I'm posting here to go over the cost and benefits of a deflationary currency.

firstly, the hourly wage would have to __decrease __over time. this means loans would be more in demand.

it also means prices would go down as well.

loaning money is riskier, but also more lucrative.

when you get paid back, assuming you do, the amount your receive is more valuable. as such interst fees could be reduced.

saving money is much more valuable, especially long term.

the government would probably raise taxes in order to pay for its functions.

]]>it encourages over spending and makes the rich too rich.

a few ideas i've had for possibly a better currency;

a time backed currency, the number of hours in a year multiplied by the population, is the upper limit. minimum wage would be a 1 hour certificate in exchange for an hour of labor.

an energy backed currency; a certificate for each kilowatt or perhaps megawatt of electricity produced, the first recipients of the currency would be the people who make electricity.

i also like the idea of cryptocurrency, using mathematics to make currency.

]]>
** subsystems and frameworks will be checked during the execution and testing stages. Last framework level tests will be performed to acknowledge the framework and exhibit the framework's status for creation administration. In any case, testing exercises won't end once the framework is in activity; testing will go on as the tasks and upkeep staff perform remedial, versatile, and other framework support exercises. What strategies are utilized to direct testing? There are Four fundamental confirmation techniques, as illustrated beneath.**

]]>

here's how i would handle the problem

so step 1, take the total population and divide by the number of seats. let's call this value D.

then assign each state 1 seat, and subtract from each, D

then again from top to bottom, assign 1 seat, subtracting D. if any are negative don't assign it a seat, obviously.

using the numbers in the video;

40075 population, 43 seats. that 932 representatives per seat.

new triangle 21878

circula 9713

squaryland 4167

octiana 3252

rhombus 1065

assign 1 seat for each state, and subtract 932.

new triangle 20946 reps 1

circula 8781 reps 1

squaryland 3,235 reps 1

octiana 2320 reps 1

rhombus 133 reps 1

---------------------

new triangle 20014 reps 2

circula 7849 reps 2

squaryland 2303 reps 2

octiana 1388 reps 2

rhombus 133 reps 1; no further reps.

---------------------

new triangle 19,082 reps 3

circula 6,917 reps 3

squaryland 1,371 reps 3

octiana 456 reps 3; no further reps.

rhombus 133 reps 1; no further reps.

---------------------

new triangle 18150 reps 4

circula 5985 reps 4

squaryland 439 reps 4 no further reps.

octiana 456 reps 3; no further reps.

rhombus 133 reps 1; no further reps.

---------------------

new triangle 442 reps 23

circula 393 reps 10

squaryland 439 reps 4

octiana 456 reps 3

rhombus 133 reps 1

with 2 reps left over. they go to the highest remaining. in this case, octiana and new triangle.

new triangle 442 reps 24

circula 393 reps 10

squaryland 439 reps 4

octiana 456 reps 4

rhombus 133 reps 1

now going from 43 to 44, we get the following

new triangle 14 reps 24

circula 603 reps 10

squaryland 523 reps 4

octiana 519 reps 3

rhombus 154 reps 1

again with 2 left over; they get assigned to curricular and squaryland.

]]>Basically the paradox is this; In terms of Special Relativity, how does spinning disk work in special relativity, if the circumference of the rotating disk undergoes length-contraction (since it's parallel to motion) while its radius does not (since it's perpendicular to motion), and this would imply that [math]\frac{circumference}{diameter} \neq \pi [/math].

I checked bunch of similar questions of the same topic, and can't find a single person giving the correct answer on Quora. Instead I find all sorts of face-palm inducing nonsense like;

- Only the atoms length contract, but the space between them does not.
- The disk would tear into smaller pieces along the radius to give shorter total circumference.
- The disk would implode under pressure from the shrinking circumference.
- Centrifugal forces would counteract Lorentz contraction.
- There's no strong enough material to build such disk because of Born rigidity and elasticity, thus no paradox.
- You need to use General Relativity to solve the paradox.
- The geometry of the spinning wheel is non-euclidean. Just accept it.

And bunch of other answers going completely off on a tangent on topics like people inside a spinning train setting their clocks. Basically every single answer I can find tells me the author probably holds serious misconceptions about Special Relativity itself.

