent variable…… export values
Independent variable ….. Date (Years)
Y – -axis……….export values
X – axis………..Years
(b) Cumulative values determination for the dates.
1947/48 40+40+30+35 = 145
1948/49 50+60+45+100 = 225
1949/50 45+46+89+110 = 290
1950/51 45+47+44+124 = 260
1951/52 47+23+50+100 = 220
1952/53 51+14+57+118= 240
(c) Vertical and horizontal scales determination
Hence; The vertical scale, 1cm represent 50 tons
Thus the graph appear as follow:-
Strengths of the compound line graph
It is useful graphical method for showing the cumulative values of more than one case.
Depending on the skill the interpreter has, from the graph, the absolute values are extracted as the value directly shown.
It has perfect replacement by compound bar graph
It is comparatively easier to assess the vertical scale to be used.
Setbacks of the compound line graph
It needs high skill to interpret the graph
It needs high skill to construct the graph
A problem may arise in the selection of the varied line textures.
These are the graphs which use bars to illustrate the values of items to give quantitative analysis.
Any bar graph has two axes
-X-axis; This is also known as the base or horizontal axis. It is used principally to show the values of independent variable like date or places.
– Y – axis; This is also known as the vertical axis. It is used show the values for the dependent variable of like output of crops, minerals etc.
Like line graphs, bar graphs are also extremely varied as differently designed to meet varied functions. With respect to this consideration, bar graphs categorized into the following:-
1. Simple bar graph
2. Divergent bar graph
3. Group bar graph
4. Compound bar graph
5. Percentage bar graph
6. Population pyramid
Simple bar graph
It is a form of bar graph, designed to have bars of similar texture to illustrate the values of one item in relation to dependent and independent variables. i.e. It is designed to show the values of
one item per varied date or places.
Construction of the simple bar graph
Consider the given data below showing cocoa purchase by areas, in 000 metric tons (1953)
Province Purchase
Ashanti 104
W-Province 39
E-Province 45
TV Togo land 22
(a) Variable identification
Dependent variable …… Purchase
Independent variable …. Provinces
Y – -axis………purchase values
X – axis………..Provinces
(b) Verticals identification
Dependent variable……..export values
Independent variable …… Date (Years)
Thus; the vertical scale: 1cm represents 20,000 tons.
Bar width – 1cm
Bar space = 0.5 cm
The graph has to be constructed accordingly.
COCOA PURCHASE BY PROVINCES (1953/54
Vertical scale; 1cm represents 20000 tons.
Strengths of the simple bar graph
It is much easier to prepare as it involves no complicated mathematical works, and also bars of similar texture established in the graph.
From the graph, the absolute values are extracted.
It is comparatively easier to read and interpret the values
It has perfect replacement by simple line graph.
Setbacks of the simple bar graph
It is a limited graphical method as only suited to represent the values for one item
Some times; it becomes difficult to assess the vertical scale if the variation between the highest and lowest values appear wider enough.
Divergent bar graph
It is a form of bar graph designed to illustrate the increase and decrease of the distribution values in relation to the mean. The graph is designed to have upper and lower sections showing positive
and negative values respectively.
The two portions are separated by the steady lien graduated with zero value along the vertical line. The steady lien also shows the average of all values.
Construction of the divergent line graph
Consider the following tabled data which show export values of coffee for country X in millions of dollars.
YEAR EXPORT VALUES (000,000 dollars)
1953 256.5
1957 330.5
(a) Variable identification
Dependent variable …… Export values
Independent variable …. Date (Years)
Y – -axis………. Export values
X – axis………..Years
(b) Computation of the arithmetic mean
345 + 256 + 283 + 300 + 335 + 330.5 + 1850
(c) Computation of the deviation values
1952 345 – 308 = 37
1953 256.5 – 308 = 52.5
1954 283 – 308 = -25
1955 300 – 308 = -8
1956 335 – 308 = 27
1957 330.5 – 308 = 22.5
(d) Estimation of the vertical scale
Thus: the vertical scale 1cm represents 15 or –15 million dollars
Bar width – 1cm
Bar space – 1cm
(e) The graph has to be redrawn accordingly as follows.
In million dollars
Vertical scale 1cm represents 15 or – 15 tons
Horizontal scale: 1cm represents 1 year
Source:- Hypothetical data
Merits of the divergent bar graph
The graphical method is useful for showing increase and decrease of the values
The graphical method shows the average of all values
It has perfect replacement by divergent line graph.
Setbacks of the divergent bar graph
The graphical method is not suited to show the increase and decrease values for more than one item, it is thus; the graphical method is limited to a single item.
It needs high skill to reveal the actual values of the item represented.
It is time consuming graphical method as its preparation involves a lot of mathematical work.
It requires high skill to construct the divergent bar graph.
Grouped bar graph
It is a form of statistical bar graph designed to have more than one bars of varied textures to illustrate the values of more than one items.
Grouped bar graph is alternatively known as composite, comparative, and multiple bar graph.
Construction of the group bar graph
Consider the given data below for cocoa purchase by provinces in Ghana (1947/48 – 1950/51)
YEAR/PROV TV Togoland E. province W. province Ashanti
1947/48 20 54 28 106
1948/49 26 80 46 126
1949/50 24 67 40 116
1950/51 22 72 45 123
(a) Variables identification
Dependent variable…… purchase values
Independent variable ….. Date
Y – -axis……… .purchase values
X – axis………..Date
(b) Vertical scale estimation
Hence; Vs, 1cm to 20,000 tons
Bar width = 1cm
Bar space = 1cm
(c) The graph should be drawn accordingly.
COCOA PURCHASE BY PROVINCES (1953/54)
Strengths of the grouped bar graph
It is much easier to prepare as it involves no complicated mathematical works
It is useful graphical method for showing the values of more than one cases.
From the graph, the absolute values are extracted as the value are directly shown
It is comparatively easier to read and interpret the values.
It has perfect replacement by group line graph.
Setbacks of the grouped graph
Some times; it becomes difficult to assess the vertical scale if the variation between the highest and lowest values appear wider enough.
A problem may arise in the selection of the varied bar textures.
Compound Bar graph
It is a bar graph designed to have bars divided proportionally showing the cumulative values of more than one items per varied dates or places
Compound bar graph is alternatively known as divided bar graph, or superimposed bar graph.
Construction of the compound bar graph
Consider the given data below showing cocoa production for the Ghana provinces in 000 tons.
Consider the given data below showing cocoa purchase by provinces (1947/48 to 1950/51)
REGION/YEAR 1947/48 1948/49 1949/50 1950/51
Ashanti 106,000 126,000 116,000 123,000
W.province 28,000 46,000 40,000 45,000
E.Province 54,000 80,000 67,000 72,000
T.Volta 20,000 26,000 24,000 22,000
(a) Variable identification
Dependent variable ….. export values
Independent variable … Date (Years).
Y – -axis………. purchase values
X – axis………..Years
(b) Cumulative values determination for the dates.
1947/48……….. 106,000 + 28,000 + 54,000 + 20,000 = 208
1948/49………… 126,000 + 46,000 + 80,000 + 26,000 = 278,000
1949/50 ………… 116,000 + 40,000 + 67,000 + 24,000 = 247,000
1950/51…………..123,000 + 45,000 + 72,000 + 22,000 = 262,000
(c) Vertical scale determination.
Thus; the VS … 1cm represents 50,000 tons.
The graph should be drawn accordingly.
COCOA PURCHASE BY PROVINCE (1947/48 – 1950/51)
Strength of the compound bar graph
It is useful graphical method for showing the cumulative values of more than one cases
Depending on the skill the interpreter has, from the graph, the absolute values are extracted as the value directly shown.
It has perfect replacement by compound line graph
It is comparatively easier to assess the vertical scale to be used.
Setbacks of the compound bar graph
It needs high skill to interpret the graph
It needs high skill to construct the graph
A problem may arise in the selection of the varied textures of the proportional segments
It is very fedious /tiresome as it involve mathematical calculation
It is time consuming in preparation
Percentage bar graph
In percentage bar graph, all bars must be drawn on the same height representing 100% and suitable scale is chosen such as 5, 10, 20 etc, and marked along the sides. The percentages of the total each
area stands for must start from zero line. Also it is advised to include the actual percentages of the face of the bars.
Construction of the Percentage bar graph
Consider the given data below showing cocoa purchase by provinces (1947/48 to 1950/51)
REGION/YEAR 1947/48 1948/49 1949/50 1950/51
Ashanti 106,000 126,000 116,000 123,000
W.province 28,000 46,000 40,000 45,000
E.Province 54,000 80,000 67,000 72,000
T.Volta 20,000 26,000 24,000 22,000
(a) Variables identification
Dependent variable ….. export values
Independent variable … Date (Years).
Y-axis………. purchase values
(b) Cumulative values determination for the dates.
1947/48……….. 106,000 + 28,000 + 54,000 + 20,000 = 208
1948/49………… 126,000 + 46,000 + 80,000 + 26,000 = 278,000
1949/50 ………… 116,000 + 40,000 + 67,000 + 24,000 = 247,000
1950/51…………..123,000 + 45,000 + 72,000 + 22,000 = 262,000
(c) The percentages by provinces in each year determination.
Hence; VS; 1 cm represents 20%
The percentage bar graph should be drawn accordingly as follow:-
COCOA PURCHASE BY PROVINCES (1947/48 – 1950/51)
Vertical scale; 1cm represents 20%
Strengths of the percentage bar graph
It is useful graphical method for showing the values of more than one cases
The data represented appear in a more simplified form as given in percentages.
It is comparatively easier to assess the vertical scale to be used
Setbacks of the percentage bar graph
It does not give the absolute values
It needs high skill to interpret the graph
It needs high skill to construct the graph
A problem may arise in the selection of the varied textures of the proportional segments
It consumes much time to be prepared.
Population pyramid graph
It is a form of bar graph designed to show population distribution by age and sex. It is a double bar chart showing the age sex structure of the population. It consists of two sets of horizontal
bars; one is for each sex showing either the p percentages or absolute numbers.
Rules for drawing the population pyramid graph
It is a principle in drawing population pyramid; the number of male population illustrated by the left set of bars; while that of females by the right set of bars.
The young population distribution is always at the bottom while that of old at the top.
Usually the last age group should be left open handled because; some people may survive beyond 100 years and their number have been omitted.
The bottom scale can be graduated as percentages or absolute numbers.
If percentages are opted to be used; the total population of both combined sexes should be used to compute the percentages.
After all the bars have been drawn, they can be shaded in one colour or separated colours for each sex.
There are two techniques of drawing the horizontal bars of an age sex pyramid. In the first
technique, the bars are drawn proportionally to the actual population numbers (absolute values).
In the second technique, the bars are drawn to represent percentages.
Age group Male Female Total
0 – 4 2291936 2242966 4534902
5-9 2000580 1962556 3963136
10-14 2034980 2003655 4038635
15-19 1681984 1721194 3403178
20-24 1328529 1504389 2832918
25-29 1094909 11664594 2259503
30-34 840692 845230 1685922
35-39 695263 723749 1419012
40-44 516502 516989 1033491
45-49 419841 418987 838828
50-54 344639 340167 684806
55-59 223691 236325 460016
60-64 194513 214715 409228
65-69 140969 160364 301333
70-74 118601 135524 254125
75-79 79166 81620 160786
80+ 95300 121038 216338
Age not stated 103487 86956 190443
All ages 14205589 14481018 28686607
The absolute value technique
The following steps are followed when constructing a population pyramid using absolute values.
Decide a suitable scale for the horizontal axis (baseline) by considering the values of the biggest and smallest age group, as well as the size of the paper on which the pyramid is to be drawn.
Horizontal scale is determined as follows.
Hence by considering the data in the table, scale of 1cm to represent 400,000 people would be suitable.
Choose a suitable scale for the vertical axis. This scale will determine how wide the bars will be and also the interval between the age groups. The width of the bars should not exceed 6mm otherwise
the pyramid will look untidy.
Take a clean graph paper and on it draw horizontal axis at least 3 cm from the bottom of the page. Draw two vertical axes of 1 cm apart and about 10 cm long, until they touch the horizontal axis.
Where the vertical axes touch the horizontal axis, mark as zero. On the horizontal axis, and at intervals of 1cm from the zero mark on the both sides, mark of the values representing the female and
male population
In the middle column, fill in the age groups starting with the youngest at the bottom. The age groups should be within the width of the horizontal bars.
Using the horizontal scale, and starting with the first age group for females, draw a bar from the vertical axis on the right hand side of the central column towards the right to represent the female
population of that group. The scale chosen in step 1 above will determine the length of the bar.
From the left hand side of the vertical axis, draw a bar representing the male population of the same age group. Steps 6 and 7 should be repeated for all the subsequent age group until the last one
has been represented.
Fig. 1.1 Kenya: Population by age and sex, 1999
The percentages technique
By this technique, the values for population distribution by age and sex given in percentages. The percentages of each female or male group over the total populations is calculated from the absolute
values in our example and a new set of data will be derived from data in the table.
This new data will be used to draw the graph. An example on how to calculate the percentage values is shown below.
The application for calculating the percentage is as follows.
For instance:
The absolute values for the females aged between 0-4 years from the table is 2242 966, while that for males is 2291936. The total populations according to the 1999 census, was 28686607. Therefore the
percentage of females is as follows:-
The percentage of male is as follows:-
The worked out percentage values from the figure in the table are given in the table next page.
Age Group 5male %female Total
1-4 8.0 7.8 15.8
5-9 7.0 6.8 13.8
10-14 7.1 7.0 14.1
15-19 5.9 6.0 11.9
20-24 4.6 5.2 9.8
25-29 3.8 4.1 7.9
30-34 2.9 2.9 5.8
35-39 2.4 2.5 4.9
40-44 1.8 1.8 3.6
45-49 1.5 1.5 3.0
50-54 1.2 1.2 2.4
55-59 0.8 0.8 1.6
60-64 0.7 0.7 1.4
65-69 0.5 0.6 1.1
70-74 0.4 0.5 0.9
75-79 0.3 0.3 0.6
80+ 0.3 0.4 0.7
After the calculation of the percentages, the following steps should be taken to come up with the age – sex pyramid.
Choose a suitable scale for the horizontal axis by considering the highest and the lowest percentages in the table. According to the values I the table, a scale of 1cm is representing 1% would be
Follow step 2 and 3 as outlined under the absolute values techniques discussed earlier.
Where the vertical axis touch the horizontal axis, mark zero and at intervals of 1cm, mark of the percentage value towards the right for females, and towards left for the males.
The age group should be indicated in the middle column just as we did when constructing an age sex pyramid using absolute values..
Using the horizontal scale and starting with age group 0-4 draw a bar on the right hand side to represent the percentage values of the female population in this age group. In our example, the
percentage is 7.8 Draw a similar bar on the left hand side to represent the value of the male population, which in our case is 0.8.
Draw bars to represent all the age groups follow steps 9 and 10 under the absolute value technique to complete the pyramid.
Kenya population by age:
Pyramid may also be for the purpose of making comparison either in terms of time or location. This can be by means of a double combined population pyramid. The double combined population pyramid
looks as follows.
Advantages of the age-sex pyramid
It is visually attractive method of presenting data.
A variety of information is shown on the same graph. The details include; age, sex and number of people
It can be used to compare the age sex structure of number of countries
It gives a clear picture and summary of the population composition of a country.
Disadvantages of the age-sex pyramid
It is tedious to construct because it involves many values.
It is difficult to tell the exact values represented because of the small scale of the horizontal axis.
Reasons for the differences in population numbers cannot be obtained from the graph directly. Therefore additional information has to be thought from elsewhere.
