text:
# Why and how are resonance frequencies of a system dependent upon the shape, mass and the way it is constrained?

I was reading a general overview on the frequencies before conducting a modal analysis for my structure using ANSYS, in hopes of obtaining the natural frequencies for my structure. I read in some article that if you change the geometrical shape of your structure, change its overall mass, and even change the locations of the supports on it while conducting the modal analysis, it is very probable you will see a different set of modal frequencies. My question is why and how? Moreover, I also want to ask that after getting the natural frequencies for my structure, does this mean that ALL the structure must be excited by an external frequency equal to natural frequency, to expect some dangerous and catistropic failure? Or even if I excite certain parts of my structure, for example some specific locations on my structure where it is connected to another part (which is essentially causing my structure to vibrate at these connecting locations) should also be dangerous?

• Have you researched natural frequency? – Solar Mike Mar 28 at 19:17
• Yes ofcourse, thats what I am aiming to get for my structure. Actually, my structure is a composite whose thickness can be changed by increasing/decreasing the number of plies. I want to know how would a change in thickness change the natural frequencies of my structure. – Rameez Ul Haq Mar 28 at 19:24
• So, if you did research the natural frequencies then you would have found the formulae... – Solar Mike Mar 28 at 19:30
• So you might find this useful: brown.edu/Departments/Engineering/Courses/En4/Notes/… – Solar Mike Mar 28 at 19:36

In order to answer that question you need to build up your knowledge in vibrational dynamics. This usually takes a whole semester in undergraduate physics at the latter stage of an engineering curriculum.

So you need to progress -at least- through the following concepts:

• The free response of the undamped Harmonic Oscillator with 1 dof (mass spring no external excitation)
• The free response of the damped Harmonic Oscillator with 1 dof (mass damper spring no external excitation)
• The forced harmonic response of the damped Harmonic Oscillator with 1 dof (mass damper spring with external sinusoidal excitation)
• The forced harmonic response of the damped Harmonic Oscillator with 1 dof (mass damper spring with external sinusoidal excitation)

Then you need to start thinking about mdof systems, and the matrix form of the problem (this is very helpful especially with the FE), and you need to do the following:

• Start with the simplest 2DOF (undamped no excitation) to see the form of the system.

The main difference from the SDOF systems, is the interactions between the position of the masses (and therefore the springs and dampers).

It is very crucial to understand at that there can be many different transformations for the same problem depending on the generalised coordinates you select. A simple example of what I mean is the following problem.

In the above problem you can describe the motion equations with coordinates $$x_1$$ and $$x_2$$ (the spring displacements), or equivalenty by $$x,\theta$$ (the motion of the center of gravity and the rotation). The result should always be the same, however the individual coordinate responses will vary.

• Then you can proceed to Modal analysis

What is important to understand is that among the infinite number of generalised coordinates , where the responses are coupled, and additionally that there is at least on set of coordinates (sometimes called principal coordinates) where the transformed coordinate is decoupled (and are solved like the sdof - hence why you need to understand the behaviour of SDOF systems).

To obtain that set of coordinates, one way is to use the eigenvalues and eigenvectors of the mass normalised stiffness matrix. The cool thing is that you can rotate the initial problem, to obtain the principal coordinates, solve for the response of each decoupled system separately and then translate back to the original coordinates. The rotational transformation does not change the eigenfrequencies, in fact in the decoupled system you have a mode shape and a sinusoidal response. So each "mass" is pulsing proportionally to the other following one eigenfrequency. When you rotate back at the original masses/coordinate system, you get a contribution from both decoupled responses (hence the seemingly chaotic behavior).

• The final step is the Forced response of MDOF systems.

The same transformation that is applied to the Masses and springs can be applied to the force matrix (both directions). If you are planning to excite at a resonant frequency the mass you can apply a resonant force on the principal coordinate system and the transform that force matrix to the original matrix. When you do that you see that in the general case, you need to apply on all of the original coordinates a portion of the force with eigenfrequency $$\omega_i$$ in order to obtain the mode shape.

The transformation is essentially a rotation matrix, which can be applied to the excitation matrix. However this means that in order to excite the if you excite the nodes with the frequency at those eigenfrequencies,

• You do describe the steps one should take in order to understand some of the mathematics behind it, but those are not strictly necessary in order to have some intuitive understanding of it. Also, it is worth mentioning that doing a finite element analysis using matrices has its limitations, since the underlying physics is actually a partial differential equation (PDE). Only the first few lowest natural frequencies obtained from those matrices are usually accurate, because the matrices are obtained by discretizing space which gives an approximation of the underlying PDE. – fibonatic Mar 29 at 15:42
• I was writing my reply, probably at the same time as you were. I do confess, my approach was more of the undergraduate textbook, for discrete systems. The way I see it you took the continuum system which applies better to real structures, and you took a more hand's on approach. To be honest, the question was very vague, and there are different interpretations. I do feel that both posts have their own merit. – NMech Mar 29 at 16:09
• I agree that both our answers have merit. And I agree that the question itself is a bit vague, probably because of the lacking theoretical knowledge. This also why I tried to keep a lot of the underlying theory out of it. I myself find it useful to first gain some intuition and maybe spark some curiosity before diving deeper into underlying mathematical concepts. But hopefully the combination of our two answers might inspire some people reading them to learn more about it. – fibonatic Mar 29 at 17:00

Natural frequencies inside structures are standing waves that are reflected and propagated throughout the structure. The frequency of such wave depends on the shape of that wave and the speed at which sound travels through the material the structure is made of. The shape of possible waves in a structure I think can be well illustrated with Chladni figures. Adding supports constraints what kind of wave shapes can appear inside a structure, which you can also see in this video where putting a finger against the Chladni plate can be seen as a support. I have to note that in that video the natural frequencies of the plate are not actually altered, the temporary support initially limits which natural frequency/mode is excited.

Those standing waves do eventually die out if no energy is supplied to excite that natural frequency/mode. This is due to ways of dissipating energy, such as sound or damping behavior the material itself. But if energy at the "right" frequency is supplied to a structure at "right" locations, then the rate of energy dissipation might be too small, causing the energy of that natural frequency/mode to build up to dangerous levels until something might break. These "right" locations means that one isn't supplying the energy at nodes of that wave shape.

Adding mass to a structure can affect the way sound is reflected near that mass in the structure and thus affect wave shapes. This also closely related to impedance, which is demonstrated in this video.

When exciting a structure, not near any of the nodes of the natural frequencies, one does not have be exactly at any of the natural frequencies in order to get a large response of the structure. Though, the closer you are to one of the natural frequencies the larger the response. This can be illustrated with frequency response functions, sometimes also referred to as Bode plots, whose core concept I think is demonstrated well in this video.


text:
This calculator determines the total path loss provided by an RF n-way power splitter.

