User talk:Nbrouard

Hi Nbrouard,

In the past I have uploaded images to the Commons, but some people got very cross at me for doing that. Some of my early animations were not good, some of my early animations did not represent the physics well. I have decided to stay away from the Commons; I have decided to concentrate on the english wikipedia, rather than spreading my attention between the english wikipedia and the Commons.

If you feel that a particular image or animation should be available at the Commons, please upload it there.

I can read french, not very well, but I manage. I haven't read the french Foucault pendulum article thoroughly yet, but I intend to do so.

I'm not at ease with notation in complex numbers. I do know the relation $$e^{ix}= sin(x) + icos(x)$$, but I'm not good at mentally translating foucault pendulum equations in that notation to other ways of representation. I'd be very grateful if you can help me in that respect. --Cleonis | Talk 00:02, 1 April 2007 (UTC)

Hum, a lesson on complex numbers! Let us try:

$$e^{i\phi}= \sin(\phi) + i \cos(\phi)\,$$ corresponds to a rotation of an angle of $$\phi_,$$ (on a circle of unit 1). Thus, $$e^{i\omega t}\,$$ corresponds to a rotation of $$\omega t$$ in time $$t\,$$.

A complex number $$z=x+iy$$ can always be written as  $$|z|e^{i\phi}$$, from which you can deduce that $$|z|=\sqrt{x^2+y^2}\,$$ and $$x=|z|\cos(\phi)\,$$ and $$y=|z|\sin(\phi)\,.$$

A nice property is that the conjugate $$\bar{z}=x-iy=|z|\cos(\phi)-i|z|\sin(\phi)=|z|e^{-i\phi}$$ and corresponds to a negative rotation ($$-\phi\,$$) on a circle of unit$$|z|\,$$.

The product of a complex by its conjugate gives a real: $$ z \bar{z}= |z|e^{i\phi}|z|e^{-i\phi}=|z|^2=x^2+y^2 \,$$ and you can also do (using $$i^2=-1\,$$) $$ z \bar{z}= (x+iy)(x-iy)=x^2+y^2\,.$$

Multiplying complex numbers on a circle of unit 1 corresponds to adding rotations: $$e^{i \omega t_1}\times e^{i \omega t_2}= e^{i \omega (t_1+t_2)}$$.

Dividing complex numbers on a circle of unit 1 corresponds to substracting rotations: $$\frac{e^{i \omega t_1}}{e^{i \omega t_2}}= e^{i \omega (t_1-t_2)}\,.$$

Thus, with complex numbers you use multiplication as in standard algebra and the geometrical interpretation is rotation.

Adding complex numbers corresponds to adding vectors which is different.

I hope that it helps.--Nbrouard 21:20, 16 April 2007 (UTC)

Complex numbers: geometrical interpretation
Thank you for your comment. It helped. I knew about the geometrical interpretation but I didn't put two and two together.

I didn't realize that for example the following expression (which applies in the case of a polar pendulum):


 * $$ e^{-i \Omega t} \left(c_1 e^{i \omega t} + c_2e^{-i \omega t}\right) $$

Can be reformulated as follows:



\begin{cases} cos (\Omega t) \left( c_1 e^{i\omega t} + c_2e^{-i \omega t}\right) \\ i sin (\Omega t) \left( c_1 e^{i\omega t} + c_2e^{-i \omega t}\right) \end{cases} $$

Then it is readily seen that the expression $$ \left(c_1 e^{i \omega t} + c_2e^{-i \omega t}\right) $$ can be interpreted as a vector that represents the swing of the pendulum, and the expression $$ e^{-i \Omega t} $$ has the interpretation that it rotates the vector that represents the swing.

Based on my physical understanding of the pendulum I had manufactured the two animations that are shown on the right. Now I can relate the notation with complex numbers to the physics taking place.



I have another question. The french wikipedia article presents the following relation:


 * $$ \omega_0 = \sqrt{\omega^2 + \Omega^2 \sin^2(\theta)} $$

How is this relation obtained?
 * It is not obtained, it is a notation to simplify the equations. Look at the mathematics which I translated into English.--Nbrouard 23:51, 18 April 2007 (UTC)


 * I see, it is in fact an announcement of a substitution.
 * The reason I asked is as follows: like a Foucault pendulum, a conical pendulum would be affected by the fact that it is circumnavigating the Earth's axis. The motion of a conical pendulum can be regarded as a linear combination of two perpendicular planar oscillations. Like the Foucault pendulum, a conical pendulum's equation of motion will contain the sine factor. Also, the ratio of &omega; to &Omega; matters for a conical pendulum. In the ranges of ratios of, say, 10:1 down to 1:1 the effect will be quite noticable.
 * So I am contemplating the possibility that the relation $$ \omega_0 = \sqrt{\omega^2 + \Omega^2 \sin^2(\theta)} $$ has a physical interpretation. --Cleonis | Talk 18:22, 19 April 2007 (UTC)