Okay, it's Quora so I shouldn't expect too much, but still I would have expected that at least someone would have given the solution to to something this simple, instead of seeing bunch of people with credentials compete with silly answers. Some of those people are citing their own book about the topic while giving a terrible answer... I mean I'm not that smart, but I solved this problem in my head, while driving. It's really that simple if you actually already understand Special Relativity properly.

What really surprised me was when I went on to check how does Wikipedia see this, and it also doesn't explain the proper solution. There is only one passing mention of the correct solution (kind of, possibly, can't really tell) in the "Brief History" section... with no actual explanation. I guess this is why no one in Quora also knows the answer, but still I'm quite dumbfounded to realize that the actual solution is apparently not very well known at all. I can't find any article actually explaining the correct solution.

Looking at all the bad answers, it seems to me that that there are few different reasons why most people get this so wrong.

One is that many people think about length contraction as something that happens to **objects**, when more accurately it's what happens to your coordinate system when you change your perspective from one inertial frame to another, and follow Einstein convention for isotropic C. If you think it happens to "objects" because they "move", you might be inclined to bring up stuff like "atoms shrink by space between them does not", and that is completely wrong perspective.

Second is that many authors start to analyze realistic materials and Born rigidity, which to some people perhaps seem like a way out of the paradox in some convoluted way. But that is also a complete red herring. The paradox is a thought experiment, and it has got nothing to do with realistic materials. It's about **geometry** in terms of special relativity, which ought to produce self-consistent results regardless of inertial frame. Solving Ehrenfest paradox by bringing up realistic materials and centrifugal forces is like solving twin paradox with "planet earth cannot produce enough energy to actually run that experiment".

Third reason is that a mathematical analysis in the framework of special relativity is easiest to do by making certain approximations, which are exactly the approximations leading into wrong answers. That misleading approximation is the idea of placing a number of **straight rods** along the circumference of the disk, and this approximation is exactly what will give you **wrong answers**. That's right, Einstein's own analysis is also flawed for the same reason, even though it led into insights that led into General Relativity.

Why that approximation produces a critically wrong answer, and what is the correct answer? I'll explain in a bit...

**The correct perspective**

First, just to convince the reader that this problem is in fact fully solvable in terms of Special Relativity without any hocus pocus about elastic materials, please be aware that the frame transformation from one inertial frame to another can be conceived as a sort of rotating / scaling of events in spacetime.

Like this;

https://en.wikipedia.org/wiki/File:Lorentz_transform_of_world_line.gif

The dots in that animation represent **events** as plotted on a spacetime diagram, and the "squishing" of the whole structure represents frame transformation from one inertial frame to another. Some events get pushed "towards the future" and some events get pushed "towards the past". Nothing actually happens to "objects" just because we choose to plot them in a different inertial frame; it's just about how we must plot events, if we are to assume isotropic C, and if we are to remain self-consistent in our mapping between frames.

It really is a good idea to view Special Relativity simply as self-consistent frame transformation rules, and you start seeing that the whole question of length contraction is **not** about how different observers "see things", or how they "measure things", or "what happens to objects", but rather about **how the universe must be plotted in spacetime diagrams when assuming different notions of simultaneity**.

In a nutshell, if we switch from one inertial frame representation to another - assuming unique simultaneity to each frame - we must plot the world state "ahead" of us as pushing towards the future (things that had not yet happened in old frame, have already happened in new frame), and conversely the world state "behind" us as pushing towards the past (something that had already happened in old frame, has not yet happened in new frame). Analyzing moving objects like this is what leads into the concept of "length contraction".

Since we are effectively molding the spacetime diagram around, but preserving the same exact light-like connections between events (the causality - the order of connected events - remains unchanged), it should be pretty easy for anyone to see that if it is possible to represent a spinning disk as a "set of events" in one frame, and it would have to also transform along with all the other events in self-consistent manner to any different frame without hiccups. From this perspective, the actual question behind the paradox is simple; **how would the spinning disk plot onto a spacetime diagram in terms of different notions of simultaneity?**

Even if you can't instantly figure out the exact solution, you should be able to already convince yourself that there is an exact solution out there which would just mold the (events making up the) spinning disk in consistent manner, along with everything else around the situation. What that exact solution is - let's get to it.**The common error**

Once the above is understood correctly, next it should be pretty easy to see how the "rigid rods along the circumference" analysis leads you down the wrong path, and at the same time get an glimpse of the correct solution.