It is a form statistical graph designed to have both bars and line to show two attributes whose values appear in varied unit. It is basically employed to show the values of rainfall and temperature
together in a year.
In the graph, the bars used to illustrate the values on amount of rainfall in mm or inch, while the line is used to illustrate the values on amount of temperature in ^C or ^F. This is also known as
climo graph.
Construction of the bar and line graph
Consider the following climatic data for Dar-win weather station Australia.
Month J F M A M J J A S O N D
Temp ^oC 28.9 27.8 28.9 29 26.7 26 25.1 26.4 28.1 29.7 29.8 29
Rain(mm) 388 330 246 114 17.8 5 2.5 2.5 12.7 53.3 132 261
(i) Identification of the variables
· Dependent variable – Rain and temperature values
· Independent variable – Data (months).
Y – -axis – Rain and temperature values
X – axis………..months
(ii) Estimation of the vertical scale to be used
Thus; the vertical scale for rainfall is 1cm to 50mm.
Thus; the vertical scale for temperature is 1 cm to 10 c
(iii) The graph has to be drawn as follows;
Strengths of the combined bar and line graph
It is useful graphical method for showing the distribution values of climate
It is more illustrative, as it provides visual idea to the users in statistics.
It allows the easy making of quantitative analysis
Setbacks of the combined bar and line graph
It is more illustrative, as it provides visual idea to the users in statistics
Needs high skill to make quantitative analysis from the graph
It is time consuming graphical method in construction
It needs high skill to construct the graph
It is tedious as it involves mathematical calculation
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short-term financial trading strategies that employ mean reversion models. Statistical Arbitrage Statistical Arbitrage is a class of short-term financial trading strategies that employ mean reversion
models, similar to a pairs. I am going to introduce a semi-variance model for statistical arbitrage. The model is compared to the standard Garch model, which is often used in daily option. Also
called statistical arbitrage. May also involve trading single stocks versus an index or an Exchange-Traded Fund (ETF) against an index. Convertible. The index is designed to provide a broad measure
of the performance of underlying hedge fund managers The index is base weighted at at Dec , does not. Further, statistical arbitrage is one of the quant trading strategies that identify statistically
significant mispricing among a very large. Definitions: Volatility: a statistical measure of the dispersion of returns for a given security or index. Beta: A measure of systematic risk of a
portfolio. What Is Statistical Arbitrage? A stat arb refers to a group of trading strategies that utilise mean reversion and analysis to invest in diverse securities.
Statistical arbitrage, often referred to as "stat arb," is a trading strategy that aims to profit from short-term price discrepancies in financial assets. This. Pairs trading and Stat Arb is still a
highly effective and lucrative strategy. All the funds you mention above have employee/ close investors. Index performance for Bloomberg Statistical Arbitrage Hedge Fund Index (BHSTA) including
value, chart, profile & other market data. Statistical arbitrage trading strategy involves buying and selling the same or similar asset in different markets to take advantage of price differences.
Essentially, stat arb seeks to exploit a mean-reverting trend in prices between related securities, or baskets of securities. A number of variations exist, also. By the time Y2K (the year software
problem) bugs were scaring everyone, hedge funds and prop trading desks were already schooling computers to do the heavy. Investment Objective and Principle Strategies. The fund exploits arbitrage
possibilities occurring in the market, supported by algorithm-based systems. , 2. 2. Λ≤. ∴. Λ. ±≤. ≤Λ∴. ≤+. =Λ∴. = = ≤Λ∴. ≤. =Λ. ∑. Investment funds. Page 5. Sharpe Ratio σ. µ. µ σ. µ r. S t. E. EE.
tN. E. Arbitrage means trading for guaranteed profit by buying in one market/asset and selling in another market or equivalent asset at the same time.
The strategy may also seek to hedge a portion of the interest rate risk. Equity Statistical Arb. Our Equity Statistical Arbitrage strategy seeks to capture. It involves the simultaneous buying and
selling of security portfolios according to predefined or adaptive statistical models. Statistical arbitrage techniques. With a sufficiently diverse portfolio of stat arb trades, the strategy can
effectively become a money printer that rarely loses. Now, although. In the context of hedge funds, a style of management that employs complex statistical models that try to capture small
abnormalities in a security's intraday. Statistical arbitrage means reverting strategies tend to be greater return contributors in times of higher volatility and can therefore provide portfolio.
The Best Growth Stock Mutual Funds | Etrade Vs Ally | {"url":"https://sswdraft.site/learn/stat-arb-funds.php","timestamp":"2024-11-14T06:46:42Z","content_type":"text/html","content_length":"14535","record_id":"<urn:uuid:f9c53df6-19a2-4b21-bc7d-ffba67e8aac6>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00738.warc.gz"} |
Uniform Circular Motion Page 3
Substituting the circumference formula for "X" and the symbol for the period of time makes the formula
since the velocity in all of these formulas is the magnitude only and the magnitude of the velocity never changes.
To define a velocity formula, we need to look at the definition of a vector. But first a question.
• Question
• Answer
Apply our vector knowledge to circular motion. To begin with we need some velocity vectors. Watch the animation below to see where we will get 8 velocity vectors.
We know that two vectors are the same if they have the same magnitude and direction.(This is what your answer to the question above illustrates.) This means that if I slide a vector anywhere on the
page and it will be the same vector as long as it has the same direction and magnitude. Watch the animation below to how we are finding 8 velocity vectors for 8 different positions.
Look for a pattern to derive a new formula. On the circle on the left the circumference represents the change in position. Because circle is traveled once around the time is the period. This gives us
the formula:
The circle on the right replaces the radius vector with the velocity vector. This means that we can make the same substitution in the formula above.
What is defined as the change in velocity over the change in time? | {"url":"https://www.mrwaynesclass.com/circular/index03.html","timestamp":"2024-11-13T03:22:04Z","content_type":"text/html","content_length":"9973","record_id":"<urn:uuid:6dfe7bbb-869b-44e9-b96d-8ea550d69ca0>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00619.warc.gz"} |
Compute additional classification performance metrics
Since R2022b
rocmetrics computes the false positive rates (FPR), true positive rates (TPR), and additional metrics specified by the AdditionalMetrics name-value argument. After creating a rocmetrics object, you
can compute additional classification performance metrics by using the addMetrics function.
UpdatedROCObj = addMetrics(rocObj,metrics) computes additional classification performance metrics specified in metrics using the classification model information stored in the rocmetrics object
UpdatedROCObj contains all the information in rocObj plus additional performance metrics computed by addMetrics. The function attaches the additional computed metrics (metrics) as new variables in
the table of the Metrics property.
If you compute confidence intervals when you create rocObj, the addMetrics function computes the confidence intervals for the additional metrics. The new variables in the Metrics property contain a
three-column matrix in which the first column corresponds to the metric values, and the second and third columns correspond to the lower and upper bounds, respectively. Using confidence intervals
requires Statistics and Machine Learning Toolbox™.
Compute Additional Metrics
Compute the performance metrics (FPR, TPR, and expected cost) for a multiclass classification problem when you create a rocmetrics object. Compute additional metrics, the positive predictive value
(PPV) and the negative predictive value (NPV), and add them to the object.
Load a sample of true labels and the prediction scores for a classification problem. For this example, there are five classes: daisy, dandelion, roses, sunflowers, and tulips. The class names are
stored in classNames. The scores are the softmax prediction scores generated using the predict function. scores is an N-by-K array where N is the number of observations and K is the number of
classes. The column order of scores follows the class order stored in classNames.
scores = flowersData.scores;
trueLabels = flowersData.trueLabels;
classNames = flowersData.classNames;
Create a rocmetrics object by using the true labels and the classification scores. Specify the column order of scores using classNames. By default, rocmetrics computes the FPR and TPR. Specify
AdditionalMetrics="ExpectedCost" to compute the expected cost as well.
rocObj = rocmetrics(trueLabels,scores,classNames, ...
The table in the Metrics property of rocObj contains performance metric values for each of the classes, vertically concatenated according to the class order. Find and display the top rows for the
second class in the table.
idx = rocObj.Metrics.ClassName == classNames(2);
ClassName Threshold FalsePositiveRate TruePositiveRate ExpectedCost
_________ _________ _________________ ________________ ____________
dandelion 1 0 0 0.045287
dandelion 1 0 0.23889 0.034469
dandelion 1 0 0.26111 0.033462
dandelion 1 0 0.27222 0.032959
dandelion 1 0 0.28889 0.032204
dandelion 1 0 0.29444 0.031953
dandelion 1 0 0.3 0.031701
dandelion 1 0 0.31111 0.031198
The table in Metrics contains the variables for the class names, threshold, false positive rate, true positive rate, and expected cost (the additional metric).
After creating a rocmetrics object, you can compute additional metrics using the classification model information stored in the object. Compute the PPV and NPV by using the addMetrics function. To
overwrite the input argument rocObj, assign the output of addMetrics to the input.
rocObj = addMetrics(rocObj,["PositivePredictiveValue","NegativePredictiveValue"]);
Display the Metrics property for the top rows.
ClassName Threshold FalsePositiveRate TruePositiveRate ExpectedCost PositivePredictiveValue NegativePredictiveValue
_________ _________ _________________ ________________ ____________ _______________________ _______________________
dandelion 1 0 0 0.045287 NaN 0.7551
dandelion 1 0 0.23889 0.034469 1 0.80202
dandelion 1 0 0.26111 0.033462 1 0.80669
dandelion 1 0 0.27222 0.032959 1 0.80904
dandelion 1 0 0.28889 0.032204 1 0.81259
dandelion 1 0 0.29444 0.031953 1 0.81378
dandelion 1 0 0.3 0.031701 1 0.81498
dandelion 1 0 0.31111 0.031198 1 0.81738
The table in Metrics now includes the PositivePredictiveValue and NegativePredictiveValue variables in the last two columns, in the order you specified. Note that the positive predictive value (PPV =
TP/(TP+FP)) is NaN for the reject-all threshold (largest threshold), and the negative predictive value (NPV = TN/(TN+FN)) is NaN for the accept-all threshold (lowest threshold). TP, FP, TN, and FN
represent the number of true positives, false positives, true negatives, and false negatives, respectively.
Input Arguments
metrics — Additional model performance metrics
character vector | string array | function handle | cell array
Additional model performance metrics to compute, specified as a character vector or string scalar of the built-in metric name, string array of names, function handle (@metricName), or cell array of
names or function handles. A rocmetrics object always computes the false positive rates (FPR) and the true positive rates (TPR) to obtain a ROC curve. Therefore, you do not have to specify to compute
FPR and TPR.
• Built-in metrics — Specify one of the following built-in metric names by using a character vector or string scalar. You can specify more than one by using a string array.
Name Description
"TruePositives" or "tp" Number of true positives (TP)
"FalseNegatives" or "fn" Number of false negatives (FN)
"FalsePositives" or "fp" Number of false positives (FP)
"TrueNegatives" or "tn" Number of true negatives (TN)
"SumOfTrueAndFalsePositives" Sum of TP and FP
or "tp+fp"
"RateOfPositivePredictions" or Rate of positive predictions (RPP), (TP+FP)/(TP+FN+FP+TN)
"RateOfNegativePredictions" or Rate of negative predictions (RNP), (TN+FN)/(TP+FN+FP+TN)
"Accuracy" or "accu" Accuracy, (TP+TN)/(TP+FN+FP+TN)
"FalseNegativeRate", "fnr", or False negative rate (FNR), or miss rate, FN/(TP+FN)
"TrueNegativeRate", "tnr", or True negative rate (TNR), or specificity, TN/(TN+FP)
"PositivePredictiveValue", Positive predictive value (PPV), or precision, TP/(TP+FP)
"ppv", "prec", or "precision"
"NegativePredictiveValue" or Negative predictive value (NPV), TN/(TN+FN)
Expected cost, (TP*cost(P|P)+FN*cost(N|P)+FP*cost(P|N)+TN*cost(N|N))/(TP+FN+FP+TN), where cost is a 2-by-2 misclassification cost matrix containing [0,cost(N|P);
cost(P|N),0]. cost(N|P) is the cost of misclassifying a positive class (P) as a negative class (N), and cost(P|N) is the cost of misclassifying a negative class as
a positive class.
"ExpectedCost" or "ecost"
The software converts the K-by-K matrix specified by the Cost name-value argument of rocmetrics to a 2-by-2 matrix for each one-versus-all binary problem. For
details, see Misclassification Cost Matrix.
"f1score" F1 score, 2*TP/(2*TP+FP+FN)
You can obtain all of the previous metrics by specifying "all". You cannot specify "all" in conjunction with any other metric.
The software computes the scale vector using the prior class probabilities (Prior) and the number of classes in Labels, and then scales the performance metrics according to this scale vector. For
details, see Performance Metrics.
• Custom metric — Specify a custom metric by using a function handle. A custom function that returns a performance metric must have this form:
metric = customMetric(C,scale,cost)
□ The output argument metric is a scalar value.
□ A custom metric is a function of the confusion matrix (C), scale vector (scale), and cost matrix (cost). The software finds these input values for each one-versus-all binary problem. For
details, see Performance Metrics.
☆ C is a 2-by-2 confusion matrix consisting of [TP,FN;FP,TN].
☆ scale is a 2-by-1 scale vector.
☆ cost is a 2-by-2 misclassification cost matrix.
The software does not support cross-validation for a custom metric. Instead, you can specify to use bootstrap when you create a rocmetrics object.
Note that the positive predictive value (PPV) is NaN for the reject-all threshold for which TP = FP = 0, and the negative predictive value (NPV) is NaN for the accept-all threshold for which TN = FN
= 0. For more details, see Thresholds, Fixed Metric, and Fixed Metric Values.
Example: ["Accuracy","PositivePredictiveValue"]
Example: {"Accuracy",@m1,@m2} specifies the accuracy metric and the custom metrics m1 and m2 as additional metrics. addMetrics stores the custom metric values as variables named CustomMetric1 and
CustomMetric2 in the Metrics property.
Data Types: char | string | cell | function_handle
Output Arguments
UpdatedROCObj — Object evaluating classification performance
rocmetrics object
Object evaluating classification performance, returned as a rocmetrics object.
To overwrite the input argument rocObj, assign the output of addMetrics to rocObj:
rocObj = addMetrics(rocObj,metrics);
Version History
Introduced in R2022b
R2024b: Additional metrics available
addmetrics has new metrics:
• "f1score", which computes the F1 score.
• "precision", which is the same as "ppv" and "prec".
• "all", which computes all supported metrics. You cannot use "all" in combination with any other metric. | {"url":"https://uk.mathworks.com/help/deeplearning/ref/rocmetrics.addmetrics.html","timestamp":"2024-11-11T09:34:59Z","content_type":"text/html","content_length":"102357","record_id":"<urn:uuid:9d4b0988-6a3a-43be-a6d2-c26e08151eb0>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00879.warc.gz"} |
Excel Formula: Weeks Between Dates
In this guide, we will learn how to calculate the number of weeks between two dates in Excel using the ROUNDUP and DAYS functions. This formula can be useful for various applications, such as project
planning, time tracking, and scheduling. By following the step-by-step explanation below, you will be able to implement this formula in your Excel spreadsheets and easily calculate the number of
weeks between any two dates.
An Excel formula
=ROUNDUP(DAYS(E4, D4) + 1, 0) / 7
Formula Explanation
This formula calculates the number of weeks between two dates and rounds up the result to the nearest whole number.
Step-by-step explanation
1. The DAYS function is used to calculate the number of days between two dates. In this case, the start date is in cell D4 and the end date is in cell E4.