(dB)

(dB)

### Overview

An N-way power divider, also known as a Wilkinson power divider, has become very popular in the RF engineering world because it allows isolation between the output ports while maintaining a matched condition on all ports. This tool will compute for the total loss in an N-way power divider given the total number of ports and the path loss. You can compute for up to 16 paths. Note that the path loss and total loss are presented in decibels (dB).

### Equations

$$Loss_{total} = -10log_{10}(\frac{1}{N}) + Loss_{path}$$

Where:

$$N$$ = number of ports

### Applications

The Wilkinson power divider has many uses, particularly in RF technology. When you feed power to one port of a power divider, the power divides equally between all ports. The bidirectional nature of power dividers means you can also use them as power combiners.

Wilkinson power dividers are mostly seen in I/Q modulators & demodulators, high power transmission systems, and signal mixers. For example, a 5-way power divider can be used in multiple-channel superheterodyne receivers, to distribute the local oscillator signal to all channels at one time from a single source. The Wilkinson power divider can also be used in high power amplifiers and T/R modules for power combining.

format: plain


text:
# NPN Transistor Analysis: How is collector current determined?

I'm trying to understand NPN transistors by reading the chapter on them in Razavi's "Fundamentals of Microelectronics", and it says that collector current (Ic) is a function of the base-emitter voltage (Vbe):

$$I_c = I_s \cdot e^{\frac{V_{be}}{V_t}}$$

where Is and Vt are constants.

In the following section, the beta value is introduced and it states that the collector current is base current times beta:

$$I_c = \beta \cdot I_b$$

I'm confused whether Ic is determined by Vbe or Ib (or both?) since both these values can be controlled independent of each other using a resistor.

Can you give an explanation of how terminal currents and/or voltages are determined?

Pic for reference:

• Vbe and Ib are not independent. If you add a resistor in front of the base, the current through base will decrease, but so will the voltage across the base-emitter junction. – avakar Apr 3 '12 at 11:46


text:
# All Resources for HPC: CU 6: 4.4: Graphs of Sine and Cosine: Sinusoids

Resource Type
Product Rating
File Type

Compressed Zip File

140 MB|8 files
Share
Product Description
- Sine and cosine functions
- Amplitude, period, frequency
- Transformation of sine and cosine functions
- Graphing sine and cosine functions
- construct a sinusoid given the amplitude and period that goes through a given point
- state the amplitude and period of the sinusoid, and (relative to the basic function) the phase shift and vertical translation
- constructing a sinusoidal model using time (tide)
- applications involving sine and cosine functions
Total Pages
8 files
Included
Teaching Duration
2 days
Report this Resource
\$8.00
List Price:
\$12.00
You Save:
\$4.00
More products from Math Hammy

Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.


text:
1. Jan 22, 2010

### 2h2o

I've come across this before, but for the life of me can't seem to get anywhere with it. I can plug this into technology and get an answer (and then move on...), but I want to know how to do it without.

An example problem:

$$4\sqrt{\sin(\pi/4)}$$

I get as far as

$$2\sqrt{2\sqrt{2}} = 2(2(2)^\frac{1}{2})^\frac{1}{2}$$

but then I run out of steam and don't see the next step. Surely it is something simple that I am missing, but I just don't see it.

Thanks for any tips. Cheers!

2. Jan 22, 2010

### Char. Limit

I don't know what you're asking, but since $$2\sqrt{2}=\sqrt{8}$$, it equals two times the quartic root of 8...

3. Jan 22, 2010

### 2h2o

I have $$\sqrt{8}$$ as an intermediate step in my work.

$$4\sqrt{\sin(\pi/4)} = \sqrt{\frac{16\sqrt{2}}{2}} = \sqrt{8\sqrt{2}} = \sqrt{4*2\sqrt{2}} = 2\sqrt{2\sqrt{2}}$$

When I plug into technology, I get the answer:

$$2*2^\frac{3}{4}$$

but I don't know how to get there. That's what I'm asking; sorry for being unclear.

4. Jan 22, 2010

### Char. Limit

Well, yeah.

Going with my two times quartic root of 8 solution, since 8 is 2 cubed, the quartic root of 8 is 2 to the three-fourths power, and the number simplifies to your number.

5. Jan 22, 2010

### 2h2o

Of course. Thank you very much! I definitely should've seen that! :)

Cheers.

6. Jan 22, 2010

### jacksonwalter

When you multiply a two numbers with the same base, the exponents add. So,

$$2(2^{1}*2^{\frac{1}{2}})^{\frac{1}{2}} = 2(2^{1+\frac{1}{2}})^{\frac{1}{2}} = 2(2^{\frac{3}{2}*\frac{1}{2}}) = 2*2^{\frac{3}{4}}$$


text:
# What is the derivative of f(t) = (t^2+1 , t-e^(1-t) ) ?

Jan 3, 2018

$f ' \left(t\right) = \left(2 t , 1 + {e}^{1 - t}\right)$

#### Explanation:

Take the derivative component-by-component:

$f \left(t\right) = \left({t}^{2} + 1 , t - {e}^{1 - t}\right)$

$f ' \left(t\right) = \left(\frac{d}{\mathrm{dt}} \left({t}^{2} + 1\right) , \frac{d}{\mathrm{dt}} \left(t - {e}^{1 - t}\right)\right)$

$= \left(2 t , 1 - {e}^{1 - t} \left(- 1\right)\right)$ (don't forget the chain rule!)

$f ' \left(t\right) = \left(2 t , 1 + {e}^{1 - t}\right)$


text:
## LaTeX forum ⇒ Fonts & Character Sets ⇒ How to make \circledR show up when using newtx?

Information and discussion about fonts and character sets (e.g. how to use language specific characters)
user49915
Posts: 58
Joined: Wed Apr 17, 2019 12:51 pm

### How to make \circledR show up when using newtx?

Compile the following code with latex+dvips:
1. \documentclass{article}
2. \usepackage{newtxtext,newtxmath}
3. \begin{document}
4. \circledR $$\circledR$$
5. \end{document}

As a result, we get an empty document (except the page number). The compilation finishes successfully, only there is a hint in the log file: "Missing character: There is no ò in font ntxsym!" How to get ® in newtx in text and in math? Any (quick) bugfix? After all, the changelog of newtx says for version 1.515 this: "Corrected definitions of [...] \circledR", so, this character should work.

The newtx author has been e-mailed on 2019-07-11, but has not reacted as of 2019-07-17. If anyone has an e-mail address of him different from the one mentioned at CTAN (http://web.archive.org/web/20170614225447/http://www.ctan.org/tex-archive/fonts/newtx), please feel free to nag him.

Crosspost: http://tex.stackexchange.com/questions/ ... p-in-newtx


text:
# How can I calculate the correct rotation output when input needs modification?