- First, imagine a wheel-of-fortune, with pins sticking out from the outer circumference.
- Then we take a spoked wheel (a bicycle wheel), just proper size to snuggly fit inside the pins of the wheel-of-fortune.
- Last, let's enclose the whole two-disk setup inside a box with a snug fit.

*The purpose of this setup is to signal us if we are doing something inconsistent with our transformation - if the inner wheel fits inside the larger wheel, and if both wheels fit inside the box in one inertial frame, this must be so in all inertial frames. If it's not, we have performed an error in our analysis.*

Now let's take

Now let's set the

Since we have rod A spinning along, let's think about what happens if we shoot

At first glance it might seem like those two rods could be setup to become

The rod that is attached to the spinning wheel is - obviously - never moving in straight line; it is rotating. It's front end is always moving in different direction than its back end (each end is moving parallel to the part of the circumference it touches). So, the first question is,

If we plot the external box of the whole setup, in terms of the inertial frame of rod B, it's easy to propose relativistic speeds where the entire box gets plotted as length contracted to shorter length than rod B. The (non-rotating) wheel-of-fortune inside the box must also be mapped inside the box in every frame, and similarly squashed in the direction of motion - snuggly fitting inside the box. And the rotating bicycle wheel must fit also inside the pins of the wheel-of-fortune. It will get plotted also as snuggly fitting inside the wheel-of-fortune. Note though, the spokes will be plotted as curved because it is actively rotating and we are mapping it by a tilted simultaneity plane - this is just the flipside of the coin same coin that makes us map it as squashed.

Basically the internal configuration of our setup cannot change based on what inertial frame we map it from - rod A does not suddenly poke through the walls of the box just because we choose to plot the situation in different inertial frame. If we think it does, we are making an error in our analysis, or using invalid frame transformation. Basically it would imply an inconsistent change in the configuration of our system (some objects transforming in different ways than others - clearly invalid)

If we investigate a moment where the exact middle points of the rods meet in the same inertial frame, and we choose to plot this in terms of rod B's simultaneity, then the "front" end of rod A (in terms of direction of rotation) has already passed the "front" end of rod B (in terms of direction of motion of rod B in lab frame) some time ago. To be more accurate, since it's attached to a

And conversely, the world state behind us is plotted as pushing towards the past; the rear ends of the rods have not yet met. And since the rod is constantly rotating, the rear end of rod A is also plotted as curving "upwards", and moving towards rod B.

This is why, if you plot down the shape of the spokes of the wheel from the perspective of rod B, the end result looks like this;

https://en.wikipedia.org/wiki/File:Relativistic_wheels.gif

This is simply a result of plotting the events making up the "supposed world state" as transformed as per the Einstein convention of clock synchronization. A convention for plotting data. Nothing more, nothing less.

The error almost everyone makes is that they view length contraction as something actually occurring to

Also, since it was attached to the spokes of the rotating wheel, you'd have to plot the wheel also as having a larger length between two spokes than can be made to fit inside the box wheel-of-fortune, or inside the box. Obviously this result would mean your analysis is completely invalid, plain and simple.

And make no mistake about this - the same error happens no matter how short measurement rods you use. Shorter rods have smaller error, but you can always propose a speed where the error becomes obvious. And with smaller rods there's more of them so the end result of any full analysis is exactly as invalid. Basically you can't have an entire rod in a single inertial frame, while also being attached to the spinning wheel. These are mutually exclusive circumstances.

The same error exists in Einstein's co-rotating observer thought experiment, albeit in more subtle manner. But the point is, the co-rotating observer cannot have a measurement rod in any single inertial frame if that rod is to be also attached to the rotating wheel. The approximation necessary to imagine that situation will always make the analysis invalid for the same reasons as described above.

**The correct solution**

First clue to understanding how this situation really gets plotted correctly is this; Rod B only shares inertial frame with an **infinitesimally thin slice of the spinning wheel**. This is true by the very definition of "spinning". Also from the perspective of the lab frame (where the hub of the wheel is stationary), each infinitesimal slice of the spinning wheel is sitting in a different inertial frame, and does not have any "length" assignable to any single inertial frame. This is a simple mathematical fact arising from the very definitions behind special relativity and "spinning wheel".