2. The result of the DAYS function is added to 1 to include the end date in the calculation. This ensures that if the start and end dates are the same, the result will be 1 week.
3. The ROUNDUP function is used to round up the result of the previous step to the nearest whole number. The second argument of the ROUNDUP function is set to 0, indicating that we want to round up
to the nearest whole number.
4. Finally, the result of the ROUNDUP function is divided by 7 to convert the number of days to weeks.
For example, if the start date in cell D4 is 01/01/2022 and the end date in cell E4 is 01/15/2022, the formula =ROUNDUP(DAYS(E4, D4) + 1, 0) / 7 would calculate the number of weeks between the two
dates as follows:
1. Calculate the number of days between the two dates: DAYS(01/15/2022, 01/01/2022) = 14
2. Add 1 to include the end date: 14 + 1 = 15
3. Round up to the nearest whole number: ROUNDUP(15, 0) = 15
4. Divide by 7 to convert days to weeks: 15 / 7 = 2.142857143
The result is approximately 2.14 weeks. | {"url":"https://codepal.ai/excel-formula-generator/query/535Judbz/excel-formula-roundup-days-weeks","timestamp":"2024-11-12T14:18:23Z","content_type":"text/html","content_length":"79260","record_id":"<urn:uuid:7933cba5-b109-45a7-9768-2eab9d47f860>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00608.warc.gz"} |
Initial Orbit Determination based on the High-order Propagation of Orbit Sets
Document details
Publishing year2017 PublisherESA Space Debris Office Publishing typeConference Name of conference7th European Conference on Space Debris Pagesn/a Volume Issue Editors
Every day thousands of detections of objects orbiting the Earth are retrieved by observatories. However, a single observation with either radar or optical sensors is in general not sufficient to
accurately determine the orbit of an unknown object. Due to the observation geometry and the uncertainty related to sensor accuracy, timing accuracy, and observer state knowledge, we obtain a
relatively large orbits set (OS) that is compatible with the single observation rather than a single orbital state. When two independent observations are linked to the same object, the orbit
knowledge can be refined by estimating a compatible initial orbit guess from the OSs obtained from the single observations. Our approach for the “linkage problem” is to look at the correlation of the
two OSs, which means that they need to be sampled, propagated in time to a common epoch and compared to decide whether they pertain to the same object. The initial effort of our project focuses on
the definition of the OS using Differential Algebra (DA). This is obtained by mapping the uncertainties from the measurement space to the state space through a high-order map. This allows us to
obtain the object state as a truncated Taylor expansion in the measurements. Once the OSs are defined, the next step is to sample them with the goal of keeping the estimated error introduced by the
truncated expansion within a certain threshold in the entire region. The approach for the sampling comes from the estimation of the radius of convergence of the truncated polynomial: outside the
convergence disk, the error of the polynomial approximation violates the threshold, which means that most probably the OS cannot be described by just one Taylor expansion but needs to be divided in
sub-regions. For this reason, the Automatic Domain Splitting (ADS) tool is introduced. The ADS splits the OS into two or more regions defined by just as many Taylor expansions when the estimated
error between the real value and the Taylor approximation exceeds a certain tolerance. This can be done when the OS is created as well as when it is propagated in time, since the error of the Taylor
expansion grows along the trajectory due to the nonlinearity of the dynamic. The accuracy of the Taylor expansion is checked by evaluating the error produced by the high-order terms. The region is
then split in the direction of the highest estimated error when the threshold is exceeded. The output of the procedure is a set of truncated Taylor expansions that represents the initial OS with the
necessary precision in every part of the region. This work presents a comparison between the proposed initial orbit determination approach and alternative methods from the literature to show the
improvements we achieved by exploiting accuracy information using DA. The OSs resulting from independent observations are then propagated to a common epoch with the ADS tool and compared to find the
probability of intersection, which is used as a measure of the likelihood of their correlation. | {"url":"https://conference.sdo.esoc.esa.int/proceedings/sdc7/paper/968","timestamp":"2024-11-14T01:53:02Z","content_type":"text/html","content_length":"25786","record_id":"<urn:uuid:8b841c19-e922-495e-9cce-4050bee64475>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00759.warc.gz"} |
Digital Signatures
Definition. A signature scheme $\mc{S} = (G, S, V)$ is a triple of efficient algorithms, where $G$ is a key generation algorithm, $S$ is a signing algorithm, and $V$ is a verification algorithm.
□ A probabilistic algorithm $G$ outputs a pair $(pk, sk)$, where $sk$ is called a secret signing key, and $pk$ is a public verification key.
□ Given $sk$ and a message $m$, a probabilistic algorithm $S$ outputs a signature $\sigma \la S(sk, m)$.
□ $V$ is a deterministic algorithm that outputs either $\texttt$ or $\texttt{reject}$ for $V(pk, m, \sigma)$.
The correctness property requires that all signatures generated by $S$ is always accepted by $V$. For all $(pk, sk) \la G$ and $m \in \mc{M}$,
\[\Pr[V(pk, m, S(sk, m)) = \texttt] = 1.\]
Properties of Digital Signatures
• Digital signatures can be verified by anyone, whereas MACs can be verified by the parties sharing the same key.
□ No need to share a key for digital signatures.
• Non-repudiation: cannot deny having created the signature.
□ Signatures can only be created by people having the secret key.
□ In cases where the secret key is leaked, then we don’t have non-repudiation.
□ In MACs, the secret key is shared by two parties, so we don’t have non-repudiation.
• Must trust the identity of the public key.
□ How do you trust that this public key is Alice’s?
□ We need public key infrastructure (PKI).
• Electronic document signing
• HTTPS/TLS certificates
• Software installation
• Authenticated email (DKIM)
• Bitcoins
Secure Digital Signatures
The definition is similar to the secure MAC. The adversary can perform a chosen message attack, but cannot create an existential forgery.
Definition. Let $\mc{S} = (G, S, V)$ be a signature scheme defined over $(\mc{M}, \Sigma)$. Given an adversary $\mc{A}$, the game goes as follows.
1. The challenger generates $(pk, sk) \la G()$ and sends $pk$ to $\mc{A}$.
2. $\mc{A}$ makes a series of signing queries to the challenger.
☆ Each query is a message $m_i \in \mc{M}$, the challenger responds with $\sigma_i \la S(sk, m_i)$.
3. $\mc{A}$ computes and outputs a candidate forgery pair $(m, \sigma) \in \mc{M} \times \Sigma$.
☆ $m \notin \left\lbrace m_1, \dots, m_q \right\rbrace$.
☆ $(m, \sigma) \notin \left\lbrace (m_1, \sigma_1), \dots, (m_q, \sigma_q) \right\rbrace$. (strong)
$\mc{A}$ wins if $V(pk, m, \sigma) = \texttt{accept}$, let this event be $W$. The advantage of $\mc{A}$ with respect to $\mc{S}$ is defined as
\[\rm{Adv}_{\rm{SIG}}[\mc{A}, \mc{S}] = \Pr[W].\]
If the advantage is negligible for all efficient adversaries $\mc{A}$, the signature scheme $S$ is (strongly) secure. $\mc{S}$ is existentially unforgeable under a chosen message attack.
• We do not make verification queries, since the adversary can always check any signature.
• The normal definition of security is sufficient. Secure signature schemes can be converted into strongly secure signature schemes. See Exercise 14.10.^1
Message Confusion
Two different messages $m, m’$ can produce the same signature $\sigma$. In this case, the scheme is vulnerable to message confusion. See Exercise 13.3.^1
In common implementations, we consider $m$, $m’$ both to be valid. But there may be situations that this is undesirable. For those cases, a signature is would be a binding commitment to the message,
and there will be no confusion.
Signer Confusion
Suppose that $(m, \sigma)$ is a valid pair with $pk$, i.e, $V(pk, m, \sigma) = \texttt{accept}$. But an attacker can generate $pk’$ different from $pk$ such that $V(pk’, m, \sigma) = \tt{accept}$. In
this cases, we have signer confusion since both can claim to have signed $m$. See Exercise 13.4.^1
Strongly Binding Signatures
Strongly binding signatures prevent both message confusion and signer confusion.
Any signature scheme can be made strongly binding by appending a collision resistant hash of $(pk, m)$ to the signature. See Exercise 13.5.^1
Extending the Message Space
We can extend the message space of a secure digital signature scheme, as we did for MACs. Let $\mc{S} = (G, S, V)$ be a signature scheme defined over $(\mc{M}, \Sigma)$ and let $H : \mc{M}’ \ra \mc
{M}$ be a hash function with $\left\lvert \mc{M}’ \right\lvert \geq \left\lvert \mc{M} \right\lvert$.
Define a new signature scheme $\mc{S}’ = (G, S’, V’)$ over $(\mc{M}’, \Sigma)$ as
\[S'(sk, m) = S(sk, H(m)), \qquad V'(pk, m, \sigma) = V(pk, H(m), \sigma).\]
This is often called the hash-and-sign paradigm, and the new signature scheme is also secure.
Theorem. Suppose that $\mc{S}$ is a secure signature scheme and $H$ is a collision resistant hash function. Then $\mc{S}’$ is a secure signature.
If $\mc{A}$ is an adversary attacking $\mc{S}’$, then there exist an adversary $\mc{B}_\mc{S}$ attacking $\mc{S}$ and an adversary $\mc{B}_H$ attacking $H$ such that
\[\rm{Adv}_{\rm{SIG}}[A, \mc{S}'] \leq \rm{Adv}_{\rm{SIG}}[\mc{B}_\mc{S}, \mc{S}] + \rm{Adv}_{\rm{CR}}[\mc{B}_H, H].\]
Proof. The proof is identical to the theorem for MACs.
Digital Signature Constructions
We can build secure signature schemes from hash functions, trapdoor permutations, or from discrete logarithms.
Textbook RSA Signatures
This is the signature scheme based on the textbook RSA. It is also insecure.
• Key generation: $pk = (N, e)$ and $sk = (N, d)$ are chosen to satisfy $d = e^{-1} \bmod \phi(N)$ for $N = pq$.
• Sign: $S(sk, m) = m^d \bmod N$.
• Verify: $V(pk, m, \sigma)$ returns $\texttt{accept}$ if and only if $\sigma^e = m \bmod N$.
Here are some possible attacks.
• No message attack
□ Just return $(\sigma^e, \sigma)$ for some $\sigma$. Then it passes verification.
• Attack using the homomorphic property.
□ Suppose we want to forge a message $m$.
□ Pick $m_1 \in \Z_N^{\ast}$ and set $m_2 = m\cdot m_1^{-1} \bmod N$.
□ Query signatures for both messages and multiply the responses.
☆ $\sigma = \sigma_1 \cdot \sigma_2 = m_1^e \cdot m^e \cdot m_1^{-e} = m^e \bmod N$.
□ Then $(m, \sigma)$ is a valid pair.
Because of the second attack, the textbook RSA signature is universally forgeable. This property is used to create blind signatures, where the signer creates a signature without any knowledge about
the message. See Exercise 13.15.^1
RSA Full Domain Hash Signature Scheme
Given a hash function $H : \mc{M} \ra \mc{Y}$, the RSA full domain hash signature scheme $\mc{S}_\rm{RSA-FDH}$ is defined as follows.
• Key generation: $pk = (N, e)$ and $sk = (N, d)$ are chosen to satisfy $d = e^{-1} \bmod \phi(N)$ for $N = pq$.
• Sign: $S(sk, m) = H(m)^d \bmod N$.
• Verify: $V(pk, m, \sigma)$ returns $\texttt{accept}$ if and only if $\sigma^d = H(m) \bmod N$.
This scheme is now secure.
Theorem. If the hash function $H$ is modeled as a random oracle, and the RSA assumptions holds, then $\mc{S}_\rm{RSA-FDH}$ is a secure signature scheme.
For any $q$-query adversary $\mc{A}$ against hashed RSA, there exists an adversary $\mc{B}$ solving the RSA problem such that
\[\rm{Adv}_{\rm{SIG}}[\mc{A}, \mc{S}_\rm{RSA-FDH}] \leq q \cdot \rm{Adv}_{\rm{RSA}}[\mc{B}].\]
Full Domain Hash Signature Scheme
The following is a description of a full domain hash scheme $\mc{S}_\rm{FDH}$, constructed from trapdoor permutation scheme $\mc{T} = (G, F, I)$.
• Key generation: $(pk, sk) \la G()$.
• Sign: $S(sk, m)$ returns $\sigma \la I(sk, H(m))$.
• Verify: $V(pk, m, \sigma)$ returns $\texttt{accept}$ if and only if $F(pk, \sigma) = H(m)$.
This scheme $\mc{S}_\rm{FDH} = (G, S, V)$ is secure if $\mc{T}$ is a one-way trapdoor permutation and $H$ is a random oracle.
Theorem. Let $\mc{T} = (G,F,I)$ be a one-way trapdoor permutation defined over $\mc{X}$. Let $H : \mc{M} \ra \mc{X}$ be a hash function, modeled as a random oracle. Then the derived FDH signature
scheme $\mc{S}_\rm{FDH}$ is a secure signature scheme.
Proof. See Theorem 13.3.^1
Schnorr Digital Signature Scheme
This one uses discrete logarithms.
The Schnorr Identification Protocol
This scheme is originally from the Schnorr identification protocol.
Let $G = \left\langle g \right\rangle$ be a cyclic group of prime order $q$. We consider an interaction between two parties, prover $P$ and a verifier $V$. The prover has a secret $\alpha \in \Z_q$
and the verification key is $u = g^\alpha$. $P$ wants to convince $V$ that he knows $\alpha$, but does not want to reveal $\alpha$.
The protocol $\mc{I}_\rm{sch} = (G, P, V)$ works as follows.
1. A secret key $\alpha \la \Z_q$ and verification key $u \la g^\alpha$ is generated. The prover $P$ has $\alpha$ and the verifier $V$ has $u$.
2. $P$ computes a random $\alpha_t \la \Z_q$, and sends $u_t \la g^{\alpha_t}$ to $V$.
3. $V$ chooses a random $c \la \Z_q$ and sends it to $P$.
4. $P$ computes $\alpha_z \la \alpha_t + \alpha c \in \Z_q$ and sends it to $V$.
5. $V$ checks if $g^{\alpha_z} = u_t \cdot u^c$. Accept if and only if it is equal.
• $u_t$ is the commitment sent to the verifier.
• $c$ is the challenge sent to the prover.
□ If $P$ can predict the challenge, $P$ can choose $\alpha_t$ and $\alpha_z$ so that verifier accepts it.
• $\alpha_z$ is the response sent to the verifier.
We must check a few things.
• Correctness: If $P$ has the correct $\alpha$, then $g^{\alpha_z} = g^{\alpha_t} \cdot (g^\alpha)^c = u_t \cdot u^c$.
• Soundness: If $P$ does not have the correct $\alpha$, it is reject with probability $1 - \frac{1}{q}$.
□ We can repeat this many times then the probability of reject is $1 - \frac{1}{q^n} \ra 1$.
□ Thus $q$ (the size of the challenge space) must be large.
• Zero-knowledge: $V$ learns no information about $x$ from the conversation.
□ This will be revisited later. See here.
Theorem. The Schnorr identification protocol is secure if the DL problem is hard, and the challenge space $\mc{C}$ is large.
Schnorr Digital Signature Scheme
We transform the above protocol to a signature scheme.^2 We need a hash function $H : \mc{M} \times G \ra \mc{C}$, modeled as a random oracle. The protocol originally involves interaction between two
parties, but a signature is computed by a single party. Intuitively, $H$ will play the role of the verifier.
The Schnorr signature scheme $\mc{S}_\rm{sch} = (G, S, V)$ is defined as follows.