I am making a GPS for vehicles in a game and I need them to turn accordingly to the rotation between the ending point and starting point. I have calculated the rotation, and before I can apply the rotation I found out I need to edit the values a bit. Not knowing how to do it but observing the roll/pitch has to be changed somewhere I have come up with the following formula, in which I input the Pitch and Yaw and I get back the almost correct pitch and roll output, but not perfect.

This is the formula:

//values are normalised in range [0,360]
Pair CorrectValues(float Pitch, float Yaw)
{
float
Roll_out = 0.0,
Pitch_out = 0.0;

//if Yaw is bigger than 315 OR Yaw is smaller than 45
if(Yaw > 315.0 || Yaw < 45.0)
Roll_out = -Pitch;

//if YAW is between 45 and 135
if(Yaw > 45.0 && Yaw < 135.0)
Pitch_out = -Pitch;

//if YAW is between 135 and 225
if(Yaw > 135.0 && Yaw < 225.0)
Roll_out = Pitch;

//if YAW is between 225 and 315
if(Yaw > 225.0 && Yaw < 315.0)
Pitch_out = Pitch;

return Pair(Roll_out,Pitch_out);
}

I think I need to do somthing with cos/sin, I have tried many combinations but all of them just make the rotation weird (which mean I just randomly tried applying sin/cos in many combinations to the out values in the form of X_out = sin or cos (Pitch) * Pitch).

Can anyone give me a push in the good direction to solve this? The main issue is that it "yaggy" / "laggy"

Edit: As for request for more detail I'm putting it here:

I Need to set the vehicle's quaternion orientation to the one which you can calculate between two positions, which means there is one rotation we will not calculate = roll.

So I did calculate the Pitch and Yaw between two 3D points (XYZ) with this formula:

const float MATH_PI = 3.14159265359;
void GetPitchYawBetweenCoords(point3D source, point3D target, point2D &output)
{
point delta3D = target-source;//X2-X1, Y2-Y1, Z2-Z1
output.x = atan2( delta3D.y, delta3D.x ) * RAD_TO_DEG;//yaw
output.y = atan2( delta3D.z, sqrt( delta3D.x * delta3D.x + delta3D.y * delta3D.y ) ) * RAD_TO_DEG;//pitch
}

Then I did use a programming library called OpenGL Mathematics to convert these rotations to a Quaternion, and apply it to the vehicle, but this wasn't working all the time so I did the above modifications which approaches the "correct rotations".

But I'm wondering if there is any correct or better way to do this?

Ad an update to my comment:

glm::eulerAngles assumes the input of Pitch,Roll,Yaw in radians which I do, It's weird how this works if the Yaw angle is 0 and the Pitch is set correctly, and if the Yaw angle is 180 degrees then the pitch is reversed.. and at other angles.. well it looks like it's being interpolated? I will upload a screenshot.

a.k.a. : When I ONLY set the Pitch and Yaw without ANY calculations/modifications and calculate it into a quaternion, the vehicle rotation is like being "interpolated" / slowly inversed, look here:

The center of All cars equals (0.0,0.0,0.0) and ALL the cars have the pitch set to a value which equals (XY In front of Car at distance 10.0, CurrentCarZ-10.0). (they all should point DOWN)

Yet you can see this is somehow producing interesting results.

Why is this "interpolating" like this?

Only the car (the closest one to me - ID 495 - pink) which is totally facing UP by some amount of degrees has actually the inverted rotation, all other not what I expected (you can see they are a bit rolled, and the one at heading 0 is totally perfect - ID 499 - green).

(Note: North is 180 degrees)

The "formula" you can see in the picture equals then to this (this is a function which just uses the GLM library):

EulerToQuat(Pitch, 0.0, Yaw, Qw, Qx, Qy, Qz);
//{Qw,Qx,Qy,Qz} = glm::quat(glm::vec3(Pitch,0.0,Yaw)*vector_to_radians)//constructs a quaternion form XYZ euler angles

• This sort of correction shouldn't be necessary. Could you please describe in a little more detail how the pitch and yaw angles are calculated, and how they're applied? I wonder if there's an error in that code. Maybe the rotations are being applied in the wrong order, or something like that. – Chris Culter Jul 31 '13 at 17:41
• I'm having a hard time understanding what your question is. Are you saying that you want a more accurate function than the one you gave? – rschwieb Jul 31 '13 at 17:41
• I just randomly tried applying sin/cos in many combinations... Usually people put the "flail about wildly with nonsense" problem solving strategy after the "try to reason carefully" strategy :) THis sort of problem should develop a fairly concrete picture which you can start reasoning from. – rschwieb Jul 31 '13 at 18:23
• @Chris Updated the question – Gizmo Jul 31 '13 at 18:43
• @Gizmo Okay, you're effectively applying pitch first, then yaw. That makes sense. Which OpenGL function are you calling? Does it assume the same order? – Chris Culter Jul 31 '13 at 18:51

//pitch,roll,yaw = all are degrees


text:
Crystalline version of the Stefan problem with Gibbs-Thompson law and kinetic undercooling

Piotr Rybka

Abstract

The author studies the modified Stefan problem in the plane with surface tension and kinetic undercooling when the interfacial curve is a polygon. Existence of local-in-time solutions is shown. Geometric properties of the flow are studied if the Wulff shape is a regular $N$-sided polygon. The author shows that an initial interface being a scaled Wulff shape with sufficiently small perimeter shrinks to a point. Moreover, at each time the interface remains a scaled Wulff shape.

Article information

Source
Adv. Differential Equations Volume 3, Number 5 (1998), 687-713.

Dates
First available in Project Euclid: 18 April 2013


text:
# Solving an equation of terms containing a variable in their denominator

I am having issues solving the following equation: $$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$

The resolution of this is 1.

It is in the section of parabolas and it should be pretty easy to solve.

My steps are as follows:

$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$ Multiply by $2x(4x-6)$

$$4x^2 -1(4x-6) = 6x$$ gets me here, remove $6x$

$$4x^2 - 10x + 6 = 0$$

Solving this with the parabola formula I ended up getting $\{4,6\}$.

-
You've misapplied the quadratic formula; if you try substituting 4 and 6 into your original equation, you do not have an equality. Post what you did with the quadratic formula and maybe we can debug from there. –  Ｊ. Ｍ. Dec 8 '10 at 3:22
@J.M.: He divided by $2$ instead of $2a$ (that is, instead of $8$, or instead of $4$ if he first divided the whole equation by $2$). –  Arturo Magidin Dec 8 '10 at 3:31
Shaharyar, you might want to post that as an answer and accept it so we have less unanswered questions in this site. :) –  Ｊ. Ｍ. Dec 8 '10 at 3:43
Done, will have to wait 2 days to be allowed to accept it :) –  Shaharyar Dec 8 '10 at 3:45

HINT $\rm\displaystyle\quad \frac{1}{2\ x}\ =\ \frac{2\ x}{4\ x-6} - \frac{3}{4\ x-6}\ =\ \frac{1}2\ \ \Rightarrow\ \ x\ =\ \ldots\quad$ Note: no quadratic formula needed.