Second key to understanding this is also associated with properly understanding length contraction as coordinate transformation. Remember when I said *"if we switch from one inertial frame representation to another, we must plot the world state "ahead" of us as pushing towards the future, and conversely the world state "behind" us as pushing towards the past."*. Note what happens in-between; **the world state in the infinitesimal slice exactly perpendicular to the motion does not transform at all!** This is btw also why the spokes at the bottom and at the top of the spinning wheel were plotted as straight in the relativistic wheel visualization above. (And I can show why the spokes are plotted as curved with another thought experiment too if anyone is interested)

This leads into the simple fact that, in the above experiment, at the moment when the middle part of rod A and rod B meet, **a non-rotating observer sitting at the hub of the wheel co-incides with this infinitesimal plane** that is cutting through the wheel, and that observer **will agree with simultaneity of all events that co-incide with that infinitesimal plane.**

We could repeat the same experiment in any direction, and get the same result, because the wheel is symmetrical. Thus we can see how

So getting back to the original Ehrenfest Paradox, the correct solution is simply to realize that the definitions of

To summarize;

The non-rotating observer (at rest with the hub of the spinning wheel) actually

If this still sounds like a strange claim to you, you are forgetting where length contraction comes from. It comes from dynamic notion of simultaneity, and only applies to how we plot

So the TL;DR solution is, the spinning wheel circumference

And as it turns out, all of the "commonly accepted" (or maybe there isn't one) solutions I can find are

Do note that this solution is all about how geometry gets plotted in terms of special relativity - it's not about how to set a wheel in rotation or about realistic materials. This solution simply arises from how objects get plotted into different inertial frames in self-consistent manner, following exactly the definitions of Special Relativity, and thus it is also exactly the correct solution to the original Ehrenfest Paradox.

Sorry about the length of this. I didn't want to just state what the correct solution is without explaining it in sufficient detail to give everyone a chance to convince themselves about this. Because it seems like the misconception here is so common that the actual end result probably just sound immediately wrong to most people until they think it through themselves.

use a water wheel sort of thing that the sand falls on, and it tightens a spring.

after an hour, the spring releases energy to flip the hour glass over.

thoughts?

]]>I was sitting silently and took in the pitter patter of rain drops upon some metal flashing.. It was harmonic, music to the ears, and purposefully synchronized as well as perfect as a drummer keeping a timed beat.

I thought to myself, why and how can this be, other than the energy formed by each drop of rain creating its own splash of wisdom.

Once upon a time I was fortunate enough to see each snow flake fall individually as if to stop time in its tracks. I use this theory to observe each interaction with life as an individual one of a kind "no two blah blah are alike philosophy... And it will always hold true.

I sort of got off topic with my thesis , but every snow flake that falls is a one of a kind experience... Mathematics cannot duplicate it...therefore. We all hold an ocean in our eyes. Six degrees of separation can help explain this.

Im Canadian, I've shovelled a lot a snow, cheer go leafs go and love wu tang. Bet I can relate to some here , some there and some not at all. But if I was to guess, everyone drinks water...to me that's like saying we are "all in together".

In closing, "be water my friend", my water is James brown type funk n jazz n smiles. Others have crocodiles for smiles. Be one with nature. I think people really really lack this.

Anyway I look forward to feedback , criticism, conflict and harmony. To quote john Lennon " and the world can live as one/be as one...

Were into an intellectual future..let's all be on the same page, or at least in the same book...

peace Bobby

]]>If you have an ftl spacecraft you won’t be able to safely go anywhere unless you can actually see what’s ahead of you and jumping requires photography light doesn’t register quickly enough

]]>The indefinite continued progress of existence and events in the past, present, and future are regarded as a whole. Fundamental Forces are Forces that act between bodies of matter and that are mediated by one or more particles.

2. STRONG NUCLEAR:

a fundamental interaction of nature that acts between subatomic particles of matter.

3. WEAK NUCLEAR:

The fundamental force that acts between leptons and is involved in the decay of hadrons.

4. ELECTROMAGNETISM:

Phenomena and the interaction of electric currents or fields and magnetic fields.

5. UNIVERSAL GRAVITY

The force that attracts a body toward the center of the earth, or toward any other physical body having mass.

6. SOUND PARTICLES:

A sound particle is an imaginary infinitesimal volume of a medium, that shares the movement of the medium in response to the presence of sound at a specified point or in a specified region.