• Key generation: a secret key $sk = \alpha \la \Z_q$ and public key $pk = u \la g^\alpha$ is generated.
• Sign: $S(sk, m)$ outputs $\sigma = (u_t, \alpha_z)$ where
□ Choose random $\alpha_t \la \Z_q$ and set $u_t \la g^{\alpha_t}$.
□ Compute $c \la H(m, u_t)$ and set $\alpha_z \la \alpha_t + \alpha c$.
• Verify: $V(pk, m, \sigma)$ outputs $\texttt{accept}$ if and only if $g^{\alpha_z} = u_t \cdot u^c$.
□ $c \la H(m, u_t)$ can be computed and $u$ is known.
Since $H$ is being modeled as a random oracle, the signer cannot predict the value of the challenge $c$. Also, $c$ must take both $m$ and $u_t$ as input, since without $m$, the signature is not
related to $m$ (the signature has no $m$ term inside it). On the other hand, without $u_t$, then the scheme is insecure since the Schnorr identification protocol is HVZK. See Exercise 19.12.^1
Theorem. If $H$ is modeled as a random oracle and Schnorr’s identification protocol is secure, then Schnorr’s signature scheme is also secure.
Proof. See Theorem 19.7.^1
Note that $\alpha \la \Z_q$ must be chosen randomly every time.
Digital Signature Algorithm
Schnorr’s scheme was protected by a patent, so NIST opted for a ad-hoc signature scheme based on a prime order subgroup of $\Z_p^{\ast}$. This algorithm eventually became the Digital Signature
Algorithm (DSA). The standard was updated to support elliptic curve groups over a finite field, resulting in ECDSA.
Public Key Infrastructure
How would you trust public keys? We introduce digital certificates for this.
Read in public key infrastructure (Internet Security). | {"url":"https://blog.zxcvber.com/lecture-notes/modern-cryptography/2023-10-26-digital-signatures/","timestamp":"2024-11-06T05:58:27Z","content_type":"text/html","content_length":"38678","record_id":"<urn:uuid:fe4e6e00-39a4-4b3d-b6b8-45728ca728cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00854.warc.gz"} |
o Girard
About me
I am a PhD student at Institut Denis Poisson, Université de Tours since September 2021. My research mostly focus on first order non-linear equations, you can find more about it in the research
section. These equations can model the pedestrian flow in evacuation contexts. You can find results of simulations in the simulation section. Apart from the theoretical research, I also create
synthesizers and scenography. These activities have their respective sections.
• Discontinuous scalar conservation law - entropy solutions, dissipative germs, finite volume schemes.
• Macroscopic pedestrian flow models - non-linear coupling, fixed point arguments, splitting methods.
• Discontinuous Hamilton-Jacobi-Bellman equations - viscosity solutions, flux-limited solutions, finite difference schemes.
• Eikonal equation - Monge solutions, fast marching methods.
Presentations and publications
I teach mathematics and computer science in the university of Tours and the Supinfo school. Here is a list of the subject I taught:
• Linear algebra (Supinfo), Algebra III (Université de Tours)
• Analysis III (Université de Tours)
• Geometry (Université de Tours)
• Python programming (Université de Tours), C programming (Supinfo)
• Internet of things and embedded systems (Supinfo)
• Mathematical reasonning (Université de Tours), Mathematical tools for biology students (Université de Tours)
One-dimensional Hughes model
In the one-dimensional setting, the Hughes' model simulate the evacuation of a corridor ( here we choose the corridor to be the domain \( [-1,1] \) ) with exits on the right and the left. It can be
rewritten as: \[\left\{ \begin{matrix} \partial_t \rho(t,x) + \partial_x \left[ \textrm{sign} (x - \xi(t)) f(\rho(t,x)) \right] = 0 \\ \int_{-1}^{\xi(t)} c(\rho(t,x)) \textrm{d} x = \int_{\xi(t)}^1 c
(\rho(t,x)) \textrm{d} x. \end{matrix} \right.\] Here \( f(\rho) = \rho v(\rho) \) is the flux function corresponding to Lighthill-Whitham-Richards (LWR) model and \(v\) corresponds to the speed of
pedestrians. Also, \( c(\rho) \) is the running cost paid by agents to travel in a location with density \( \rho \). The unknown \( \xi \) named the "turning curve" corresponds to the point in space
where pedestrian pay the same total cost to leave the corridor through the left or the right exit. Below, two simulations of this model using a finite volume scheme. The turning curve \(\xi\) is in
red and the density is plotted on \([-1.5,1.5]\) even if the dynamics of the turning curve only depends on what's inside \([-1,1]\).
If you can't see the video, you can download it here.
With a cost function \(c(\rho) = 1 + 2\rho\).
If you can't see the video, you can download it here.
With constrained evacuation at the exit \(x=1\) and a cost function \(c(\rho) = 1 +4\rho^2\).
2D Hughes' model
In the two-dimensional setting, the Hughes' model takes the following form: \[\left\{ \begin{matrix} \partial_t \rho(t,x) + \textrm{div} \left( \vec{V}(x) f(\rho(t,x)) \right) = 0 \\ \left|\nabla u
(x) \right| = c(\rho(t,x)) \\ \vec{V}(x) = - \frac{\nabla u(x)}{|\nabla u (x)|}. \end{matrix}\right.\] Here \( u \) is a solution of the Eikonal equation with a Dirichlet condition \(u=0\) at the
exits. Below is a simulation of the 2D Hughes model where the exits are represented in green and the density is represented by a gradient from red (high dentity) to black (low density).
If you can't see the video, you can download it here.
The musical instruments I make are synthesizers inspired by wind instruments. The Polyclav is a numeric modular synth made to be compatible with the DM48 midi harmonica from Lekholm instruments. The
Soubasynth is a synth inspired by the sousaphone. The Brassynth is an improved and portable version of the Soubasynth compatible with trombone and trumpet mouthpieces. The Soubasynth and the
Brassynth both use a Bela board. This constructor made a post about my instruments on their blog.
You can contact me by e-mail at theo.girard /@/ univ-tours.fr
Bureau : E1350 bâtiment E2
Faculté Sciences et techniques, Campus Grandmont
40 avenue Monge 37200 Tours, France | {"url":"https://theorgirard.github.io/","timestamp":"2024-11-12T05:45:26Z","content_type":"text/html","content_length":"17504","record_id":"<urn:uuid:3e59e9a2-59fc-44cc-b8da-7c94b0c02a70>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00681.warc.gz"} |
Cite as
Michael D. Ernst, Dan Grossman, Jon Jacky, Calvin Loncaric, Stuart Pernsteiner, Zachary Tatlock, Emina Torlak, and Xi Wang. Toward a Dependability Case Language and Workflow for a Radiation Therapy
System. In 1st Summit on Advances in Programming Languages (SNAPL 2015). Leibniz International Proceedings in Informatics (LIPIcs), Volume 32, pp. 103-112, Schloss Dagstuhl – Leibniz-Zentrum für
Informatik (2015)
Copy BibTex To Clipboard
author = {Ernst, Michael D. and Grossman, Dan and Jacky, Jon and Loncaric, Calvin and Pernsteiner, Stuart and Tatlock, Zachary and Torlak, Emina and Wang, Xi},
title = {{Toward a Dependability Case Language and Workflow for a Radiation Therapy System}},
booktitle = {1st Summit on Advances in Programming Languages (SNAPL 2015)},
pages = {103--112},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-939897-80-4},
ISSN = {1868-8969},
year = {2015},
volume = {32},
editor = {Ball, Thomas and Bodík, Rastislav and Krishnamurthi, Shriram and Lerner, Benjamin S. and Morriset, Greg},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
address = {Dagstuhl, Germany},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.SNAPL.2015.103},
URN = {urn:nbn:de:0030-drops-50208},
doi = {10.4230/LIPIcs.SNAPL.2015.103},
annote = {Keywords: Synthesis, Proof Assistants, Verification, Dependability Cases, Domain Specific Languages, Radiation Therapy} | {"url":"https://drops.dagstuhl.de/search/documents?author=Wang,%20Xi","timestamp":"2024-11-05T15:18:52Z","content_type":"text/html","content_length":"161169","record_id":"<urn:uuid:19fefc23-d5bf-4d07-bc4c-5f9d78bc8434>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00275.warc.gz"} |
Crushingvalue Aggregate For Concrete
Concrete Calculator. Our online tools will provide quick answers to your calculation and conversion needs. On this page, you can calculate material consumption viz., cement, sand, stone gravel for
the following concrete mix ratios - 1:1.5:3, 1:2:4, 1:3:6, 1:4:8, 1:5:10. Once, the quantities are determined, it is easy to estimate the cost of a ... | {"url":"https://www.elektryk-waw.pl/20139+crushingvalue-aggregate-for-concrete.html","timestamp":"2024-11-13T15:56:47Z","content_type":"text/html","content_length":"35560","record_id":"<urn:uuid:9f8d05c9-b113-46f5-8b6c-de0f2a42515f>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00652.warc.gz"} |
4-3 journal: self analysis: behavioral competency: the leadership
39) Sam Lewis owns a firm in New York City's garment district. If Sam keeps adding workers to use the same number of sewing machines, eventually the workplace will become so crowded that workers will
get in each other's way. At this point
A) the marginal product of labor in Sam's business would be negative and his total output would decrease.
B) Sam should encourage his workers to share their sewing machines.
C) Sam's business will be in violation of safety rules that have been established by the New York City government.
D) Sam should begin using a division of labor in his business.
40) In his book The Wealth of Nations, Adam Smith employed the example of a pin factory in order to explain what economic concept?
A) the relationship between the marginal and average product of labor
B) the law of diminishing returns
C) why no firm would want to hire so many workers as to experience a negative marginal product of labor
D) the division of labor
41) The total output produced by a firm divided by the quantity of workers employed by the firm is the definition of
A) the marginal product of labor.
B) the division of labor.
C) the average product of labor.
D) the average cost of production.
42) After Suzie, owner of Suzie's Sweet Shop, hires her 8th worker the average product of labor declines. Which of the following statements must be true?
A) The marginal product of the 8th worker is negative.
B) The marginal product of the 8th worker is less than the average product of labor before the 8th worker was hired.
C) Suzie's profits would be greater if she did not hire the 8th worker.
D) The average product of labor is negative.
43) Which of the following statements is true?
A) The average product of labor is at its maximum when the average product of labor equals the marginal product of labor.
B) The average product of labor is at its minimum when the average product of labor equals the marginal product of labor.
C) The average product of labor tells us how much output changes as the quantity of workers hired changes.
D) Whenever the marginal product of labor is greater than the average product of labor the average product of labor must be decreasing.
44) The marginal product of labor is calculated using the formula
A) L/Q.
B) ΔL/ΔQ.
C) ΔQ/ΔL.
D) Q/L.
45) Which of the following describes how output changes in the short run? Because of specialization and the division of labor, as more workers are hired
A) output will first increase at an increasing rate, then output will increase at a decreasing rate.
B) output will first decrease at an increasing rate, then increase at a decreasing rate.
C) the marginal product of labor will first decrease, then increase at a decreasing rate.
D) the marginal product of labor will first be negative and then will be positive.
46) If 11 workers can produce 53 units of output while 12 workers can produce 56 units of output, what is the marginal product of the 12th worker?
A) 0.16
B) 3
C) 4.67
D) 36 | {"url":"https://summitessays.com/2023/03/09/4-3-journal-self-analysis-behavioral-competency-the-leadership/","timestamp":"2024-11-12T13:16:52Z","content_type":"text/html","content_length":"146602","record_id":"<urn:uuid:372c1a4b-d0c6-439b-a71d-12a2104a520c>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00752.warc.gz"} |
Enough injectives with just Zorn’s lemma and not the axiom of choice:
draft paper (see appendix)
About me
I’m a mathematician working in applied topos theory. I obtained my PhD in October 2017 under the supervision of Marc Nieper-Wißkirchen at the University of Augsburg. Besides research, I love teaching
mathematics at all levels and am passionate about doing mathematics with school students. Currently I’m back in Augsburg. During the academic year 2018/2019, I was working at the Università di Verona
under the supervision of Peter Schuster. In the winter term 2017/2018, I was substituting a junior professor at the University of Augsburg, and during the summer of 2018, I was a guest at the Max
Planck Institute for Mathematics in the Sciences in Leipzig.
I explore applications of the internal language of toposes, particularly in commutative algebra and in algebraic geometry.
I contribute to the nLab and to the Stacks Project, and encourage everyone to consider Eugenia Cheng’s manifesto for inclusivity in category theory and mathematics at large.
Recent results by Matthias Ritter (formerly Matthias Hutzler): The infinitesimal topos classifies the theory of quotients of local algebras by nilpotent ideals (master’s thesis), Syntactic
presentations for glued toposes and for crystalline toposes (PhD thesis)
Recent result by Johannes Riebel: The undecidability of BB(748)
Travel plans
Illustration: Carina Willbold (CC BY-SA) | {"url":"https://www.ingo-blechschmidt.eu/","timestamp":"2024-11-08T19:08:03Z","content_type":"text/html","content_length":"17322","record_id":"<urn:uuid:95d7d8eb-50fd-4768-b7c2-d82d18543030>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00110.warc.gz"} |
Rapid developments in the fields of control engineering and in microprocessor and semiconductor technology have resulted in the widespread use of electronically controlled systems in every branch of
industry today. This has created a need for sensors that are inexpensive but, at the same time, sufficiently robust, both electrically and mechanically, to withstand a wide range of temperatures
(e.g. from -40 to +160 degrees centigrade), particularly in applications involving large quantities, such as the automobile industry. Fig. 1 provides a summary of the various types of sensors for
angular and linear motion that are in use today. Such potentiometers essentially comprise the following components:
1. The resistance element (support material + a resistance track of conductive plastic)
2. A wiper (precious metal alloy)
3. A drive shaft or acutating rod
4. Bearings (ball bearing or plain bearing)
5. Housing
Fig. 1
Terminology / Terms and Definitions
When we refer nowadays to a potentiometer as a sensor, it is important to bear in mind the statements made here only apply if the potentiometer is connected as voltage divider rather than as a
variable resistor (rheostat) (Fig. 2). The wiper voltage must be connected, free of load, to an operational amplifier such as a 741, OP 07 or some other component with a high input impedance. Fig. 3
explains the terms used, such as electrical and mechanical travel. L1 indicates the defined electrical travel. L2 indicates the continuity travel which also includes the non-linear connection fields
Fig. 4. L3 indicates the total electrical contact travel of the potentiometer. L4 indicates the mechanical travel. An electrical potential need not be defined for the whole of this travel. In case of
nothing otherwise is defined, the fields L1, L2, L3 and L4 are nominally designed symmetrically.
Of all the quality features mentioned, linearity and conformity are the values most often defined. These terms express the extent to which the voltage output from a potentionmeter, and also other
types of angular of linear movement sensor, differs from a prescribed theoretical function. In by far the majority of cases, the desired output function is directly proportional to the angel or
linear movement that is input.
Formula: Fig. 5.
Whereby m characterizes the gradient, the offset voltage of the potentiometer and the linear or angular travel. Where there is a linear relationship, deviation is referred to as linearity. Where the
relationship is nonlinear.
U = f (x) + a + b
the deviation is referred to as conformity.