-

@J. M. :

After checking out the formula I found out that I dividing by $2$ instead of $4a$/$8$

$$\frac{10 \pm \sqrt{100-96}}{2}$$

With the fixed form I get the results $1.5$ and $1$ of which $1$ is matching result.

Thanks for the kind support!

-
And do you see why you need to discard the solution $x=\frac{3}{2}$? –  Arturo Magidin Dec 8 '10 at 4:35
Yup, evil division by $0$ of doom. –  Shaharyar Dec 8 '10 at 7:14


text:
Article Text

Sutureless cataract surgery with nucleus extraction: outcome of a prospective study in Nepal
Free
1. A Hennig1,
2. J Kumar1,
3. D Yorston2,
4. A Foster3
1. 1Lahan Eye Hospital, Nepal
2. 2Moorfields Eye Hospital, London, UK
3. 3London School of Hygiene and Tropical Medicine, London, UK
1. Correspondence to: Dr Albrecht Hennig, Lahan Eye Hospital, Nepal; ahennig{at}mos.com.np

## Abstract

Aim: To report the short and medium term outcome of a prospective series of sutureless manual extracapsular cataract extractions (ECCE) at a high volume surgical centre in Nepal.

Methods: Cataract surgery was carried out, on eyes with no co-existing diseases, in 500 consecutive patients who were likely to return for follow up. The technique involved sclerocorneal tunnel, capsulotomy, hydrodissection, nucleus extraction with a bent needle tip hook, and posterior chamber intraocular lens (PC-IOL) implantation according to biometry findings. Surgical complications, visual acuity at discharge, 6 weeks, and 1 year follow up, and surgically induced astigmatism are reported.

## DISCUSSION

Many patients travel considerable distances to obtain treatment at Lahan, and they usually cannot return for postoperative examinations. In order to obtain adequate follow up in such difficult circumstances, only 500 patients from closer districts out of a total of 14 500 were selected. All patients with known pre-existing eye disease were excluded. The authors believe that the selection of local patients has not biased the results, as there is no evidence that district of residence influences the outcome of cataract surgery.

### Surgical outcome

Sutureless cataract surgery is more difficult than conventional sutured ECCE and PC-IOL. The preparation of a relatively long and narrow sclerocorneal tunnel, handling the instruments, and removing the entire nucleus all require good surgical skills and additional training. However, once the technique is mastered the surgery is faster and less expensive, because no sutures are required. When performed by an experienced surgeon, the complication rate is low—there was only one posterior capsule rupture in this study. In Japan, posterior capsule rupture occurred in 5.9% of eyes, and this was thought to be the result of learning the new technique.9

Bleeding from the wound into the AC occurred in 9.4% of operations, which prolonged the surgery by another 5 minutes. Hyphaema was the most common complication resulting in four additional intraocular procedures and 36 additional patient days in hospital.

Two eyes suffered persistent corneal oedema. Removal of the nucleus through a scleral tunnel can be traumatic. Endothelial damage can be minimised by nucleus extraction from within the capsular bag rather than displacing the nucleus into the AC. This reduces manipulation in the AC, and avoids contact between the nucleus and the endothelium. Endothelial trauma can be reduced further by the use of intracameral viscoelastics before extraction of the nucleus.19

Eighty (16.0%) of cases had pre-existing central posterior capsule plaques; however, only 10 of the 80 cases had an acuity less than 6/18 at discharge attributable to the plaque. The frequency and location of this type of posterior capsule opacification corresponds with a previous histopathological study on human crystalline lenses from our centre.20

### Visual outcome

Before surgery nearly all eyes (96.8%) were severely visual impaired or blind. This was reduced to less than 1% by discharge.

In comparison with sutured manual ECCE and PC-IOL, sutureless surgery provides fast visual recovery. In our patients, 76.8% had an unaided acuity of 6/18 or better at discharge (Table 2). By comparison, in a randomised controlled trial of standard ECCE and PC-IOL and intracapsular cataract extraction and aphakic glasses, only 57.5% of the ECCE and PC-IOL eyes achieved an unaided acuity of 6/18 or better by 2 months after surgery.15 However, another study of sutureless ECCE in Nepal showed that only 58.3% of eyes obtained an uncorrected vision of 6/18 or better at 8 weeks.10

The low incidence of poor visual outcome (less than 6/60) is due to the low rate of operative complications, and to the exclusion of eyes with known comorbidity.

There was a statistically significant reduction (6.3%) in the number of eyes achieving 6/18 or better without correction between discharge and 6 weeks’ follow up. A further slight decrease (5.6%) between 6 weeks and 1 year was not significant. The most likely cause for the worsening uncorrected acuity is an increase in surgically induced astigmatism (Table 4).

In another study of sutured ECCE and PC-IOL, the proportion of eyes with an uncorrected vision of 6/18 or better increased from 57.5% at 2 months to 83.7% by 1 year.15

Although keratometry was not performed at discharge, it seems likely that the worsening of unaided acuity is the result of increasing against the rule astigmatism. Between 6 weeks and 1 year, the increase in astigmatism and the corresponding decrease in unaided visual acuity is small, suggesting that, in most cases, the wound is stable by 6 weeks after surgery.

In nearly two thirds of the eyes with an unaided acuity less than 6/18, the cause of reduced vision was astigmatism. We tried to limit the astigmatism by placing the incision as far behind the limbus as possible, and by keeping the width of the sclerocorneal tunnel to a minimum. However, in eyes with large nuclei, the incision had to be 8 mm wide in order to extract the nucleus. Even these larger tunnel openings remained self sealing without sutures.

The mean induced astigmatism of 1.4D is similar to other reports from industrialised countries.9,19

It is known that uncorrected refractive error is a common cause of poor outcome following cataract surgery in developing countries. The cost of glasses, and, in some countries, the lack of expertise to prescribe and manufacture high quality spectacles, limits their use. If the outcome of cataract surgery globally is to improve, it is imperative to find means of improving uncorrected postoperative visual acuity, as well as best corrected vision. We believe that the technique of sutureless ECCE may help to increase the number of patients with good uncorrected vision.

Refinements in the surgical technique may further reduce refractive error as a cause of poor vision after this type of surgery. For example, a single radial stitch may reduce against the rule astigmatism in sutureless ECCE.21 Alternatively, a temporal or oblique incision may be performed in eyes that have against the rule astigmatism preoperatively. In this study, 65.4% of eyes had against the rule astigmatism before surgery (Table 4).