7. LIGHT PARTICLES:

a particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass. ]]>

**.Indeed after setting up, they were extremely hard to use. What would be a realizable result to this? I am sure you guessed it – erecting an affordable CRM software and delivering it entirely online as a service. This was the main idea behind Salesforce. Started as a Software as a Service( SaaS) company, Salesforce has grown into the fifth- largest software company in the world. Salesforce course in pune**

**As seen in the below image, Salesforce provides you with the fastest path from Idea to App. let me introduce you to Salesforce and answer the question on your mind What is Salesforce? Below is an image which shows the power of Salesforce in moment’s tech-smart world. From tech elephants like Google and Facebook to your near call center, all of them use Salesforce services and products to break their problems.**

But, once these regions reach quantum-mechanically relevant scales, down to Planck scale, the quantum uncertainty of the total energy content WITHIN any such hypervolume climbs astronomically, until there is at least as much needed to start a new Big Bang. And since time become unmeasurable within any such hypervolume, the probability of the energy being able to start a new Big Bang rises to unity, ins ALL such hypervolumes. So we'd get perhaps an infinite number of new, causally isolated universes for the price of the death of one. But now INFLATION kicks in. It expands the universe at much greater than the speed of light, which was how regions within the PREVIOUS universe all became causally isolated from each other in the first place. But now we're looking at what happens 'between' universes- I've been wondering whether inflation of these individual microuniverses is fast enough to reconnect them all causally. So we consolidate until there eventually is just one big universe.

An analogy can be found in the way that water droplets or soap bubbles behave. Tiny droplets and soap bubbles have very large surface tensions, and will merge together if in contact. This reduces the surface tension (and contact angle if any), making the resulting object happier (lower energy). But at the other end of the continuum, we have very large water drops and bubbles, with very low surface tension and low contact angle. These often will break up, which increases the surface tension and contact angle. So apparently these objects are happiest in the middle of the range of surface tensions and contact angles. Perhaps a cyclic universe based on my Big Rip scenario above is something like what we see with water drops and soap bubbles. Energy content is comparmentalized between normal and dark matter, photons and other bosons, and dark energy (which may be related to a form of surface tension) And these continually remix- the total content never changes, only that of the compartmentalized components- but they're all in equilibrium.

]]>
I have always wondered why they cut the larger stones with so flat surfaces already in the stone quarry.

This is why......

Just give me 100 liter of mercury and I will build a narrow fit channel for any large megalithic stone

and build the channel so that front of the channel have room for the stone to be able to float

and be moved the larger amount of mercury under the stone as the mercury is transfered under and beside the stone to the back.

Then I gather the mercury, fill in behind the stone and dig up in front of it to be able to make the stone to float and pushed another bit in my channel.

I am sick of all this nonsens of magic or aliens.

I say it again.

I can move any megalithic stone my self if I just get 100 liters or more, to make the stone to float in the mercury.

Master of Science in Engineering Physics.

Magnus Ivarsson

]]>
Today I realized that I could get these numbers by resorting to the (2,1)-sided Pascal Triangle. As you know, you can generate Fibonacci numbers by summing terms along the shallow diagonals of the Triangle.

But there are infinite numbers of 'generalized' Pascal Triangles with edge values more complex than the simple 1's running down those of the classical Pascal Triangle.

The (2,1)-sided Pascal Triangle has 1's running down one side and 2's down the other (with the identity of the apex term ambivalent between 1 and 2). Summing terms along the shallow diagonal upwards towards the 2's side generates the Lucas numbers, and summing terms upwards towards the 1's side generates the Fib numbers (but upshifted one move, that is, lacking the first 1 of 1,1,2,3,5,8...).

Anyway, the doubled Fibonacci numbers can be gotten by adding Lucas and Fibonacci numbers together. So 2+0=2; 1+1=2; 3+1=4; 4+2=6; 7+3=10; 11+5=16; 18+8=26....

Now, the shallow diagonals of any generalized Pascal Triangle aren't all identical. They differ in whether they teminate in an edge term, or not (that is, BETWEEN edge terms). You can generate the doubled Fibonacci numbers (as described above) by INTERSECTING the shallow diagonals that give the Lucas and Fibonacci numbers on the 2's side of the (2,1)-sided generalized Pascal Triangle. The ones that have terminal 1's along the shallow diagonals summing to Fib numbers are the ones that are used by String Theories, and the ones that DON'T have terminal ones are those NOT used.