Can you tell me in what way it is related to what is represented in the animations?
 * It is the new pulsation of the pendulum in its oscillation plan as well as in the plan orthogonal to the oscillation plan, thus generating an ellipse in that easy case of launching.--Nbrouard 23:51, 18 April 2007 (UTC)

What is represented in the animation that gives the straight-above view is that the overall rotation affects the vibration; the dynamics falls in the category of coupled oscillators.
 * True, the pendulum is not free, the Coriolis effect varies during the motion even if the rotational speed of the Earth is constant. --Nbrouard 23:51, 18 April 2007 (UTC)

In my sandbox article about the Foucault pendulum, I present that the Foucault pendulum precession and its sine law are readily understood in terms of coupling of oscillators. --Cleonis | Talk 17:14, 18 April 2007 (UTC)

Case of the Foucault pendulum
The Earth being rotating within an inertial frame of reference, the Coriolis effect must be taken into account. The Coriolis acceleration can be written $$2 \Omega (\vec{v} \wedge \vec{k})$$ where $$\vec{v}$$ is the velocity of the pendulum, $$\vec{k}$$ is a unit vector along the rotation axis and $$\Omega\,$$ is the angular velocity of the rotating system. This angular velocity Ω is much lower than the own pulsation ω of the pendulum.

If the pendulum is let at a latitude θ, then the vector $$\Omega \vec{k}$$ has as components in the reference mark Oxyz $$\begin{pmatrix} 0\\ \Omega \cos{\theta} \\ \Omega \sin{\theta} \end{pmatrix}$$. $$\vec{v}$$ has as components $$\begin{pmatrix} \dot{x}\\ \dot{y} \\ 0 \end{pmatrix}$$, so that the acceleration of Coriolis will have as components $$\begin{pmatrix} 2\dot{y} \Omega \sin{\theta}\\ - 2\dot{x} \Omega \sin{\theta} \\ 2\dot{x} \Omega \cos{\theta} \end{pmatrix}$$.

The equations of the movement in the plan Oxy become: $$ \left\{\begin{matrix} \ddot{x} = - \omega^2 x + 2\dot{y} \Omega\sin{\theta}\\ \ddot{y} = - \omega^2 y - 2\dot{x} \Omega \sin{\theta}\end{matrix}\right.$$.

Switching to complex coordinates $$z=x+iy$$, the equations read: $$ \ddot{z}+2i\Omega\sin{\theta} \dot{z} + \omega^2 z = 0 $$

Let us propose a traditional solution of the form $$z(t)=e^{rt}$$, one deduces from it that the complex $$r$$ must check the quadratic equation: $$r^2+2i\Omega\sin(\theta)r + \omega^2=0$$ which reads: $$\left(r+i\Omega\sin(\theta)\right)^2- i^2\omega^2\left(1+\sin^2(\theta)\frac{\Omega^2}{\omega^2}\right)=0$$

By noting $$\omega_0 = \sqrt{\omega^2 + \Omega^2 \sin^2(\theta)}$$, the two solutions of the quadratic equation are: $$r=-i(\Omega\sin(\theta) \pm \omega_0)$$ and one can then deduce from it that the general solution of the system is:

$$z(t)= e^{-i \Omega \sin{\theta} t} \left(c_1 e^{i\omega_0t} + c_2e^{-i\omega_0t}\right) \qquad\qquad(1)$$

where $$c_1$$ and $$c_2$$ are two constants, possibly complex, which one can determine by two initial conditions such as for example, the position the pendulum and its velocity at the date $$t=0$$ which lead to the two equations:

$$ \left \{ \begin{align} z(0) = & c_1 +c_2 \,,\\ \dot{z}(0) = &-i \left( \Omega\sin(\theta) z(0) - \omega_0 (c_1-c_2) \right) \end{align}\right. \,.$$

By replacing the expressions found for the two constants in the equation (1), one can then write an equation more easily interpretable: $$ z(t) e^{i \Omega \sin{\theta} t} =z_0\left(1\times\cos(\omega_0 t) +i \frac{\Omega \sin(\theta)}{\omega_0}\sin(\omega_0 t) \right) + \frac{\dot{z}_0}{\omega_0} \sin(\omega_0 t) \qquad\qquad(2)$$

Thus, if $$\dot{z}_0$$ is null and $$z_0$$ a real number, the trajectory on the ground of the pendulum in a reference mark turning according to a pulsation $$\Omega \sin(\theta)$$ is an ellipse traversed in one period of $$\frac{2\pi}{\omega_0}$$.