Independent linearityIf a voltage U0 is applied to a potentiometer with a linear characteristic as in Fig. 5
and the wiper is moved in direction Alpha (standardized movement, angle 0;1 ) then the relationship illustrated in Fig. 6 will exist between the output voltage and the mechanically input value. The
maximum deviation of the potentiometer curve from an ideal straight line is referred to as the independent linearity error. The slope and axis intercept of this straight line can be so chosen that
the error f within the travel L1 is minimized. The error ±f isindicated as a deviation in percentage terms of the output voltage from the theoretical in relation to the input voltage. Since direct
measurement of the potentiometer characteristic does not make it possible to assess the extent of such an error, only the difference between the potentiometer characteristic and that of an
essentially perfect master potentiometer is plotted as in the practical example given in Fig. 7. Typical values for independent linearity nowadays lie between 0.2 % and 0.02 %.
Fig. 7
Absolute linearity
With the ever increasing automation of assembly lines, users are finding that values for absolute linearity are steadily gaining importance. Unlike independent linearity, for absolute linearity the
reference slope is fully defined (Fig. 8) so that there is no need for subsequent system trimming. The definition of an index point establishes a relationship between the mechanical input value
(travel or angle) and the output voltage.
Potentiometers whose linearity is defined by these criteria can be installed without a need for subsequent adjustment. As with independent linearity, it is best to determine the absolute linearity of
a potentiometer by comparing its output with that of a master potentiometer. With absolute linearity, it is frequently necessary for the tolerance fields to be stepped. Fig. 9 shows a practical
Fig. 9
Absolute conformity
As already indicated already, conformity is a more general concept than linearity. The definition of absolute conformity is similar to that of absolute linearity. It is essential for an index point
to be defined. The functional relationship can either be determined mathematically to be defined. The functional relationship can either be determined mathematically or by plotting a number of points
to establish a curve with the aid of suitable interpolation. It is also possible with a potentiometer to achieve steadily increasing or steadily decreasing functions such as logarithmic, exponential,
sinusoidal or cosinal functions.
Contact resistance
Contact resistance is the resistance between the wiper terminal and the wiper's immediate point of contact on the potentiometer's resistive track. As will be subsequently explained, this contact
resistance affects all the important quality features of a potentiometer. Contact or transition resistance can be broken down into three components. The first component, describes the integral
voltage drop between the current-carrying track and the contact surface. This component is largely dependent on technology factors and amounts to several hundred Ohm. The second component, the
external component, is far more difficult to master than the first. This external transition resistance has much in common with the contact resistances occurring in switches and plug- and
socket-connectors. It is caused by the transition between the wiper and the potentiometer track not being ideal from an electrical viewpoint. Metal oxides, chlorides and sulphides, mixed with various
organic substances, can result in the formation of thin nonconductive facings at the interface. If not kept within bounds, this external transition resistance can, under unfavourable conditions, lead
to complete failure to keep within a tolerance range. It is absolutely essential that the materials used in potentiometer manufacutre be subjected to stringent quality control and be matched one with
another. The third component, the dynamic component, is related to dynamic drive forces acting on the wiper at high speeds of actuation. With the aid of damped wipers, actuation speeds of up to 10 m/
sec can be achieved without any appreciable increase in the dynamic component of contact resistance.
Linearity errors due to electrical circuitry
From here onwards, we are only concerned with the linear characteristics (linearity). Relationships must be suitably adapted for applications with non-linear characteristics (conformity) but there
are no essential differences. As already mentioned, the required linearity values can only be utilized so long as the signal output by the sensor "potentiometer" carries no current. We have now to
consider the effect of wiper current on linearity. Fig. 10a illustrates the functional relationship between wiper current, contact resistance and linearity error. As is shown by the example in Fig.
10b, with a wiper current of 10 µA and a contact resistance of 10 kOhm, a potentiometer which has a resistance of 2 kOhm already has linearity error of 1.1 %. A similar situation arises with an ohmic
load. This clearly shows how important are the roles played by both wiper current and contact resistance.
Fig. 10b
Linearity errors resulting from mechanical coupling
If there is axial misalignment (eccentricity) between the drive shaft and the shaft of a potentiometer used to sense angular, motion, this will cause a linearity error that increases as the coupling
radius decreases in relation to the degree of eccentricity. The following equation determines the maximum relative error
Fmax = E/Pi · rk
where E = Eccentricity
und rk = the coupling radius.
It is only possible to take full advantage of the linearity or conformity of any rotationary sensor system, if coupling alignment errors (offset and angular misalignment) are avoided or at least
reduced to a mimimum. This means that with highly accurate measurement systems, due allowance must be made for any coupling misalingment in accordance with the above equation.
Smoothness is a measure of the deviations from perfect regularity that appear in the output voltage of a potentiometer. This irregularity is measured over a specified travel increment, for example 1
%, and is expressed as a percentage of the applied voltage. For the measurement of smoothness, the VRCI definition calls for a bandpass filter to be used as a means of suppressing any linearity error
and for the potentiometer to be operated with a load resistance (e.g. 100 . Rp). This method has certain disadvantages:
a) The use of a filter causes both the absolute wiper velocity and any changes in such velocity to affect the smoothness values. Since the filter partly integrates and partly differntiates, the
chart-recorded smoothness curve does not accurately indicate the variations in the output signal.
b) The load applied to the potentiometer also contributes to error by causing variation in the contact resistance which is greatest with the wiper at the voltage application end and lowest at the
grounded end of the potentiometer track.
c) The use of a 1 % evaluation window is not accurate enough for many of today's applications.
d) The sometimes arbitrary selection of a filter type, load resistance and travel increment results in-smoothness values not being directly comparable.
Microlinearity, which is defined as the maximum linearity variation within a travel or angular increment that amounts - as with smoothness measurement - to 1 % of the electrical range if nothing to
the contrary is defined. Microlinearity is indicated as a percentage of the absolute voltage that is applied. Fig. 11 illustrates the characteristic obtained for a potentiometer with a microlinearity
error. This was evaluated by a computer-supported system while making linearity measurements. The travel increments are super-imposed on the linearity curve and have an overlap of at least 50 %.
Contrary to a smoothness measurement, the error here is pureley a linearity error that describes the maximum error within a defined increment. Microlinearity does not, however, make it possible to
determine whether a potentiometer will be suitable for a particular applications because any variations in gradient (sensitivity) can only be determined with considerable difficult.
Relative gradient variation (RGV)
If, in a highly sensitive control system, the amplification should, for example, be so arranged that the control circuit will be stable with the mean slope (gradient) of the sensor, then it is
important to be aware of any variations there may be in that slope (Fig. 12a, Fig. 12b). If, at any point, the gradient is appreciably steeper than the mean gradient, then there will be a higher
closed-loop gain in this position and this could lead to feedback oscillation. If, on the other hand, the gradient is less steep at some point than the mean gradient, then repeatability would be
reduced and there would be less control accuracy. If we relate this type of local gradient variation gl to the mean gradient go of the potentiometer, then this criterion is independent of the
potentiometer length and can be used for the direct comparison of various potentiometers.
local gradient g[l]
RGV = ————————— = —
mean gradient g[o]
(Fig. 13)
The RGV is indicated as a ± deviation in percentage terms from L (standardized mean gradient).
Interpretation of RGV
Measurements have shown that with conductive-plastic potentiometers, gradient fluctuations are statistically distributed to travel increments of less than 1 µm, i.e. there is no periodicity or
regularity. Fig. 14a shows the RGV curve and Fig. 14b the RGV values with step widths of 0.1, 0.2 and 0.3°. The form of these distributions more or less in conformity with the normal distribution to
be expected in view of the central limit value principle. The mean distribution value is around 1 (mean gradient), the variance (STDEV) decreases as the step width increases. Since each individual
RGV value represents a mean value, it is to be expected that the variance of such mean values will decrease with a root function in proportion to the increase in step size, since each such increase
amounts to an increase in the size of the random sample.
RGV(X) Y
———— = √ — applies for STDEV
RGV(Y) X
In Fig. 15 this variance has been plotted as a function of the step width. RGV variance can thus be considered as a characteristic quality feature of a potentiometer. The functional relationsship
also serves to indicate the maximum resolution of a potentiometer which is not infinite, as many potentiometer manufacturers would imply. Maximum RGV values have also been plotted in Fig. 15. This
curve, like the mimimum value curve, does not of course, obey statistical laws but arises from the inclusion of defective positions and faults in the system "potentiometer" as a whole. These values
are vital criteria for assessing the stability and repeatability to be expected of a control system.
When measuring RGV values of potentiometers up to synchro size 20 from series production, with step widths of 0.1° RGV values of ± 10 % are obtained. If we specify the RGV value of ± 100 % as the
resolution limit, from the equation in 7.4 we obtain with 10 % a resolution of 1/1000°. The degree of resolution is primarily determined by the homogeneity and grain-size distribution in the
conductive plastic layer, by the wiper contact surface running parallel to the equipotential lines (Fig. 16a, 16b) and by the wiper current.
The hysteresis value specifies the signal differential resulting if a prescribed position is approached from one side, the point is travelled over and the same position approached from the other
side. Hysteresis is mainly affected by mechanical factors such as the bearings, th stiffness of the wiper system and the coefficient of friction between the conductive layer and the wiper. For this
reason, attention must be paid to ensure a backlash-free, rigid mechanical coupling. This can be achieved using, for example, a spring-borne concial pin or lever. Fig. 18 shows the recorded
hysteresis of a standard potentiometer. The measuements were taken with clockwise and counterclockwise rotation being repeated three times. While the curves recorded in one direction almost coincide
(indicating goog resolution), in the opposite direction they indicate an hysteresis of around four thousands of a degree. The fact that the curves in one direction almost coincide and that there is
an constant hystersis in the opposite direction indicates a stable displacement of the wiper contact line which means there is no stick-slip effect.
Fig. 18
This term is taken to mean any optional approach movement towards a prescribed position from various directions. It represents the sum of 2 x resolution + hyteresis.
Temperature and humidity coefficients
In many data sheets issued by potentiometer manufacturers, reference is frequently made to the temperature coefficient (Tk) and humidity coefficient (Fk) of the nominal resistance. Where
potentiometers are used as voltage dividers (Fig. 2), these values are irrelevant. For this application, it is the Tk and Fk applicable to the voltage divider behaviour which are of significance. It
often also happens that the humidity is not kept constant while making Tk measurements, with a result that a mixture of Tk and Fk is often given as a temperature coefficient. Detailed measurements
have show that the Tk and Fk of the nominal resistance in conductive-plastic potentiometers (without housing) are of an order of magnitude somewhat less than 200 ppm/°C and 500 ppm/% RH respectively.
The Tk and Fk of the voltage-dividing behaviour are some two orders of magnitdue lower which means that here changes within a range of less than 5 ppm/°C and 5 ppm/% RH can be expected, ensuring
constancy over a wide temperature and humidity range. However, this advantage can only be utilized given a suitably designed housing and if, for example, no trimming resistances are used in the
potentiometer circuit.
Service life
The magnitude of the contact resistance and the wear to which the resistance track is subjected and the resulting change in electrical characteristics dertermine the number of operations to which a
potentiometer can be subjected and thus also its service life. Although of considerable importance for industrial applications, no standard has so far been issued that specifies a service-life
definition or a particular method of testing. It is, of course, most difficult to specify a value for wear or for an increase in contact resistance for a given number of actuation cycles since such
values are markedly influenced by such external factors as temperature and humidity, and by mechanical and chemical influences. Such values need to be established for each particular application.
This applies to a lesser extent for the method of testing, and here the establishment of standard method would facilitate comparing the service life of various potentiometers. The first is a
practical test in which extremely small wiper movements are simulated such as frequently occur in feedback control systems. Typical values are: Wiper travel 2° Test frequency 100 Hz. This dither test
permits a relatively rapid result to be obtained concerning contact reliability and any change in gradient within a micro range since, at such a high frequency, some 8.6 million cycles can be
effected daily. The second test, the half-stroke test, gives information concerning linearity changes, zero-point shift and wiper wear. This test is performed at a frequency of 10 Hz (0.86 million
cycles per day) over 50 % of the track length. As is shown in Fig. 19, this results in a maximum linearity change. A criterion for rejection here might be a doubling of linearity in relation to the
state when new and a maximum contact resistance value.
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Machine learning practice 1
1. (Model Selection, [ISL] 6.8, 25 pt) In this exercise, we will generate simulated data, and will then use this data to perform model selection.
1. Use the rnorm function to generate a predictor \(X\) of length \(n = 100\), as well as a noise vector \(\epsilon\) of length \(n = 100\).
X <- rnorm(100)
epsilon <- rnorm(100)
2. Generate a response vector \(Y\) of length \(n = 100\) according to the model \[ Y = \beta_0 + \beta_1 X + \beta_2 X^2 + \beta_3 X^3 + \epsilon, \] where \(\beta_0 = 3\), \(\beta_1 = 2\), \(\
beta_2 = -3\), \(\beta_3 = 0.3\).
b0 <- 3 ; b1 <- 2 ; b2 <- -3 ; b3 <- 0.3
Y <- b0 + b1*X + b2*X^2 + b3*X^3 + epsilon
3. Use the regsubsets function from leaps package to perform best subset selection in order to choose the best model from the set of predictors \((X, X^2, \cdots, X^{10})\). What are the best models
obtained according to \(C_p\), BIC, and adjusted \(R^2\), respectively? Show some plots to provide evidence for your answer, and report the coefficients of the best model obtained.
df <- data.frame(X,X^2,X^3,X^4,X^5,X^6,X^7,X^8,X^9,X^10,Y)
best_model <- regsubsets(Y ~ ., data = df, nvmax=10)
summary_model <- best_model %>% summary
#the best models according to C_p, BIC, and adjusted R^2
min.cp <- which.min(summary_model$cp)
min.bic <- which.min(summary_model$bic)
min.adjr2 <- which.max(summary_model$adjr2)
# plots that show the best models
par(mfrow = c(2,2))
plot(summary_model$cp, xlab = "Number of Poly(X)",
ylab ="Cp", type="l")
points(min.cp, summary_model$cp[min.cp], col = "red", pch = 4, lwd = 5)
plot(summary_model$bic, xlab = "Number of Poly(X)",
ylab = "BIC", type="l")
points(min.bic, summary_model$bic[min.bic], col = "red", pch = 4, lwd = 5)
plot(summary_model$adjr2, xlab = "Number of Poly(X)",
ylab = "Adjusted R^2", type="l")
points(min.adjr2, summary_model$adjr2[min.adjr2], col = "red", pch = 4, lwd = 5)
coef(best_model, min.cp)
coef(best_model, min.bic)
coef(best_model, min.adjr2)
4. Repeat (c), using forward stepwise selection and also using backward stepwise selection. How does your answer compare to the results in (c)?
# forward stepwise selection.
best_model_f <- regsubsets(Y~., data=df, nvmax=10, method='forward')
summary_model_f <- best_model_f %>% summary
# refer to C_p, BIC, and adjusted R^2
min.cp <- which.min(summary_model_f$cp)
min.bic <- which.min(summary_model_f$bic)
min.adjr2 <- which.max(summary_model_f$adjr2)
coef(best_model, min.cp)
coef(best_model, min.bic)
coef(best_model, min.adjr2)
# backward stepwise selection.
best_model_b = regsubsets(Y~., data=df, nvmax=10, method='backward')
summary_model_b <- best_model_f %>% summary
#the best models according to C_p, BIC, and adjusted R^2
min.cp <- which.min(summary_model_b$cp)
min.bic <- which.min(summary_model_b$bic)
min.adjr2 <- which.max(summary_model_b$adjr2)
coef(best_model, min.cp)
coef(best_model, min.bic)
coef(best_model, min.adjr2)
We found that both forward and backward method provided same results as part (c).