The (2,1)-sided generalized Pascal Triangle also motives terms in equations that give the powers of the Metallic Means (see https://en.wikipedia.org/wiki/Metallic_mean). A decade ago I realized, while working with both these issues at the same time, that the numerical coefficients of the terms in the equations for the powers of the Metallic Means were IDENTICAL to those along the shallow diagonals of the (2,1)-sided generalized Pascal Triangle leading to the Lucas numbers, and furthermore, that the powers that these terms were raise to were IDENTICAL to the dimensional labels of the edge-parallel diagonals that intersected the shallow diagonals at the terms as well. In the classical (1,1)-sided Pascal Triangle, the outer 1's have a 0D label, as they don't change. The natural numbers have label 1D, since they change linearly/monotonically. The triangular numbers are about close-packing of circles in a plane, so 2D. The tetrahedral numbers are motivated by close-packing of spheres in a volume, so 3D, and so on. This works as well for any generalized Pascal Triangle.

Jess Tauber

]]>Some while back I came up with a geometrical model able to deal with the relative charges of ALL the Standard Model fermions with a single figure- a cube embedded in a reference plane along a body diagonal (a diagonal that runs through the cube's center and two diametrically opposed vertices). If instead of thinking of the charges in terms of thirds (the usual way), we think of them as whole numbers, then the neutrinos have charges of +0 or -0, the lighter quarks +1 or -1, the heavier quarks +2 or -2, and the electrons +3 or -3, in sets of alternating polarity, so +0, -1, +2, -3 for normal matter, and -0, +1, -2, +3 for antimatter. Now we use the vertices of the cube to mark each of these relative charges. If we use the body diagonal as a rotational axis, we can rotate the cube until the sines OR cosines (for any particular angle) fall into these relative absolute values. The default angle is arctan(sqrt27), or 79.10660535.... Two of the vertices are IN the reference plane along the body diagonal, and these represent the two neutrinos. Then the three other vertices closest to these neutrino vertices represent triplets that will have the lengths 1, 2, and 3x the minimum nonzero value.

But one can go beyond the default angle. You can add or subtract ANY multiple of 30 degrees and you still get the correct charge magnitude sequences. And since there are 12 such angles around a circle you get exactly 12 variants on this theme- interesting because there are 12 Standard Model fermions for each of matter and antimatter. A coincidence? Somehow I don't think so. For the default angle, you have to use the SINES to get the triplets- but for that angle PLUS (or MINUS) 30 degrees, you switch to the COSINES. The next increment, plus or minus, of 30 degrees brings you back to sines, and so on.

When you sum the charge triplets (which occupy the vertices of an equilateral triangle that is normal to the reference plane) that sum is always ZERO.

And, when you SQUARE the sines or cosines of the various utilized angles, you always get multiples of 1/28. This is no accident either- I worked out that the denominator of this fraction is always ONE MORE than the value under the square root sign in the default angle (arctan(sqrt27). Anyway, for any set of sines or cosines for any particular angle in this system, the sum of the sines and cosines squared is always exactly 1. Curiously the sums of the numerators of these fractions across triplets of squared sines or cosines is always 42.

So now, having dealt with the relative charges, I'll now move on to the MASSES. The Koide formula relates the different masses along the electron, muon, and tauon sets perpendicular to the charges in tables of these properties. See https://en.wikipedia.org/wiki/Koide_formula. While it has been challenged many times over the years, it still has some interesting aspects to it.

I mentioned above the triplets of squares of sines and cosines that sum to 42 re the charge model. The particular ones that come from the sines (or cosines) relevant to the charges are always 27, 3, and 12. Those that are NOT used for the charges always have numerators of the squares that are 1, 25, and 16. Notice that 1, 25, and 16 are squares of 1, 5 and 4. If you look at the WIki page on the Koide formula you'll see a square figure in the upper right that is supposed to be a representation of the square roots of the masses of the electron, muon, and tauon. I measured these and found that the edges of these component squares are (relative to the smallest, representing the square root of the electron mass), 1, 5, and 4, just as the square roots of the numerators of the squares of the sines or cosines not used for the charge series in the cube-in-plane model.

I divided the mass of the tauon by 16 and it comes out to a bit larger than 111. If 1776 MeV/C^2 exactly, then you get 111 exactly. I also took the value of the mass of the muon and divided that by 25, and if you take this as 105 exactly, the result is 4.2 exactly. Then, if you multiply 4.2 by 10, and divide 111 by 42 you get almost exactly the square of the Golden Ratio, for some reason.