### Keratometry and A-scan

For keratometry we used an automated keratometer, which requires the patient’s cooperation to fixate on an illuminated target. Many patients with advanced cataract were unable to focus on this symbol, making it difficult to obtain reproducible keratometer readings. In such cases we performed keratometry on 162 fellow eyes with less cataract, and assumed that the average corneal power was similar in both eyes.

The time required for preoperative biometry was greater than for the surgery itself. In this population axial ametropia appears to be relatively uncommon, although patients with known high myopia were excluded. At least 78% of eyes would have been within 1D of emmetropia with a 22D lens. Further study is required to weigh the visual benefits of biometry against the extra costs and time required in rural settings with a high surgical volume.

### Cost of surgery

It has been shown that, in Europe, phacoemulsification with a foldable IOL is less costly than manual ECCE and IOL.4 However, this analysis did not include the capital cost of a phacoemulsifier, and used a PMMA IOL costing £38.00. In developing countries, high quality, locally made IOLs can be purchased for less than £3.00, but high quality foldable IOLs must be imported, and are much more expensive. The use of locally produced consumables minimises the cost of surgery. Until phacoemulsification consumables, and foldable IOLs, can be manufactured in developing countries, phacoemulsification will be too costly for the majority of cataract patients.

This study shows that sutureless manual cataract surgery can be a safe and effective technique in the hands of an experienced surgeon in areas with a high demand for cataract surgery, including advanced cataract with large nuclei.

The surgical technique is more demanding than sutured ECCE and PC-IOL and requires additional surgical training. However, once this technique is mastered it provides a good visual outcome and fast visual recovery. Further work is required to reduce the incidence of peroperative hyphaema and postoperative astigmatism.

## Footnotes

• Series editors: W V Good and S Ruit


text:
# Tutorial¶

## Model Building¶

Consider a simple system of chemical reactions given by:

$\begin{split}A \xrightarrow[]{k_1} B\\ B \xrightarrow[]{k_2} C\\\end{split}$

Suppose k0.11 = 1, k0.1 = 1 and there are initially 100 units of A. Then we have the following model string

>>> model_str = """
const compartment comp1;
comp1 = 1.0; # volume of compartment

r1: A => B; k1; # differs from antimony
r2: B => C; k2; # differs from antimony

k1 = 0.11;
k2 = 0.1;
chem_flag = false;

A = 100;
B = 0;
C = 0;
"""

The format of the model string is based on a subset of the antimony modeling language, but with one key difference. Antimony allows the user to specify custom rate equations for each reaction. cayenne automagically generates the rate equations behind the scenes, and user need only supply the rate constants. The format is discussed below:

### Model format¶

const compartment comp1;

This defines the compartment in which the reactions happen.:

comp1 = 1.0;

This defines the volume of the compartment in which reactions happen. For zero and first order reactions, this number does not matter. For second and third order reactions, varying compartment volume will affect the kinetic outcomes even when the rest of the model is not changed. A blank line after this separates these definitions from the reactions.:

r1: A => B; k1; # differs from antimony
r2: B => C; k2; # differs from antimony

Here r1 and r2 refer to the names of the reactions. This is followed by a colon and the reactants in that reaction. In r1 there is only one reactant, A. Additional reactants or stoichiometries can be written like A + 2B. This is followed by a => which separates reactants and products. Products are written in a fashion similar to the reactants. A semi-colon indicates the end of the products. This is followed by a symbol depicting the rate constant e.g. k1, and the reaction ends with a second semi-colon. A blank line after this separates these reactions from rate-constant assignments.:

k1 = 0.11;
k2 = 0.1;

The rate constants are assigned one per line, with each line ending in a semi-colon. Every rate constant defined in the reactions must be assigned a numerical value at this stage, or cayenne will throw a cayenne.model_io.RateConstantError.:

chem_flag = false;

An additional element that is included at this stage is the chem_flag boolean variable. This is discussed more in detail in the documentation of cayenne.Simulation class under the notes section. Briefly, if

1. the system under consideration is a chemical system and the supplied rate constants are in units of molarity or M or mol/L, chem_flag should be set to true
2. the system under consideration is a biological system and the supplied rate constants are in units of copies/L or CFU/L, chem_flag should be set to false

A blank line after this separates rate constants from initial values for the species.:

A = 100;
B = 0;
C = 0;

The initial values for species are assigned one per line, with each line ending in a semi-colon. Every species defined in the reactions must be assigned an integer initial value at this stage, or cayenne will throw a cayenne.model_io.InitialStateError.

Warning

Antimony has a set of reserved keywords that cannot be used as species, compartment or variable names, Eg. formula, priority, time, etc. Refer to the antimony documentation for more information.

Note

cayenne only accepts zero, first, second and third order reactions. We decided to not allow custom rate equations for stochastic simulations for two reasons:

1. A custom rate equation, such as the Monod equation (see here for background) equation below, may violate the assumptions of stochastic simulations. These assumptions include a well stirred chamber with molecules in Brownian motion, among others.
$\mu = \frac{\mu_{max}S}{K_S + S}$
1. An equation resembling the Monod equation, the Michaelis-Menten equation, is grounded chemical kinetic theory. Yet the rate expression (see below) does not fall under 0-3 order reactions supported by cayenne. However, the elementary reactions that make up the Michaelis-Menten kinetics are first and second order in nature. These elementary reactions can easily be modeled with cayenne, but with the specification of additional constants (see examples). A study shows that using the rate expression of Michaelis-Menten kinetics is valid under some conditions.
$\frac{dP}{dt} = \frac{\mu_{max}S}{K_S + S}$

Note

The chem_flag is set to True since we are dealing with a chemical system. For defintion of chem_flag, see the notes under the definition of the Simulation class.

These variables are passed to the Simulation class to create an object that represents the current system

>>> from cayenne import Simulation

class cayenne.simulation.Simulation(species_names: List[str], rxn_names: List[str], react_stoic: numpy.ndarray, prod_stoic: numpy.ndarray, init_state: numpy.ndarray, k_det: numpy.ndarray, chem_flag: bool = False, volume: float = 1.0)[source]

A main class for running simulations.

Parameters: species_names (List[str]) – List of species names rxn_names (List[str]) – List of reaction names react_stoic ((ns, nr) ndarray) – A 2D array of the stoichiometric coefficients of the reactants. Reactions are columns and species are rows. prod_stoic ((ns, nr) ndarray) – A 2D array of the stoichiometric coefficients of the products. Reactions are columns and species are rows. init_state ((ns,) ndarray) – A 1D array representing the initial state of the system. k_det ((nr,) ndarray) – A 1D array representing the deterministic rate constants of the system. volume (float, optional) – The volume of the reactor vessel which is important for second and higher order reactions. Defaults to 1 arbitrary units. chem_flag (bool, optional) – If True, divide by Na (Avogadro’s constant) while calculating stochastic rate constants. Defaults to False.
results

The results instance

Type: Results
Raises: ValueError – If supplied with order > 3.