If $$\dot{z}_0$$ is nonnull and has only an imaginary part, the elliptic movement is slightly modified by an oscillation perpendicular to the principal plan of oscillation and of the same frequency $$\omega_0\,$$.

Then let us examine two manners of launching the pendulum:


 * Let us suppose that the pendulum could be propelled since the position of balance($$z(0)=0$$ to the east with an inital speed $$\dot{z}(0)=V_0$$ and we obtain the movement described by the equation:

$$ z(t)= \frac{V_0}{\omega_0} \sin(\omega_0 t) e^{-i\Omega\sin(\theta)t }$$
 * The pendulum passes exactly by its position of balance to each oscillation as shown in the above animation. Animation corresponds to a very fast rotation of the Earth and the pendulum is located in the southern hemisphere.
 * But one does not see how such a movement could be initiated in a simple way.


 * Let us suppose now, as Foucault made it, that the pendulum is isolated of its position of balance by a cord tended (for example towards the east: $$z_0$$ meters) and that is burned it in order to release the pendulum with a null initial speed ($$\dot{z}_0=0$$), the following solution reads:

$$ z(t)= z_0 e^{-i\Omega\sin(\theta)t} \left[ \cos(\omega_0t) + i \frac{\Omega\sin(\theta)}{\omega_0} \sin(\omega_0 t) \right]\qquad\qquad(3)$$


 * and it is enough to plot the curve parameterized by the real part (longitude west to east) and the imaginary part (latitude south to north) to obtain the animation A much is more realistic (click on the figure to read the program of the animation written using the Gnuplot language) even if the number of revolutions of the Earth is still fictitious and about a rotation in 110 seconds instead of a rotation by 24 hours. If one puts a camera in the oscillation plan of the pendulum, one obtains animation B where the terrestrial reference frame turns. One can notice that the pendulum does not oscillate rigorously in the plan but on both sides of the plan according to the ellipse described above. The length of the wire being constant, one calculates for each projection on the ground $$x(T), y(T)$$ the corresponding distance $$h(T)$$ of the pendulum to the ground. [[Image: Foucault-rotz.gif|thumb|right|B Pendulum of Foucault of 67 meters released in the Pantheon of Paris at a distance of 50 meters in the east of the point of balance with a null initial speed. The number of revolutions of the Earth is exaggerated (1 turn in 110 seconds). Sight of a camera related to the plan of oscillation.]]

It is also possible to see the same pendulum from the sun, i.e. from a camera fixed on stars.

The pendulum of Foucault of the Pantheon in Paris oscillates on our real Earth with a pulsation $$\omega_0$$ very close to the pulsation of a simple pendulum $$\omega$$ (the first 8 digits of both numbers are identical) since $$\Omega$$ is very small in comparison to $$\omega$$.

The period of oscillation, $$\ frac{2\pi} {\Omega} =2 \pi\sqrt{\frac{L}{G}}$$ is worth, if the length of the wire makes 67 meters, 16,42 seconds. The ratio of the small side of the ellipse to its large side can be written $$\frac{\Omega\sin(\theta)}{\omega_0}$$ and is very small. The pendulum of Foucault thus oscillates almost in a plan which turns because of the rotation of the Earth. But the plan carries out a full rotation of 24 hours only at the North and South poles. At a latitude $$\theta$$, the period, $$\frac{2\pi}{\Omega\sin(\theta)}$$, inversely proportional to the sine of the latitude, is longer. The sine of 30° being 1/2, a pendulum of Foucault established at a latitude of 30° would carry out a full rotation in 48 hours. The duration of a complete rotation of a pendulum of Foucault located at a latitude other than the equator thus makes it possible to determine this latitude independently of any other measurement.

At the northern latitude of 48°52' of the Pantheon in Paris, the plan thus turns of $$-\frac{2\pi}{\Omega\sin(\frac{(48+52/60)\times 2\pi}{360})}\frac{360}{2\pi}3600=$$-11°19' in one hour.

Because the Earth does not turn only on itself, but also around the Sun and other stars influencing it, the rotation of the terrestrial reference frame is not 24 hours per day, but 23 houts 56 minutes per sidereal day. We thus represented on the figure hereafter the first 3 oscillations of the Foucault Pendulum being let from a distance of 6 meters in the east of the center of the dome of the Pantheon in Paris. Being given the very short deviations South-North compared to the East-West displacement of the pendulum during these first three oscillations, the scale of the ordinate (South-North) is multiplied per 1000 and corresponds to a displacement in millimeters.