5. Now fit a LASSO model with glmnet function from glmnet package to the simulated data, again using \((X,X^2,\cdots,X^{10})\) as predictors. Use cross-validation to select the optimal value of \(\
lambda\). Create plots of the cross-validation error as a function of \(\lambda\). Report the resulting coefficient estimates, and discuss the results obtained.
# Lasso model
lasso = glmnet(df[, -11]%>%as.matrix, df[, 11], alpha=1)
# cross-validation
cv.out = cv.glmnet(df[, -11]%>%as.matrix, df[, 11], alpha=1)
# cross-validation error
# the optimal value of lambda
best_lam = cv.out$lambda.min
# Coefficients from lasso model with best lambda.
predict(lasso, s=best_lam, type="coefficients")
According the results, we see that estimated coefficients of b1~b3 from lasso model with best lambda are largely consistent with the true value of b1~b3. Nevertheless, lasso model included X^7 and X^
9 by mistake, with small coefficients.
6. Now generate a response vector \(Y\) according to the model \[Y = \beta_0 + \beta_7 X^7 + \epsilon,\] where \(\beta_7 = 7\), and perform best subset selection and the LASSO. Discuss the results
# generate y2
b7 <- 7
Y2 <- b0 + b7*X^7 + epsilon
df <- data.frame(X,X^2,X^3,X^4,X^5,X^6,X^7,X^8,X^9,X^10,Y2)
# best subset selection
best_model <- regsubsets(Y2 ~ ., data=df, nvmax=10)
summary_model <- summary(best_model)
# refer to C_p, BIC, and adjusted R^2
min.cp <- which.min(summary_model$cp)
min.bic <- which.min(summary_model$bic)
min.adjr2 <- which.max(summary_model$adjr2)
# check results
coef(best_model, min.cp)
coef(best_model, min.bic)
coef(best_model, min.adjr2)
# LASSO model
lasso = glmnet(df[, -11]%>%as.matrix, df[, 11], alpha=1)
# cross-validation
cv.out = cv.glmnet(df[, -11]%>%as.matrix, df[, 11], alpha=1)
# cross-validation error
# the optimal value of lambda
best_lam = cv.out$lambda.min
# Coefficients from lasso model with best lambda.
predict(lasso, s=best_lam, type="coefficients")
According to the results, we found that the best subset selection (referring to the BIC value) and Lasso regression provides the relatively more accurate estimates than that of best subset selection
(referring to C_p and adjusted R^2).
2. (Prediction, [ISL] 6.9, 20 pt) In this exercise, we will predict the number of applications received (Apps) using the other variables in the College data set from ISLR package.
1. Randomly split the data set into equal sized training set and test set (1:1).
train_id <- sample(dim(College)[1], dim(College)[1]/2)
train_set <- College[train_id, ]
test_set <- College[-train_id, ]
2. Fit a linear model using least squares on the training set, and report the test error obtained.
# linear model
modl <- lm(Apps~., data = train_set)
pred1 <- predict(modl, newdata = test_set[,-2], type = "response")
# test error
MAE(pred1, College$Apps[-train_id])
MSE(pred1, College$Apps[-train_id])
RMSE(pred1, College$Apps[-train_id])
R2(pred1, College$Apps[-train_id], form = "traditional")
err1 <- data.frame(MAE = MAE(pred1, College$Apps[-train_id]),
MSE = MSE(pred1, College$Apps[-train_id]),
RMSE = RMSE(pred1, College$Apps[-train_id]),
R2 = R2(pred1, College$Apps[-train_id], form = "traditional"))
3. Fit a ridge regression model on the training set, with \(\lambda\) chosen by 5-fold cross-validation. Report the test error obtained.
# prepare data
train_set <- model.matrix(Apps~.-1, data=College[train_id, ])
test_set <- model.matrix(Apps~.-1, data=College[-train_id, ])
# ridge regression with cross-validation
modr_cv <- cv.glmnet(train_set %>% as.matrix,
College$Apps[train_id] %>% as.matrix,
nfolds = 5, alpha = 0)
lambda <- modr_cv$lambda.min
pred2 <- predict(modr_cv, s = lambda, newx = test_set)
# test error
MAE(pred2, College$Apps[-train_id])
MSE(pred2, College$Apps[-train_id])
RMSE(pred2, College$Apps[-train_id])
R2(pred2, College$Apps[-train_id], form = "traditional")
err2 <- data.frame(MAE = MAE(pred2, College$Apps[-train_id]),
MSE = MSE(pred2, College$Apps[-train_id]),
RMSE = RMSE(pred2, College$Apps[-train_id]),
R2 = R2(pred2, College$Apps[-train_id], form = "traditional"))
4. Fit a LASSO model on the training set, with \(\lambda\) chosen by 5-fold cross-validation. Report the test error obtained, along with the number of non-zero coefficient estimates.
# LASSO regression with cross-validation
modr_cv <- cv.glmnet(train_set %>% as.matrix,
College$Apps[train_id] %>% as.matrix,
nfolds = 5, alpha = 1)
lambda <- modr_cv$lambda.min
pred3 <- predict(modr_cv, s = lambda, newx = test_set)
# test error
MAE(pred3, College$Apps[-train_id])
MSE(pred3, College$Apps[-train_id])
RMSE(pred3, College$Apps[-train_id])
R2(pred3, College$Apps[-train_id], form = "traditional")
err3 <- data.frame(MAE = MAE(pred3, College$Apps[-train_id]),
MSE = MSE(pred3, College$Apps[-train_id]),
RMSE = RMSE(pred3, College$Apps[-train_id]),
R2 = R2(pred3, College$Apps[-train_id], form = "traditional"))
# the number of non-zero coefficient estimates (intercept term included)
coef <- predict(modr_cv, s=best_lam, type="coefficients")
coef[which(coef != 0)] %>% length
5. Comment on the results obtained. How accurately can we predict the number of college applications received? Is there much difference among the test errors resulting from these three approaches?
data.frame(Model = c("Linear", "Ridge", "Lasso"),
rbind(err1, err2, err3) %>% round(2))
We used MAE, MSE, RMSE and R-squared to access the model test error.
• MAE (Mean absolute error) represents the difference between the original and predicted values extracted by averaged the absolute difference over the data set.
• MSE (Mean Squared Error) represents the difference between the original and predicted values extracted by squared the average difference over the data set.
• RMSE (Root Mean Squared Error) is the error rate by the square root of MSE.
• R-squared (Coefficient of determination) represents the coefficient of how well the values fit compared to the original values. The value from 0 to 1 interpreted as percentages. The higher the
value is, the better the model is. source
Generally speaking, all of models showed high R2 (\(\geq0.9\)), indicating that the high accuracy of the predictions. According to four test errors, we found that Ridge regression provides the
lowerst MSE, RMSE and the highest R-squared, which suggested that Ridge regression performed the best. Nevertheless, the test errors differences are very small. | {"url":"https://zianzhuang.com/2021/05/machine-learning-practice-1/","timestamp":"2024-11-03T11:58:16Z","content_type":"text/html","content_length":"27570","record_id":"<urn:uuid:3ec00747-fa2f-405a-8ecb-b139b2b2bf5c>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00600.warc.gz"} |
how to find the degree of a vertex
I directrix at y = - Don’t stop learning now. Ellipses: Find the center, foci, vertices, and co-vertices of each ellipse Ellipses: Write the standard equation of each ellipse Ellipses: Use the
information about the vertex, co-vertex, and focus to write a standard equation (center is 0,0) Ellipses: Use the information about the vertex, co-vertex, focus, and center to write a standard
equation The vertex in astrology is a senstive angle. It is also necessary to remember that Vertex is one of the biggest used motor oil processors in the United States at this point. Inorder Tree
Traversal without recursion and without stack! Murray says: 24 Mar 2013 at 11:45 am [Comment permalink] @Simon: You'll need to use the "Vertex Method" as detailed in the article. And we talk about
where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. In astrology, it is considered an auxiliary Descendant. generate link and
share the link here. The idea is based on below fact.If there exist mother vertex (or vertices), then one of the mother vertices is the last finished vertex in DFS. With a = 3 we find y = x2-the
graph is magnified by 3. code. We need to output anyone of them. Tour the campus. In two steps we have reached the model parabola opening upward. The sustainability degree program includes 12
courses, with at least one, three-week course taken on campus. Experience. By using our site, you close, link The height is the distance from the vertex opposite to the base at a 90 degree angle to
the base. edit When someone else’s planet or point contacts your vertex, this indicates a fated relationship that will serve an important purpose in your life. Please write comments if you find
anything incorrect, or you want to share more information about the topic discussed above. What is a Mother Vertex? 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, Printing all solutions in N-Queen Problem, Warnsdorff’s algorithm for Knight’s tour problem,
The Knight’s tour problem | Backtracking-1, Count number of ways to reach destination in a Maze, Count all possible paths from top left to bottom right of a mXn matrix, Print all possible paths from
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above algorithm. Find the Degree of a Particular vertex in a Graph. Algorithm:- 1. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become
industry ready. For instance seven and a half degrees is now usually written 7.5°. Some astrologers refer to the Vertex as the “third angle” of a chart. We strongly recommend you to minimize your
browser and try this yourself first.How to find mother vertex? Y = 3X2, with the vertex moved to the origin. ONE DAY replacement of a natural gas furnace and AC unit, plus outside condenser….the
install technicians were just as amazing; thorough, worked steady, and got it done, even with all the new challenges they faced with a new system. Probably the easiest, there's a formula for it. Now,
there's many ways to find a vertex. Alternatively, you can use 3D modeling software (such as 3ds Max, Maya, Blender, 3D Coat or Modo) to assign vertex colors. Can we do better? We would like to show
you a description here but the site won’t allow us. Radians We will find these special numbers or degree points embedded all throughout our charts. Apply to the program. Below is the implementation
of the approach. 08, Feb 21. We can find a mother vertex in O(V+E) time. generate link and share the link here. Inorder Tree Traversal without recursion and without stack! How To Find The Vertex Of A
Parabola Method 1. How does the above idea work? Print Postorder traversal from given Inorder and Preorder traversals, Construct Tree from given Inorder and Preorder traversals, Construct a Binary
Tree from Postorder and Inorder, Construct Full Binary Tree from given preorder and postorder traversals, Dijkstra's shortest path algorithm | Greedy Algo-7, Primâ s Minimum Spanning Tree (MST) |
Greedy Algo-5, Kruskalâ s Minimum Spanning Tree Algorithm | Greedy Algo-2, Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph), Travelling Salesman Problem | Set 1 (Naive and
Dynamic Programming), Minimum number of swaps required to sort an array, Write Interview The idea is based on Kosaraju’s Strongly Connected Component Algorithm. Please use ide.geeksforgeeks.org,
Recursive DFS call is made for v before u. Create the graphs adjacency matrix from src to des 2. This approach takes O(V(E+V)) time, which is very inefficient for large graphs. Attention reader! So
the x-coordinate of the vertex … To zoom in on the vertex Rescale X and Y by the zoom factor a: Y = 3x2 becomes y/a = 3(~/a)~. This is your opportunity to demonstrate your commitment and ability to
perform well as a Harvard student. brightness_4 For this vertex sign calculator, an exact birth time is required. Using the Poly Brush package, we can directly paint vertex colors inside the editor.