Examples

>>> V_r = np.array([[1,0],[0,1],[0,0]])
>>> V_p = np.array([[0,0],[1,0],[0,1]])
>>> X0 = np.array([10,0,0])
>>> k = np.array([1,1])
>>> sim = Simulation(V_r, V_p, X0, k)

Notes

Stochastic reaction rates depend on the size of the system for second and third order reactions. By this, we mean the volume of the system in which the reactants are contained. Intuitively, this makes sense considering that collisions between two or more molecules becomes rarer as the size of the system increases. A detailed mathematical treatment of this idea can be found in [3] .

In practice, this means that volume and chem_flag need to be supplied for second and third order reactions. volume represents the size of the system containing the reactants.

In chemical systems chem_flag should generally be set to True as k_det is specified in units of molarity or M or mol/L. For example, a second order rate constant could be = 0.15 mol / (L s). Then Avogadro’s constant ($$N_a$$) is used for normalization while computing k_stoc ($$c_\mu$$ in [3] ) from k_det.

In biological systems, chem_flag should be generally be set to False as k_det is specified in units of copies/L or CFU/L. For example, a second order rate constant could be = 0.15 CFU / (L s).

References

 [3] (1, 2) Gillespie, D.T., 1976. A general method for numerically simulating the stochastic time evolution of coupled chemical reactions. J. Comput. Phys. 22, 403–434. doi:10.1016/0021-9991(76)90041-3.

## Running Simulations¶

Suppose we want to run 10 repetitions of the system for at most 1000 steps / 40 time units each, we can use the simulate method to do this.

>>> sim.simulate(max_t=40, max_iter=1000, n_rep=10)

Simulation.simulate(max_t: float = 10.0, max_iter: int = 1000, seed: int = 0, n_rep: int = 1, n_procs: Optional[int] = 1, algorithm: str = 'direct', debug: bool = False, **kwargs) → None[source]

Run the simulation

Parameters: max_t (float, optional) – The end time of the simulation. The default is 10.0. max_iter (int, optional) – The maximum number of iterations of the simulation loop. The default is 1000 iterations. seed (int, optional) – The seed used to generate simulation seeds. The default value is 0. n_rep (int, optional) – The number of repetitions of the simulation required. The default value is 1. n_procs (int, optional) – The number of cpu cores to use for the simulation. Use None to automatically detect number of cpu cores. The default value is 1. algorithm (str, optional) – The algorithm to be used to run the simulation. The default value is "direct".

Notes

The status indicates the status of the simulation at exit. Each repetition will have a status associated with it, and these are accessible through the Simulation.results.status_list.

status: int

Indicates the status of the simulation at exit.

1: Succesful completion, terminated when max_iter iterations reached.

2: Succesful completion, terminated when max_t crossed.

3: Succesful completion, terminated when all species went extinct.

-1: Failure, order greater than 3 detected.

-2: Failure, propensity zero without extinction.

## Plotting¶

To plot the results on the screen, we can simply plot all species concentrations at all the time-points using:

>>> sim.plot()

A subset of the species can be plotted along with custom display names by supplying additional arguments to Simulation.plot as follows:

>>> sim.plot(species_names = ["A", "C"], new_names = ["Starting material", "Product"])

By default, calling the plot object returns the matplotlib figure and axis objects. To display the plot, we just do:

>>> sim.plot()
>>> import matplotlib.pyplot as plt
>>> plt.show()

Instead to save the figure directly to a file, we do:

>>> sim.plot()
>>> import matplotlib.pyplot as plt
>>> plt.savefig("plot.png")

Simulation.plot(species_names: list = None, new_names: list = None)[source]

Plot the simulation

Parameters: species_names (list, optional) – The names of the species to be plotted (list of str). The default is None and plots all species. new_names (list, optional) – The names of the species to be plotted. The default is "xi" for species i. fig (class ‘matplotlib.figure.Figure’) – Figure object of the generated plot. ax (class ‘matplotlib.axes._subplots.AxesSubplot) – Axis objected of the generated plot.

Note

1. The sim.plot method needs to be run after running sim.simulate
2. More detailed plots can be created manually by accessing the sim.results object

## Accessing the results¶

The results of the simulation can be retrieved by accessing the Results object as

   >>> results = sim.results
>>> results
<Results species=('A', 'B', 'C') n_rep=10algorithm=direct sim_seeds=[8325804 1484405 2215104 5157699 8222403 7644169 5853461 6739698  374564
2832983]>

The Results object provides abstractions for easy retrieval and iteration over the simulation results. For example you can iterate over every run of the simulation using

>>> for x, t, status in results:
...     pass

You can access the results of the n th run by

>>> nth_result = results[n]

You can also access the final states of all the simulation runs by

>>> final_times, final_states = results.final

# final times of each repetition
>>> final_times
array([6.23502469, 7.67449057, 6.15181435, 8.95810706, 7.12055223,
7.06535004, 6.07045973, 7.67547689, 9.4218006 , 9.00615099])

# final states of each repetition
>>> final_states
array([[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100],
[  0,   0, 100]])

You can obtain the state of the system at a particular time using the get_state method. For example to get the state of the system at time t=5.0 for each repetition:

>>> results.get_state(5.0)
[array([ 1,  4, 95]),
array([ 1,  2, 97]),
array([ 0,  2, 98]),
array([ 3,  4, 93]),
array([ 0,  3, 97]),
array([ 0,  2, 98]),
array([ 1,  1, 98]),
array([ 0,  4, 96]),
array([ 1,  6, 93]),
array([ 1,  3, 96])]

Additionally, you can also access a particular species’ trajectory through time across all simulations with the get_species function as follows:

>>> results.get_species(["A", "C"])

This will return a list with a numpy array for each repetition. We use a list here instead of higher dimensional ndarray for the following reason: any two repetitions of a stochastic simulation need not return the same number of time steps.

class cayenne.results.Results(species_names: List[str], rxn_names: List[str], t_list: List[numpy.ndarray], x_list: List[numpy.ndarray], status_list: List[int], algorithm: str, sim_seeds: List[int])[source]

A class that stores simulation results and provides methods to access them

Parameters: species_names (List[str]) – List of species names rxn_names (List[str]) – List of reaction names t_list (List[float]) – List of time points for each repetition x_list (List[np.ndarray]) – List of system states for each repetition status_list (List[int]) – List of return status for each repetition algorithm (str) – Algorithm used to run the simulation sim_seeds (List[int]) – List of seeds used for the simulation

Notes

The status indicates the status of the simulation at exit. Each repetition will have a status associated with it, and these are accessible through the status_list.

1: Succesful completion, terminated when max_iter iterations reached.

2: Succesful completion, terminated when max_t crossed.

3: Succesful completion, terminated when all species went extinct.

-1: Failure, order greater than 3 detected.