I'd done work some years back relating to the decimalized expansions of terms in straight-line relations in the Pascal Triangle (that is, the edge-parallel diagonals, the columns, the shallow diagonals, and the rows). Ratios between these decimalized expansions are, respectively, 1/9 (or, 1/(5+sqrt16)), 1/(5+sqrt15), and 1/(5+sqrt35) and 1/11. (or 1/(5+sqrt36). Note that there is a difference of exactly 1 between the terms under the square roots in the denominators between the pairs of ratios for the edge-parallel diagonals and the columns, on the one hand, and those for the shallow diagonals and the rows. A pair of related pairs.

Now, we know that you can get the Golden ratio by summing terms along shallow diagonals, which generates the Fibonacci numbers, and then taking the limit ratio for pairs of contiguous Fib numbers. If you take the reciprocal of the Golden Ratio, you get that number minus one, exactly- that is 1.618033989... versus 0.618033989... When you square the Golden Ratio, you instead ADD one, so 2.618033989.

I found that for the values of the the ratios of consecutive columns and shallow diagonals in the Pascal Triangle, if you take THEIR reciprocals, you don't subtract or add 1 to the former. Instead you have to add or subtract 10x the original nonreciprocal value.

I've wondered whether this has any relation, in what I wrote above about the masses of the Standard Model fermions, and the ratios of their reduced values (that is, divided by 16 for the Tauon, and 25 for the Muon), in having to multiply the latter by 10 when taking the ratio that is so close to the square of the Golden Ratio.

Anyway, that's about it.

Jess Tauber

]]>
I got the following assignment in college.

In parallelogram PQSR, what is PQ?

(2x + 5) cm

a) 2 cm

b) 5 cm

c) 6 cm

d) 9 cm

I will be very grateful for your help in solving it.

]]>Please forgive my lack of proper terminology, but:

Take a Torus. There is a nice analytic formula for a Torus.

Now suppose that we rotate the Torus in the 4th Dimension. There seems to me to be at east 3-ways to go about this.

A.} We can rotate the Torus around its own axis—analogous to how a circle can be rotated to form a sphere.

B.} We can rotate the whole Torus at a distance—analogous to how circle is rotated through space to become a Torus.

And,

C.} I Think that it matters—though I may be wrong—what attitude the Torus is in when it is rotated in 4-Space.

I mean, wouldn't a Horizontal Torus with the center hidden from our 3-D perspective, form a different shape if it was rotated in the 4th Dimension that way, than a Torus that was rotated while facing us head-on like a Bullseye? Or is my power of visualization lacking? Are the two cases the same?

Now, suppose that instead of the Torus staying in the same orientation, it could Rotate around its own axis as it generates what was a "Simple" 4-D Toroidal analog. It can rotate at all sorts of frequencies—making a full Rotation while it makes a full circle around its center, or 2-Rotations, or 100-Rotations—not to mention Fractional Numbers of Rotations.

How can one find the equations of these various—well, what do you call them? They aren't Polytopes are they?

IF we decide to go up to 5-Dimensions, the possibilities become even more staggering.

Now, one way to get a Torus analog is to take a Sphere and rotate it in the 4th Dimension similar to how a circle is rotated to create a torus. Is this yet another 4-Dimensional Toroid, or is it the same as one of the cases we've already considered?

BUT, Suppose that instead of a Sphere, we use a Dodecahedron, an Icosahedron or one of the Stellated Polyhedra instead, and rotate it in 4-D?

Is there an exhaustive, systematic way that I can answer such questions for myself?

Thank You!

Saxon Violence

]]>Paper, link = https://ui.adsabs.harvard.edu/abs/2018JPCM...30s5602B/abstract

Physics Forums discussion, link = https://www.physicsforums.com/threads/how-do-you-break-quantum-entanglement-bonds.1015342/

Can entangled particles become disentangled?

Would a EMP(https://en.wikipedia.org/wiki/Electromagnetic_pulse) disrupt electron quantum entanglement bonds?

[Math][/math] - renders mathematical equations

[Math][/math]- renders inline math

[Ce][/Ce] - renders chemical equations.