i find just a little problem solving a problem. There can be two possibilities. A Naive approach : A trivial approach will be to perform a DFS/BFS on all the vertices and find whether we can reach
all the vertices from that vertex. While doing traversal keep track of last finished vertex ‘v’. code, // This code is contributed by rishabhdeepsingh98. Do DFS traversal of the given graph. Talk
with professors, coaches, staff, and current students. One way to understand the vertex is to see the quadratic function expressed in vertex form. Recursive DFS call is made for u before v. If an
edge u-→v exists, then v must have finished before u because v is reachable through u and a vertex finishes after all its descendants. By using our site, you Parts of a degree are now usually
referred to decimally. Come and visit us! Don’t stop learning now. Since a vertex with a loop (i.e. Finding minimum vertex cover size of … This equation makes sense if you think about it. This step
takes O(V+E) time. Let the last finished vertex be v. Basically, we need to prove that there cannot be an edge from another vertex u to v if u is not another mother vertex (Or there cannot exist a
non-mother vertex u such that u-→v is an edge). It includes both a circular chart and a rectangular Indian style chart, but both indicate the same. Get started. Find dependencies of each Vertex in a
Directed Graph. Vertex Colour Mask. In a graph of strongly connected components, mother vertices are always vertices of source component in component graph. Use Vertex Form. Attention reader! a
connection directly back to itself) could never be properly colored, it is understood that graphs in this … Print Postorder traversal from given Inorder and Preorder traversals, Construct Tree from
given Inorder and Preorder traversals, Construct a Binary Tree from Postorder and Inorder, Dijkstra's shortest path algorithm | Greedy Algo-7, Prim’s Minimum Spanning Tree (MST) | Greedy Algo-5,
Kosaraju’s Strongly Connected Component Algorithm, Kruskal’s Minimum Spanning Tree Algorithm | Greedy Algo-2, Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph), Travelling
Salesman Problem | Set 1 (Naive and Dynamic Programming), Minimum number of swaps required to sort an array, Find the number of islands | Set 1 (Using DFS), Write Interview Please use
ide.geeksforgeeks.org, You begin by completing two stipulated degree courses as outlined on the degree requirements page. Can you help me with the problem please. Writing code in comment? Java
Program to Find a Good Feedback Vertex Set in a Graph. close, link Check if v is a mother vertex by doing DFS/BFS from v. This step also takes O(V+E) time. 11, Jan 18. The vertex form of a quadratic
function can be expressed as: Vertex Form: ƒ(x) = a(x−h) 2 + k. Where the point (h, k) is the vertex. Each vertex stores a unit of Vector3 information that we refer to as Vertex Colour. In this case
also, if an edge u-→v exists, then either v must finish before u (which contradicts our assumption that v is finished at the end) OR u should be reachable from v (which means u is another mother
vertex). Sit in on a class. When a single angle is drawn on a xy-plane for analysis, we’ll draw it in standard position with the vertex at the origin (0,0), one side of the angle along the x-axis,
and the other side above the x-axis. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Find a Mother vertex in a Graph
using Bit Masking, Check if incoming edges in a vertex of directed graph is equal to vertex itself or not, Check if vertex X lies in subgraph of vertex Y for the given Graph, Check if every vertex
triplet in graph contains two vertices connected to third vertex, Maximum number of edges that N-vertex graph can have such that graph is Triangle free | Mantel's Theorem, Find the Degree of a
Particular vertex in a Graph, Java Program to Find a Good Feedback Vertex Set in a Graph, Find dependencies of each Vertex in a Directed Graph, Finding minimum vertex cover size of a graph using
binary search, k'th heaviest adjacent node in a graph where each vertex has weight, Add and Remove vertex in Adjacency Matrix representation of Graph, Add and Remove vertex in Adjacency List
representation of Graph, All vertex pairs connected with exactly k edges in a graph, Topological Sort of a graph using departure time of vertex, Minimum Manhattan distance covered by visiting every
coordinates from a source to a final vertex, Print all possible paths in a DAG from vertex whose indegree is 0, Sum of Nodes and respective Neighbors on the path from root to a vertex V, Queries to
count connected components after removal of a vertex from a Tree, Maximum difference of count of black and white vertices in a path containing vertex V, Vertex Cover Problem | Set 1 (Introduction and
Approximate Algorithm), Largest subtree sum for each vertex of given N-ary Tree, Sum of shortest distance on source to destination and back having at least a common vertex, Finding the path from one
vertex to rest using BFS, Maximize the number of uncolored vertices appearing along the path from root vertex and the colored vertices, 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. A mother vertex in a graph G = (V,E) is a vertex v such that all other vertices in G
can be reached by a path from v.Example : There can be more than one mother vertices in a graph. (Or a mother vertex has the maximum finish time in DFS traversal).A vertex is said to be finished in
DFS if a recursive call for its DFS is over, i.e., all descendants of the vertex have been visited. Experience. edit Writing code in comment? Vertex coloring. brightness_4 Okay, this is a little
tricky, but it makes more sense looking at the picture below. For the given vertex then check if a path from this vertices to other exists then increment the degree. 29, May 20. If there exist mother
vertex (or vertices), then v must be one (or one of them). I am to find a equation of a parablo given the vertex (7,-2) and one x-intercept (4,0). If we discover a planet or important angle
(Ascendant (1 st house cusp), IC (4 th house cusp), Descendant (7 th house cusp) or Midheaven (10 th house cusp) living in the powerful 29 th degree it’s always a special marker. Calculate your
vertex sign with this Vertex Sign Calculator. For example, in the below graph, vertices 0, 1 and 2 are mother vertices. Within days (over the weekend) Vertex got the equipment they needed and
installed on the following Monday. This article is contributed by Rachit Belwariar. A directed graph has an Eulerian trail if and only if at most one vertex has − = 1, at most one vertex has
(in-degree) − (out-degree) = 1, every other vertex has equal in-degree and out-degree, and all of its vertices with nonzero degree belong to a single connected component of the underlying undirected
graph. Joshua Boger`73 Founder, Vertex Pharmaceuticals BA Chemistry, Philosophy Experience Wesleyan. Find the Degree of a Particular vertex in a Graph, https://www.youtube.com/watch?v=iNCLqZkxl_k,
Check if incoming edges in a vertex of directed graph is equal to vertex itself or not, Check if vertex X lies in subgraph of vertex Y for the given Graph, Check if every vertex triplet in graph
contains two vertices connected to third vertex, Maximum number of edges that N-vertex graph can have such that graph is Triangle free | Mantel's Theorem, Find if a degree sequence can form a simple
graph | Havel-Hakimi Algorithm, Nodes with prime degree in an undirected Graph, Difference Between sum of degrees of odd and even degree nodes in an Undirected Graph, Find a Mother vertex in a Graph
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k'th heaviest adjacent node in a graph where each vertex has weight, Add and Remove vertex in Adjacency Matrix representation of Graph, Add and Remove vertex in Adjacency List representation of
Graph, All vertex pairs connected with exactly k edges in a graph, Topological Sort of a graph using departure time of vertex, Print the degree of every node from the given Prufer sequence, Print the
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final vertex, Print all possible paths in a DAG from vertex whose indegree is 0, Data Structures and Algorithms â Self Paced Course, Ad-Free Experience â GeeksforGeeks Premium, We use cookies to
ensure you have the best browsing experience on our website. 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, Finding in and out degrees of all vertices in a graph, Graph implementation using STL for competitive programming | Set 2 (Weighted graph), Graph
implementation using STL for competitive programming | Set 1 (DFS of Unweighted and Undirected), Printing all solutions in N-Queen Problem, Warnsdorff’s algorithm for Knightâ s tour problem, The
Knight’s tour problem | Backtracking-1, Count number of ways to reach destination in a Maze, Count all possible paths from top left to bottom right of a mXn matrix, Print all possible paths from top
left to bottom right of a mXn matrix, Unique paths covering every non-obstacle block exactly once in a grid, Tree Traversals (Inorder, Preorder and Postorder). It was uplisted from the over the
counter markets to the NASDAQ back in 2013 and at this point commands a market capitalization of $126.642 million. To find your Sun, Moon or Ascendant sign and degree or those of any other planet,
scroll to the planetary detail table This birth chart calculator is set for tropical zodiac. Given a graph G(V,E) as an adjacency matrix representation and a vertex, find the degree of the vertex v
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adjacency matrix from src to des 2 this code is contributed rishabhdeepsingh98. From the vertex DSA Self Paced Course at a 90 degree angle to the base needed and installed on degree! Recommend you to
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Connected component Algorithm equation of a chart vertex in graph. Within days ( over the weekend ) vertex got the equipment they and. The site won ’ t allow us for v before u as a student... Of
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outlined! Both indicate the same a path from this vertices to other exists then the. Just a little problem solving a problem degrees is now usually written 7.5° discussed above in vertex.... Idea is
based on Kosaraju ’ s strongly Connected components, mother vertices are always vertices of source component component. Vertex stores a unit of Vector3 information that we refer to as vertex Colour
Mask the following Monday vertices,... Become industry ready Experience Wesleyan you begin by completing two stipulated degree courses as outlined on following... | {"url":"http://mtable.com.ua/i78cr/how-to-find-the-degree-of-a-vertex-8d6195","timestamp":"2024-11-07T19:10:21Z","content_type":"text/html","content_length":"39313","record_id":"<urn:uuid:248dfb2a-93cc-462e-affa-ff8b92204d20>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00576.warc.gz"} |
Rectangular 3513 - math word problem (3513)
Rectangular 3513
Dad wants to draw a rectangular garden on the plan for this house, which measures 23m and 20m. What will be the sides of this rectangle on the plan in scale M 1:1000?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
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You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Grade of the word problem:
We encourage you to watch this tutorial video on this math problem:
Related math problems and questions: | {"url":"https://www.hackmath.net/en/math-problem/3513","timestamp":"2024-11-06T10:17:30Z","content_type":"text/html","content_length":"47801","record_id":"<urn:uuid:7b878460-54f6-41b5-912c-de3e71a7e6a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00803.warc.gz"} |
DC–DC Power Converters
Direct current–direct current (dc–dc) power converters are employed in a variety of applications, including power supplies for personal computers, office equipment, spacecraft power systems, laptop
computers, and telecommunications equipment, as well as dc motor drives. The input to a dc–dc converter is an unregulated dc voltage Vg. The converter produces a regulated output voltage V, having a
magnitude (and possibly polarity) that differs from V[g].
For example, in a computer off-line power supply, the 120 V or 240 V ac utility voltage is rectified, producing a dc voltage of approximately 170 V or 340 V, respectively. One or more dc-dc
converters then reduce the voltage to the regulated several volts required by the processor integrated circuits (ICs). High efficiency is invariably required because cooling of inefficient power
converters is difficult and expensive. The ideal dc–dc converter exhibits 100% efficiency; in practice, efficiencies of 70% to 95% are typically obtained.
This is achieved using switched-mode, or chopper, circuits whose elements dissipate negligible power. Pulse-width modulation (PWM) allows control and regulation of the total output voltage. This
approach is also employed in applications involving alternating current, including high efficiency dc–ac power converters (inverters and power amplifiers), ac–ac power converters, and some ac–dc
power converters (low-harmonic rectifiers).
Power Stage Operation
A basic dc–dc converter circuit known as the buck converter is illustrated in Fig. a.
A single-pole double throw (SPDT) switch is connected to the dc input voltage Vg as shown. The switch output voltage v[s](t) is equal to Vg when the switch is in position 1 and is equal to
zero when the switch is in position 2.
The switch position varies periodically, such that vs(t) is a rectangular waveform having period Ts and duty cycle D. The duty cycle is equal to the fraction of time that the switch is connected in
position 1, and hence, 0 ≤ D ≤ 1. The switching frequency fs is equal to 1/Ts.
In practice, the SPDT switch is realized using semiconductor devices such as diodes, power metal-oxide-semiconductor field-effect transistors (MOSFETs), insulated-gate bipolar transistors (IGBTs),
bipolar junction transistors (BJTs), or thyristors.
Typical switching frequencies lie in the range 1 kHz to 1 MHz, depending on the speed of the semiconductor devices. The switch network changes the dc component of the voltage. By Fourier analysis,
the dc component of a waveform is given by its average value. The average value of vs(t) is given by
The integral is equal to the area under the waveform or the height Vg multiplied by the time DTs. It can be observed that the switch network reduces the dc component of the voltage by a factor equal
to the duty cycle D. As 0 ≤ D ≤ 1, the dc component of Vs is less than or equal to Vg.
The power dissipated by the switch network is ideally equal to zero. When the switch contacts are closed, then the voltage across the contacts is equal to zero and hence the power dissipation is
zero. When the switch contacts are open, then there is zero current and the power dissipation is again equal to zero. Therefore, the ideal switch network can change the dc component of voltage
without dissipation of power. In addition to the desired dc voltage component Vs, the switch waveform vs(t) also contains undesired harmonics of the switching frequency. In most applications, these
harmonics must be removed, such that the converter output voltage v(t) is essentially equal to the dc component V = Vs. A low-pass filter is employed for this purpose. The converter of Fig. 1
contains a single-section L-C low-pass filter. The filter has corner frequency f0 given by,
The corner frequency f0 is chosen to be sufficiently less than the switching frequency fs, so that the filter essentially passes only the dc component of vs(t).
To the extent that the inductor and capacitor are ideal, the filter removes the switching harmonics without dissipation of power. Thus, the converter produces a dc output voltage whose magnitude is
controllable via the duty cycle D, using circuit elements that (ideally) do not dissipate power. The conversion ratio M (D) is defined as the ratio of the dc output voltage V to the dc input voltage
Vg under steady state conditions:
For the buck converter, M(D) is given by,
It can be observed that the dc output voltage V is controllable between 0 and Vg, by adjustment of the duty cycle D. Figure 3 illustrates one way to realize the switch network in the buck converter,
using a power MOSFET and diode. A gate drive circuit switches the MOSFET between the conducting (on) and blocking (off) states, as commanded by a logic signal δ(t).
When δ(t) is high (for 0 < t < DTs), then MOSFET Q1 conducts with negligible drain-to-source
voltage. Hence, vs(t) is approximately equal to Vg, and the diode is reverse-biased. The positive inductor current iL(t) flows through the MOSFET.
At time t = DTs, δ(t) becomes low, commanding MOSFET Q1 to turn off. The inductor current must continue to flow; hence, iL(t) forward-biases diode D1 and vs(t) is now approximately equal to zero.
Provided that the inductor current iL(t) remains positive, then diode D1 conducts for the remainder of the switching period. Diodes that operate in the manner are called freewheeling diodes. | {"url":"https://university.listenlights.com/2017/10/10/dc-dc-power-converters/","timestamp":"2024-11-09T14:11:16Z","content_type":"text/html","content_length":"45146","record_id":"<urn:uuid:6287e592-090b-45b2-bc50-848b77e9bc66>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00361.warc.gz"} |
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Aggregate Points (Map Viewer Classic)
This tool is now available in Map Viewer, the modern map-making tool in ArcGIS Enterprise. To learn more, see Aggregate Points (Map Viewer).
Tornadoes are the most violent type of storm that occurs in the United States. You want to know the effect of tornadoes, including loss of life, injuries, property damage, and financial loss, in each
state and county. You have access to tornado locations across the United States, but you need a better way to visualize your data within the boundaries of your choice. You can aggregate your tornado
data into state and county boundaries and normalize your data by population to find the areas most affected by tornadoes.
Usage notes
Aggregate Points is designed to collect and summarize point features within a set of boundaries. The input parameters must include points to be aggregated and aggregation areas.
You can provide the area layer to use for analysis, or you can generate bins of a specified size and shape (hexagon or square) into which to aggregate. The bin size specifies how large the bins are.
If you are aggregating into hexagons, the size is the height of each hexagon and the width of the resulting hexagon will be 2 times the height divided by the square root of 3. If you are aggregating
into squares, the bin size is the height of the square, which is equal to the width.
You can add a layer that is not in Map Viewer Classic to the tool pane by selecting Choose Analysis Layer on the drop-down menu.
The Keep areas with no points box is checked by default. When the box is checked, all of the areas used in the analysis will be included in the result, regardless of the location of points. Areas
with no points will be empty and have a count of 0. When the box is unchecked, the areas with no points are completely removed from the result layer. Unchecking the box may have a significant effect
on the result areas.
The images show the difference in result layers when areas with no points are kept (left) or not kept (right).
The most basic aggregations will calculate a count of the number of points in each boundary. Basic statistics (sum, minimum, maximum, average, and standard deviation) can also be calculated on
numerical fields. The statistics will be calculated on each area separately.
Statistical calculations can also be grouped using a field with categorical values. When statistics are grouped by a field, the statistics are calculated for both the area as a whole and for each
group. Both statistics can be viewed in the result layer's pop-up. The overall statistics are given numerically and the grouped statistics are given in chart form within the pop-up. A summary table
listing each feature and statistic by group by field value will also be created. The Add minority, majority check box will return the groups with the lowest and highest counts and the Add percentages
check box will return the percentage of points in the minority and majority groups.
If Use current map extent is checked, only the features in the input point layer and area layer visible within the current map extent will be analyzed. If unchecked, all features in both the input
point layer and the area layer will be analyzed, even if they are outside the current map extent.
Inputs must include a point layer and an area layer. Lines and areas cannot be aggregated within boundaries using the Aggregate Points tool.
How Aggregate Points works
The sections below describe the functionality of the Aggregate Points tool.
Generating bins
Square and hexagonal bins can be generated for aggregation areas rather than aggregating points into an input area layer. The bins are generated in a custom, area-preserving projected coordinate
system using the specified size dimensions to ensure the sizes are equal and appropriate for the area of interest. An appropriate equal-area projection and parameters are chosen based on the
geographic extent of the input layers. Once bins are created, they are projected back to the coordinate system of the input data before being used in the analysis.
After the analysis is complete, the result is projected to Web Mercator for display (the default) or to the projection of your custom basemap. A Web Mercator projection may cause your results to
appear distorted, especially for large bins or bins near the polar regions. These distortions are part of the display only and do not reflect an inaccurate analysis.
Standard deviation is calculated using the following equation:
• N = Number of observations
• x[i] = Observations
• x̄ = Mean
Point layers are summarized using only the point features within the input boundary. The results will be displayed using graduated symbols.
The figure and table below explain the statistical calculations of a point layer within a hypothetical boundary. The Population field was used to calculate the statistics (Sum, Minimum, Maximum,
Average, and Std Deviation) for the layer.
Point layers are summarized using only points located within the boundary layer. An example attribute table is displayed above with values to be used in hypothetical statistic calculations.
Statistic Result District A
Sum 280 + 408 + 356 + 361 + 450 + 713 = 2,568
Minimum of:
[280, 408, 356, 361, 450, 713] = 280
Maximum of:
[280, 408, 356, 361, 450, 713] = 713
Average 2,568/6 = 428
Std Deviation = 150.79
Null values are excluded from all statistical calculations. For example, the mean of 10, 5, and a null value is 7.5 ((10+5)/2).
A real-life scenario in which this analysis could be used is in determining the total number of students in each school district. Each point represents a school. The Type field gives the type of
school (elementary, middle school, or secondary) and a student population field gives the number of students enrolled at each school. The calculations and results are given for District A in the
table above. From the results, you can see that District A has 2,568 students. When running the Aggregate Points tool, the results would also be given for District B.
Similar tools
Use Aggregate Points to summarize points within areas. Other tools may be useful in solving similar but slightly different problems.
Map Viewer Classic analysis tools
If you are trying to summarize lines or areas, use the Summarize Within tool. | {"url":"https://enterprise-k8s.arcgis.com/en/latest/analyze/aggregate-points.htm","timestamp":"2024-11-03T16:13:47Z","content_type":"text/html","content_length":"26443","record_id":"<urn:uuid:1e2cfd65-09e8-4ca9-8622-ba373894f3af>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00146.warc.gz"} |
WadRay library for Cairo
This library implements two types of fixed-point decimal numbers, "wad" (18 decimals of precision) and "ray" (27 decimals of decimal numbers), available in signed (SignedWad and SignedRay) and
unsigned (Wad and Ray) versions, written in Cairo for Starknet.