-2: Failure, propensity zero without extinction.

 Results.__iter__() Iterate over each repetition Results.__len__() Return number of repetitions in simulation Results.__contains__(ind) Returns True if ind is one of the repetition numbers Results.__getitem__(ind) Return sim. Results.final Returns the final times and states of the system in the simulations Results.get_state(t) Returns the states of the system at time point t.

## Algorithms¶

The Simulation class currently supports the following algorithms (see Algorithms):

You can change the algorithm used to perform a simulation using the algorithm argument

>>> sim.simulate(max_t=150, max_iter=1000, n_rep=10, algorithm="tau_leaping")


text:
# Is thermodynamic work always defined, even for irreversible processes?

Suppose we have a thermodynamic system, and we are capable of doing pressure-volume $(PV)$ work on the system. For a infinitesimal reversible process (where the only type of work done is pressure-volume work), the incremental amount of work done on the system is $\delta W = -P dV$. For an irreversible process of the same type, done slowly enough so that variables like pressure and volume are still easily measurable, the work is now $\delta W = -P_{ext} dV \leq -P dV$, where $P_{ext}$ is the external pressure exerted by the system's environment. Let me know if I'm mistaken about any of this.

In cases like these, and in many other situations, the amount of work done can be easily calculated, as long as pressure and volume measurements have been made on the system during each process. However, I can imagine much more complicated scenarios where it's less obvious how work would be calculated. For example, suppose we have a violent expansion of a gas and a corresponding shrinking of the gas's environment, which is so rapid that the external pressure is no longer uniform across the gas-environment boundary, and where the densities of gas and environment vary so much that volume is hard to measure or even define. (Ok, in this process, maybe we could argue that it occurs so rapidly that no heat exchange occurs, and so the work is just $W= \Delta E$, the change in energy of the system. But suppose I could come up with a better example, where heat flow is possible but quantities like $P$ and $V$ are still ill-defined in this way.)

In such a situation, is thermodynamic work still defined? Even if we can't calculate it directly with a formula like $\delta W = -P_{ext} dV$, I'm wondering if there's a way to get at it indirectly. Or are there certain processes which just don't have a definite value of work, from an operational point of view?

• @Countto10 Thanks. Modelling with reversible processes seems like it could work in some cases, but since many reversible processes can correspond to the same initial and final states, I don't see how this would always give you a unique value of work. Also, I agree that for an adiabatic process, the work will just be equal to the change in energy of the system, and is therefore easy to calculate. So my example isn't the best in that regard; to get my point across I really need to think of an example which isn't adiabatic. – Wade Hodson Feb 28 '17 at 16:46
• Thanks Wade, I learnt from your reply, so that's something;) +1. – user146020 Feb 28 '17 at 16:52
• In complex scenario, you can integrate to get the work $\delta W = -\int P_{ext} dV$. However, you still need to measure P and V as a function of space. It would be more difficult. If you cannot measure them, you wouldn't get the work because that is its definition. You may derive the work value, for example, assuming it is a reversible process with measured heat flow. Now you see you get in a loop. – user115350 Feb 28 '17 at 18:25
• I would imagine that as long as you know $P, V$ along a path parameterized by $t$, in principle you should be able to calculate work. Of course, one would run into issues if $P, V$ are no longer well defined, so it seems to me more of question of whether or not your thermodynamic state variables are well-defined. The question of the well-definedness of work is determined by the well-definedness of the state variables. – Aaron Feb 28 '17 at 18:48
• @Aaron and user115350: Ok thanks! So it seems that in the general case, without well-defined state variables and/or external parameters, it's not possible to assign a value to the work done on a system. – Wade Hodson Feb 28 '17 at 20:25

The amount of work done on the system is always the integral of $-P_{ext} dV$ (where $P_{ext}$ represents the force per unit area applied to the gas at the piston face), irrespective of whether the process is reversible or irreversible. But, if the process is reversible, then the gas pressure and temperature are uniform spatially within the cylinder, and thus $P_{ext}=P$. Under these circumstances, one can use the ideal gas law (or other equation of state) to calculate the work.
If the process is irreversible (involving, say, a very rapid deformation), the pressure and temperature within the cylinder are not typically uniform spatially, so the equation of state can not be applied globally. In addition, there are viscous stresses present within the gas that contribute to the force per unit area at the piston face. This too prevents using an equation of state to determine $P_{ext}$ and the work. So, using only thermodynamics, unless you can manually control $P_{ext}$ from the outside, you cannot determine the work.


text:
Advertisement Remove all ads

# In a Mixture of 126 Kg of Milk and Water, Milk and Water Are in Ratio 5 : 2. How Much Water Must Be Added to the Mixture to Make this Ratio 3 : 2? - Mathematics

In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?

Advertisement Remove all ads

#### Solution

Quantity of milk : Quantity of waer = 5:2

∴ Quantity of milk = 126 x 5/7 = 90 kg

=> Quantity of water = 126 - 90 = 36 kg

New ratio = 3 : 2

Let the quantity of water to be added be x kg

Then milk : water  = 90/(36 + 6)

90/(36 + x) = 3/2

=> 180 = 108 + 3x

=> 3x = 72

=> x = 24

Thus quanitiy of water to be added is  24 kg

Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?


text:
## Thursday, April 24, 2014

### SRM 618: When python failed

This is the eighth of ten matches in which I promised to use python exclusively. This is also the first match in which I feel like I can claim I lost plenty of points because of this restriction.

## Div1 250: The one with heteronormativity

So I am supposed to be solving this problem, and all I can think of after reading for the first time is that the problem's definition of family allows incest but is all about not allowing same-gender pairings :/ okay. ..

So a "family" is defined as a graph, in which each edge either has zero incident edges or exactly two incident edges. The two incident edges MUST be from two nodes with different genders. Also the graph must be topsorted. Given a graph of up of 100 nodes return if it could be a family or not. Basically, you have to assign genders to each node. It turns out that the input format guarantees that the graph is topsorted and that each node has 0 or 2 incident nodes. So all that we need to do is ask if we can assign the genders correctly.

If two nodes appear as the parents of a node, then their genders must be different. If we imagine a different graph, in which each of these pairs of nodes that must have a different gender are connected by a single edge. Then the graph must be bipartite. If the graph is bipartite, then the answer is that Yep, the graph could be a "family". Else not. Checking if a graph is bipartite is an easy application of the Depth-first search.

class Family:
def isFamily(self, parent1, parent2):
n = max(max(parent1),max(parent2)) + 1
# Make a set of edges in the hopefully bipartite graph, they are undirected
E = set(zip(parent1, parent2)) | set( zip(parent2, parent1) )
E.remove( (-1,-1) )

# is the graph bipartite?
color = [-1] * n
result = [ "Possible" ]
def dfs(x, c):
if color[x] == -1:
color[x] = c
for (u,v) in E:
if u == x:
dfs(v, 0 if c == 1 else 1)
elif color[x] != c:
result[0]  = "Impossible"

for i in range(n):
if color[i] == -1:
dfs(i, 0)

return result[0]

I took a bit of time mostly because of the above-mentioned distraction :P, but python helped me keep the code clean. I think it should pass unless 100x100 is suddenly too slow (I doubt it).