The force of Coriolis, perpendicular to displacement and proportional to the velocity, deviates the pendulum from its initial plan of oscillation towards north; it is maximum when that velocity is maximum i.e. when the pendulum passes close to the point of balance, that it exceeds to 0,86 mm ($$z_0\frac{\Omega\sin(\theta)}{\omega_0}$$). The pendulum stops at the end of a half period (thus 8,21 seconds) on the other side from the center of balance but was still slightly deviated towards north. With the return, the direction of the velocity is reversed and the force of Coriolis makes move the pendulum towards the south. Now, it passes to 0,86 mm to the south of the point of balance then stops to 5,4 mm in the south of the point of launching at the end of the period of oscillaton i.e. after 16,42 seconds. The speed of the pendulum compared to our terrestrial reference mark being then null, the force of Coriolis is thus null and the pendulum sets out again in the same direction with a cusp.



My preference: to show how the equation of motion is derived
I copy and paste from above:

What is represented in the animation that gives the straight-above view is that the overall rotation affects the vibration; the dynamics falls in the category of coupled oscillators.
 * True, the pendulum is not free, the Coriolis effect varies during the motion even if the rotational speed of the Earth is constant. --Nbrouard 23:51, 18 April 2007 (UTC)

In my sandbox article about the Foucault pendulum, I present that the Foucault pendulum precession and its sine law are readily understood in terms of coupling of oscillators. --Cleonis | Talk 17:14, 18 April 2007 (UTC)

There is a difference of approach between my sandbox article and the english and french wikipedia articles (I intend to integrate the material in my sandbox article into the english wikipedia article.)

I believe that an encyclopedic article that features a mathematical treatment should focus on the derivation. In the material that I have prepared the mathematical treatment starts with first principles, and subsequently works its way towards the equation of motion:



\begin{cases} \ddot{x} = - \omega^2 x + 2\dot{y} \Omega\sin{\theta} \\ \ddot{y} = - \omega^2 y - 2\dot{x} \Omega \sin{\theta} \end{cases} $$

Once the equation of motion is formulated the physics has been taken care of, and I proceed no further.

I find the treatment in the current english wikipedia article and in the current french wikipedia article unsatisfactory in the following way: the mathematical treatment merely states the equation of motion; the physics taking place (coupling of oscillations) is not discussed. --Cleonis | Talk 18:32, 19 April 2007 (UTC)

TUSC token f33e6f4a2924653949499bdb40b85fd7
I am now proud owner of a TUSC account!

Disambiguation link notification for August 13
Hi. Thank you for your recent edits. Wikipedia appreciates your help. We noticed though that when you edited List of information graphics software, you added a link pointing to the disambiguation page Ipe. Such links are almost always unintended, since a disambiguation page is merely a list of "Did you mean..." article titles. Read the FAQ* Join us at the DPL WikiProject.

It's OK to remove this message. Also, to stop receiving these messages, follow these opt-out instructions. Thanks, DPL bot (talk) 09:23, 13 August 2014 (UTC)

June 2022
Welcome to Wikipedia and thank you for your contributions. I am glad to see that you are discussing a topic. However, as a general rule, talk pages (including user talk pages) such as Talk:Byju's are for discussion related to improving the article in specific ways based on reliable sources and Wikipedia's policies and guidelines. They are not for general discussion about the article topic or unrelated topics, or statements based on your thoughts or feelings. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. Thank you. — DaxServer (t · m · c) 15:47, 8 June 2022 (UTC)


 * Not permitting internet traffic to Byju’s website
 * Here is what I got, trying to access a google reference on "Comma operator in C" language! Any idea why this restriction? Should it be mentioned?
 * As a result of the EU’s General Data Protection Regulation (GDPR).
 * We are not permitting internet traffic to Byju’s website from countries within European Union at this time. Nbrouard (talk) 16:00, 8 June 2022 (UTC)

You could have answer in the Talk Page instead of erasing the contribution. And explain in the main page why the access to Byju's website is forbidden. As you mention on your personal page relation with India and Germany, you could try to access the website from Germany and confirm (or not) that you can't access to Byju's.-- Nbrouard (talk) 16:03, 8 June 2022 (UTC)


 * NOTAFORUM - Read it again please. This is not customer care — DaxServer (t · m · c) 22:25, 8 June 2022 (UTC)