Wad and Ray are implemented as structs with a single u128 member for the value, while SignedWad and SignedRay are implemented as structs with a u128 member for the value and bool member for the sign
(i.e. if the sign is true, then the value is negative).
This library includes arithmetic, comparison and conversion functions.
Addition and Subtraction
Addition and subtraction can be performed via the Add, AddEq, Sub and SubEq traits as follows:
• a + b
• a += b
• a - b
• a -= b
where both a and b are of the same type.
Multiplication and Division
Multiplication and division can be performed via the the Mul, MulEq, Div and DivEq traits as follows:
• a * b
• a *= b
• a / b
• a /= b where both a and b are of the same type.
There is also a set of functions for operations involving a Wad and a Ray:
• wmul_wr/wmul_rw: Multiply a Wad value with a Ray value, and divide the product by one Wad scale to return a Ray
• wdiv_rw: Scale up a Ray value by one Wad scale and divide the scaled value by a Wad value to return a Ray
• rmul_wr/rmul_rw: Multiply a Wad value with a Ray value, and divide the product by one Ray scale to return a Wad
• rdiv_wr: Scale up a Wad value by one Ray scale and divide the scaled value by a Ray to return a Wad
• rdiv_ww: Scale up a Wad value by one Ray scale and divide the scaled value by a Wad to return a Ray
For multiplication, the prefixes w and r denote whether the product is divided (i.e. scaled down) by a Wad or Ray respectively. For division, the prefixes w and r denote whether the first operand is
multiplied (i.e. scaled up) by a Wad or Ray before the division respectively.
As these are fixed point operations, do take note that there will be a loss of precision.
Zero and one values
The following values and functions for both Wad and Ray, and SignedWad and SignedRay, are available via the Zeroable and Oneable traits.
• WadZeroable::zero()/RayZeroable::zero(): Returns the zero value for Wad and Ray respectively
• is_zero(): Returns true if the Wad or Ray value is zero
• is_non_zero() Returns true if the Wad or Ray value is not zero
• WadOneable::one()/RayOneable::one(): Returns the scale value for Wad (i.e. 10^18) and Ray (i.e. 10^27) respectively
• is_one(): Returns true if the Wad or Ray value is the scale value (i.e. 10^18 and 10^27 respectively)
• is_non_one() Returns true if the Wad or Ray value is not the scale value (i.e. 10^18 and 10^27 respectively)
• SignedWadZeroable::zero()/SignedRayZeroable::zero(): Returns the zero value for Wad and Ray respectively
• is_zero(): Returns true if the SignedWad or SignedRay value is zero, regardless of the sign.
• is_non_zero() Returns true if the SignedWad or SignedRay value is not zero
• SignedWadOneable::one()/SignedRayOneable::one(): Returns the positive scale value for "wad" (i.e. 10^18) and "ray" (i.e. 10^27) respectively
• is_one(): Returns true if the SignedWad or SignedRay value is the positive scale value (i.e. 10^18 and 10^27 respectively)
• is_non_one() Returns true if the SignedWad or SignedRay value is not the positive scale value (i.e. 10^18 and 10^27 respectively)
Bound values
The bound values for both Wad and Ray can be obtained via the BoundedInt trait.
• BoundedWad::max(): Returns the maximum Wad value of 2^128 - 1
• BoundedWad::min(): Returns the minimum Wad value of 0
• BoundedRay::max(): Returns the maximum Ray value of 2^128 - 1
• BoundedRay::min(): Returns the minimum Ray value of 0
Comparison for both Wad and Ray, and SignedWad and SignedRay, can be performed via the PartialEq and PartialOrd traits as follows:
• a == b
• a != b
• a > b
• a >= b
• a < b
• a <= b
where both a and b are of the same type.
Type Casting
Any type that can be converted to a u128 via the Into trait can similarly be converted to a Wad or Ray via the Into trait.
Additionally, the following conversions from integer types are supported for SignedWad and SignedRay via the Into trait`:
• u128 -> SignedWad: Convert a u128 to a SignedWad without modifying the value, with the sign set to false
• u128 -> SignedRay: Convert a u128 to a SignedRay without modifying the value, with the sign set to false
The following conversions from this library's types can also be performed via the Into trait:
• Wad -> felt252: Convert a Wad to a felt252 without modifying the value
• Wad -> u256: Convert a Wad to a u256 without modifying the value
• Wad -> SignedWad: Convert a Wad to a SignedWad without modifying the value, with the sign set to false
• Ray -> SignedRay: Convert a Ray to a SignedRay without modifying the value, with the sign set to false
• SignedWad -> felt252: Convert a SignedWad to a felt252 without modifying the value
• SignedRay -> felt252: Convert a SignedRay to a felt252 without modifying the value
The following conversions can be performed via the TryInto trait:
• u256 -> Option::<Wad>: Returns Option::Some<Wad> if the value is within bounds of u128 or Option::None otherwise
• SignedWad -> Option::<Wad>: Returns Option::Some<Wad> if sign is false or Option::None otherwise
• SignedRay -> Option::<Ray>: Returns Option::Some<Ray> if sign is false or Option::None otherwise
The following functions can be used to scale between Wad and Ray:
• fn ray_to_wad(x: Ray) -> Wad: Divide the Ray value by 10^9 and return a Wad
• fn wad_to_ray(x: Wad) -> Option::<Ray>: Multiply the Wad value by 10^9 and return Option::Some<Ray> if there is no overflow or Option::None otherwise
Additional notes on conversion between
Starting from v0.5.0 of this library, we have made significant changes to how Wad values are converted to Ray values and vice versa. This aims to improve type safety and align with the semantics of
Rust's Into trait.
Key Changes
1. Previously, Into<Wad, Ray> scaled the value by 10^9 while Into<Ray, Wad> scaled the value by 10^-9. Both now perform direct type cast without value modification
2. Introduced wad_to_ray() and ray_to_wad() functions for value-preserving conversions
Recommended Usage
1. Prefer wad_to_ray() and ray_to_wad() for most conversions between Wad and Ray.
2. Use the Into trait only when a simple type cast is required (expected to be rare).
The following functions are available for SignedWad and SignedRay via the Signed trait:
• is_negative(): Returns true if the value is less than zero
• is_positive(): Returns true if the value is greater than zero
To use this library, add the repository as a dependency in your Scarb.toml:
wadray = "0.5.0"
then add the following line in your .cairo file
use wadray::Wad;
fn add_wad(a: Wad, b: Wad) -> Wad {
a + b
You can also refer to the test file src/test_wadray.cairo for another example.
Run tests
To run the tests:
scarb test
Formal Verification
This library has been formally verified using our tool Aegis. Specifications and their correctness proof can be found in this repository, the verification currently refers to revision 30f7664 of this
We welcome contributions of any kind! Please feel free to submit an issue or open a PR if you have a solution to an existing bug.
This library is released under the MIT License. | {"url":"https://scarbs.xyz/packages/wadray","timestamp":"2024-11-09T03:55:37Z","content_type":"text/html","content_length":"49739","record_id":"<urn:uuid:e59e2ddc-8497-42cd-9866-3e333a740ae4>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00812.warc.gz"} |
Modern Algebra Hand Written Note By SKM Academy - HUNT4EDU
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Q3. The beam AB, shown in Figure 3a), has the cross-sectional area as
Q3. The beam AB, shown in Figure 3a), has the cross-sectional area as shown in Figure 3b).The loaded beam is a “1"-shape steel beam reinforced with a horizontal rectangular-shape wooden
beam throughout its length AB (shown in black in Figure 3b). Do the following: (a) Derive the centroid position from the top of the cross-section, y (b) Calculate the second moment of area about axis
zz for the overall (steel and wooden component) beam cross-section. The second moment of area of the steel section of the beam about axis zz is Isteel=4.24 x 10-5 (m4)[5 marks] (c) Calculate maximum
tensile and compressive bending stresses in the beam at a position8 m away from the support A.[8 marks] (d) Determine the bending stress at the point located 62 mm above the bottom edge of the
cross-section (point C), at a distance of 8 m away from the support A. Also calculate the percentage increase of bending stress at this point if the steel beam was not reinforced with the wooden beam
as shown in Figure 3b.[5 marks]
Fig: 1
Fig: 2
Fig: 3
Fig: 4
Fig: 5
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Calculate the Kinetic Frictional Force
• Thread starter SRooney
• Start date
The frictional force is the force that opposes the motion of the fireman down the pole. It is equal in magnitude and opposite in direction to the net force. Therefore, the magnitude of the frictional
force exerted on the fireman as he slides down the pole is 19.8 N.
Member warned to use the homework template for posts in the homework sections of PF.
The alarm at a fire station rings and a 71.3-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.63 m). Just before landing, his speed is 1.42 m/s. What is the
magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
m = 71.3 kg
d = 3.63 m
Vf = 1.42 m/sd = Vf(t)/2
a = Vf/t
F = ma
3.63 m = 1.42 m/s (t) /2
7.26 m = 1.42 m/s (t)
t = 5.113 s
a =1.42 m/s /5.113 s
a = .278 m/s^2
F = 71.3 kg (.278 m/s^2)
F = 19.8 N
But that's not the answer
Homework Helper
Gold Member
SRooney said:
The alarm at a fire station rings and a 71.3-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.63 m). Just before landing, his speed is 1.42 m/s. What is the
magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
m = 71.3 kg
d = 3.63 m
Vf = 1.42 m/sd = Vf(t)/2
a = Vf/t
F = ma
3.63 m = 1.42 m/s (t) /2
7.26 m = 1.42 m/s (t)
t = 5.113 s
a =1.42 m/s /5.113 s
a = .278 m/s^2
F = 71.3 kg (.278 m/s^2)
F = 19.8 N
But that's not the answer
I didn't check your calculations but the force you have calculated is the "net force" on the fireman's body. You are asked to find the frictional force.
FAQ: Calculate the Kinetic Frictional Force
1. What is kinetic friction?
Kinetic friction is the force that opposes the relative motion of two objects that are in contact with each other. It is caused by the microscopic roughness of the surfaces in contact, which results
in interlocking and resistance to motion.
2. How is kinetic friction different from static friction?
Kinetic friction occurs when two objects are in motion relative to each other, while static friction occurs when there is no relative motion between the objects. In other words, kinetic friction acts
to slow down a moving object, while static friction acts to prevent an object from moving.
3. How do you calculate the kinetic frictional force?
The kinetic frictional force can be calculated using the formula Fk = μk * N, where Fk is the kinetic frictional force, μk is the coefficient of kinetic friction, and N is the normal force between
the two objects in contact.
4. What factors affect the magnitude of kinetic friction?
The magnitude of kinetic friction depends on the coefficient of kinetic friction, the normal force between the objects, and the surface properties of the materials in contact. Rougher surfaces and
higher normal forces result in a higher kinetic frictional force.
5. How is the coefficient of kinetic friction determined?
The coefficient of kinetic friction is determined experimentally by measuring the force required to keep an object in motion at a constant velocity on a given surface. It is a dimensionless value
that varies depending on the materials in contact and the surface properties. | {"url":"https://www.physicsforums.com/threads/calculate-the-kinetic-frictional-force.901170/","timestamp":"2024-11-06T05:45:31Z","content_type":"text/html","content_length":"78873","record_id":"<urn:uuid:9d9a1af5-e659-448e-84f7-12d4fe878b69>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00083.warc.gz"} |
559 Attometer/Second Squared to Knot/Millisecond
Attometer/Second Squared [am/s2] Output
559 attometer/second squared in meter/second squared is equal to 5.59e-16
559 attometer/second squared in centimeter/second squared is equal to 5.59e-14
559 attometer/second squared in decimeter/second squared is equal to 5.59e-15
559 attometer/second squared in dekameter/second squared is equal to 5.59e-17
559 attometer/second squared in femtometer/second squared is equal to 0.559
559 attometer/second squared in hectometer/second squared is equal to 5.59e-18
559 attometer/second squared in kilometer/second squared is equal to 5.59e-19
559 attometer/second squared in micrometer/second squared is equal to 5.59e-10
559 attometer/second squared in millimeter/second squared is equal to 5.59e-13
559 attometer/second squared in nanometer/second squared is equal to 5.59e-7
559 attometer/second squared in picometer/second squared is equal to 0.000559
559 attometer/second squared in meter/hour squared is equal to 7.24464e-9
559 attometer/second squared in millimeter/hour squared is equal to 0.00000724464
559 attometer/second squared in centimeter/hour squared is equal to 7.24464e-7
559 attometer/second squared in kilometer/hour squared is equal to 7.24464e-12
559 attometer/second squared in meter/minute squared is equal to 2.0124e-12
559 attometer/second squared in millimeter/minute squared is equal to 2.0124e-9
559 attometer/second squared in centimeter/minute squared is equal to 2.0124e-10
559 attometer/second squared in kilometer/minute squared is equal to 2.0124e-15
559 attometer/second squared in kilometer/hour/second is equal to 2.0124e-15
559 attometer/second squared in inch/hour/minute is equal to 4.7537007874016e-9
559 attometer/second squared in inch/hour/second is equal to 7.9228346456693e-11
559 attometer/second squared in inch/minute/second is equal to 1.3204724409449e-12
559 attometer/second squared in inch/hour squared is equal to 2.8522204724409e-7
559 attometer/second squared in inch/minute squared is equal to 7.9228346456693e-11
559 attometer/second squared in inch/second squared is equal to 2.2007874015748e-14
559 attometer/second squared in feet/hour/minute is equal to 3.9614173228346e-10
559 attometer/second squared in feet/hour/second is equal to 6.6023622047244e-12
559 attometer/second squared in feet/minute/second is equal to 1.1003937007874e-13
559 attometer/second squared in feet/hour squared is equal to 2.3768503937008e-8
559 attometer/second squared in feet/minute squared is equal to 6.6023622047244e-12
559 attometer/second squared in feet/second squared is equal to 1.8339895013123e-15
559 attometer/second squared in knot/hour is equal to 3.9117926718e-12
559 attometer/second squared in knot/minute is equal to 6.519654453e-14
559 attometer/second squared in knot/second is equal to 1.0866090755e-15
559 attometer/second squared in knot/millisecond is equal to 1.0866090755e-18
559 attometer/second squared in mile/hour/minute is equal to 7.5026843235505e-14
559 attometer/second squared in mile/hour/second is equal to 1.2504473872584e-15
559 attometer/second squared in mile/hour squared is equal to 4.5016105941303e-12
559 attometer/second squared in mile/minute squared is equal to 1.2504473872584e-15
559 attometer/second squared in mile/second squared is equal to 3.4734649646067e-19
559 attometer/second squared in yard/second squared is equal to 6.1132983377078e-16
559 attometer/second squared in gal is equal to 5.59e-14
559 attometer/second squared in galileo is equal to 5.59e-14
559 attometer/second squared in centigal is equal to 5.59e-12
559 attometer/second squared in decigal is equal to 5.59e-13
559 attometer/second squared in g-unit is equal to 5.7002136305466e-17
559 attometer/second squared in gn is equal to 5.7002136305466e-17
559 attometer/second squared in gravity is equal to 5.7002136305466e-17
559 attometer/second squared in milligal is equal to 5.59e-11
559 attometer/second squared in kilogal is equal to 5.59e-17 | {"url":"https://hextobinary.com/unit/acceleration/from/ams2/to/knms/559","timestamp":"2024-11-04T12:11:12Z","content_type":"text/html","content_length":"97653","record_id":"<urn:uuid:e03a91d9-3356-4875-b02c-7482e58d44f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00615.warc.gz"} |