## Div1 500: The one with strings.

You should count the number of maximal-length strings that use an alphabet of n letters and follow these properties:

• No consecutive equal letters.
• For each pair of (possibly equal) characters X and Y, string "XYXY" is NOT a subsequence of the string.

The first issue is how to know what is the maximum length possible. After plenty of analysis, I found that the maximum length possible is 2n-1. However, there are two methods to achieve it:

• One way, is to surround a string of 2(n-1)-1 length using (n-1) alphabet characters, using two instances of one character different to those (n-1) ones. For example: A????????????A, where ????? contains no A.
• The other way has 3 instances of the extreme character. A??????A!!!!!!!A, but also ?????? and !!!!!! cannot share characters.

So the rest is to count the number of valid patterns. It is not necessary to worry about which character goes to which position of the pattern, because we can just multiply the result by n!.

The rest is an O(n^2) dynamic programming, unfortunately, n is up to 5000 (I wish it was 1000), so no matter what I do, the thing will take at least 6 seconds. That's too much.

MOD = 1000000007

def modFactorial(x):
r = 1
for i in xrange(1, x + 1):
r = (r * i) % MOD
return r

class LongWordsDiv1:
def count(self, n):
# The maximum number of letters is 2*n-1, although there are multiple ways
# to reach that maximum
# Example, for n = 5
# 012343210
# 010234320
# 010232420
# 012103430
# 012321040
# 012131040
# there should be 9, but can't think of the other 3

dp = [0] * 5001
dp[0] = 1
dp[1] = 1
for i in xrange(2,n+1):
#1) 2n - 1 = 2 + (2(n-1) - 1)
res = dp[i-1]
#2) find a,b: 2n - 1 = 3 + 2a + 2b - 2
#                 b  = (2n - 1 - 3 - 2a + 2) / 2
#                 b  = (2n - 2a - 2) / 2
#                 b  = n - a - 1
a = 1
while a < i:
b = i - a - 1
if b > 0:
res += (dp[b] * dp[a]) % MOD
else:
break
a += 1
dp[i] = res

return ( dp[n] * modFactorial(n) ) % MOD

It is possible I missed something in this analysis, though.

ffao said...

"there should be 9, but can't think of the other 3"

012134310
012321410
012324210

*flies away*

d07RiV said...

If you were so determined to use python for the 500, guess you should've tried hardcoding the answers ;p

vexorian said...

Now I just feel stupid.


text:
# What is the experimental uncertainty of an ensemble measurement? [duplicate]

Let's say you measure the time it takes for 10 oscillations of a mass undergoing simple harmonic motion to within ± 0.01s, what is the uncertainty of the period of one oscillation?

• possible duplicate of Multiple measurements of the same quantity - combining uncertainties – DavePhD May 15 '14 at 15:25
• I don't think it's a duplicate. The article linked to is about measuring some quantity, say 10 different times and computing the mean. I was referring to taking one measurement of 10 oscillations and dividing by 10. – 11Kilobytes May 15 '14 at 15:32
• When you divide a random variate by a constant, the standard deviation is also divided by the same amount. So the uncertainty in the period of one oscillation becomes 0.001 second, ie, you divide the error by 10. The link DavePhD provided isn't an exact duplicate for the reason you cited, but I'm sure this has been asked before, and in any case it's more of a statistics question than a physics question. – DumpsterDoofus May 15 '14 at 15:41
• @DumpsterDoofus that's good as a WAG, but the actual uncertainly is probably more like $\frac{error} {\surd{10}}$ , since you don't know for certain that each oscillation is the same as the others (at least, not in a chaotic situation) – Carl Witthoft May 15 '14 at 15:47

If we assume that you know the oscillations are regular and your time for ten of them is $T \pm \Delta T$, then the time for a single oscillation $\tau$ is:
$$\tau = \frac{T \pm \Delta T}{10} = \frac{T}{10} \pm \frac{\Delta T}{10}$$
So you divide your error by 10 and the error in the time for a single oscillation is $\pm 0.001$ seconds.
Carl's point arises if you measure a single oscillation ten times in separate measurements. In that case if $T$ is now the sum of all ten times, and $\Delta T$ is the error in any single measurement the time for a single oscillation is:
$$\tau = \frac{T}{10} \pm \frac{\Delta T}{\sqrt{10}}$$


text:
# Structure and Enumeration of $K4$-minor-free links and link diagrams

3 ALGCO - Algorithmes, Graphes et Combinatoire
LIRMM - Laboratoire d'Informatique de Robotique et de Microélectronique de Montpellier
Abstract : We study the class L of link types that admit a K 4-minor-free diagram, i.e., they can be projected on the plane so that the resulting graph does not contain any subdivision of K 4. We prove that L is the closure of a subclass of torus links under the operation of connected sum. Using this structural result, we enumerate L and subclasses of it, with respect to the minimal number of crossings or edges in a projection of L ∈ L. Further, we enumerate (both exactly and asymptotically) all connected K 4-minor-free link diagrams, all minimal connected K 4-minor-free link diagrams, and all K 4-minor-free diagrams of the unknot.
Keywords :
Document type :
Journal articles
Domain :
Complete list of metadata

Cited literature [10 references]

https://hal-lirmm.ccsd.cnrs.fr/lirmm-01890505
Contributor : Dimitrios Thilikos <>
Submitted on : Monday, October 8, 2018 - 4:33:48 PM
Last modification on : Saturday, December 26, 2020 - 1:46:07 PM
Long-term archiving on: : Wednesday, January 9, 2019 - 4:36:54 PM

### File

spknots_DMC.pdf
Files produced by the author(s)

### Citation

Juanjo Rué, Dimitrios M. Thilikos, Vasiliki Velona. Structure and Enumeration of $K4$-minor-free links and link diagrams. Electronic Notes in Discrete Mathematics, Elsevier, 2018, 68, pp.119-124. ⟨10.1016/j.endm.2018.06.021⟩. ⟨lirmm-01890505⟩

Record views


text:
## Calculus (3rd Edition)

The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ (2n+1)!}$, converges.
For the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ (2n+1)!}$, the positive term $b_n=\frac{1}{ (2n+1)!}$ The terms are decreasing and positive. Evaluate the limit: $$\lim_{n\to \infty}b_n=\lim_{n\to \infty}\frac{1}{ (2n+1)!}=\frac{1}{\infty}=0$$ Thus, by the alternating series test, the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ (2n+1)!}$, converges.