Some details can be seen from the following link:

https://mytutorsource.com/blog/1-20-divisibility-rules-in-mathematics/

In arithmetic, divisibility rules area unit a collection of ways or rules to see whether or not the given range or {integer whole range number} is divisible by the opposite number wherever the rest are going to be zero. Rules of numbers but five area units are quite simple to understand; but, the quality rules of seven, 11, and thirteen area units are complicated and arduous to know. All numbers or integers area units are separable by one. Their equality takes a look at maybe a technique to see if the given rare one is separable by a tested division, maybe not a reminder. Learning visibility rules are important as they're handy to unravel word issues, perform fast calculations, and check prime numbers.

]]>

And speed of sun light is infinity.

Look at the scale below. Correct me if i m wrong.

1000 km= 1 cm .

10000km=10 cm.

13000km=13 cm.

15000km=15 cm.

100000km=100 cm (1meter).

200000km=200 cm (2meter).

696000km=696 cm (6.96 meter).

1000000km=1000 cm (10 meter).

10000000km (1 CR)=10000cm (100 meter).

100000000km (10cr)=100000cm (1km)

150000000km(15 cr)=150000cm(1.5 km)

Year 1: Calculus 1 = https://www.scienceforums.com/topic/38852-calculus-1-lectures/

Year 2: Calculus 2 = https://www.scienceforums.com/topic/38853-calculus-2-lectures/

Year 3: Calculus 3 = https://www.scienceforums.com/topic/38854-calculus-3-lectures/

Year 4: Differential Equations = https://www.scienceforums.com/topic/38855-differential-equations-lectures/

P.S. This is not a accredited program if you learn this material you would probably do well in a actual college program for mathematics.

]]>

]]>

This is mainly to check all the variables in the differential Equation to make sure that they all solve correctly and to make sure the Quaternion is anomaly free and solve the equation for (x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,μ,Y,q,a,β)

∇'(x,y,z,t,ω_{s},ω_{p,}M,I,k,φ,**S,X,Z,μ,Y,q,a,**β**) = (d ^{2}/**

(d^2/∇') - (Ctp)^2 = ds^2

https://www.wolframalpha.com/input/?i=(d^2+%2F+∇')+-+(C+t)^2

(Universe Volumetric Planck State @ size of universe in radius) =(4/3)π((RUniverse/(tpC))^3 Luniverse

https://www.wolframalpha.com/input/?i=(4%2F3)+π+L+((R%2F(t+C))+)^3

https://www.wolframalpha.com/input/?i=∇+d+(4%2F3)+π+((R%2F(t+C))^3

Luniverse = (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy)

https://www.wolframalpha.com/input/?i=(∇'Q%2C+∇'u%2C+∇'F+%2C∇'g++-+∇'D)

https://www.wolframalpha.com/input/?i=∇+d

https://www.wolframalpha.com/input/?i=(∇+g)+-+(∇+d)

Charge possible states per point (1,2/3, 1/3, 0,-1/3,-2/3,-1)

Color Possible states per point(R,B,G,0,antiG,antiB,antiR)

Flavour possible states per point (I,II,III,0,darkIII,darkII,darkI)

Gravity/Dark Energy possible states per point of space (Energy,Mass,Spin,0,-spin,-mass,-Energy)

Atleast the graphing equation and Equivalence principal are in working order having A.I. do the work.

I have decided to use this equation for a proton instead of the entire universe as it would be too much data to ever complete.

(Universe Volumetric Planck State @ size of universe in radius) =(4/3)π((RUniverse/(tpC))^3 (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy)

RUniverse = RProton = 10^-15 meters

The Equation Yields a Planck State of 9.9023511969154288921026543960449 * 10^59 (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy)

So a Field with 9.9023511969154288921026543960449 * 10^59 cubes that are a Planck length with states of (+1/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) if the proton is at rest.

The Strong Nuclear Force or color Map will look something like this which is the only thing over the 3-D field that varies in a proton.

If the Proton is in motion let's say moving in a particle accelerator at 8 Tev then the State is 7.6171932283964837631558879969576 * 10^57 (+1/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

All of the Information being within the equation with a smaller color field of the same picture being less Planck Lengths within the particle due to length contraction.

The Graphing Equation displays all possible properties of the particle or substance to an quantized amount of a Planck Length being exact without error, I could write the entire Tensor for each substance but it would take the big number amount of states. These were done assuming Dark Energy was not existent and a non expanding universe which are the zero terms. There is only one unknown in these equations which is the Spin number of Dark Energy particles being the final zero in the spin term, the graph is over d/dx + d/dy + d/dz the big number shows the number of planck lengths that the fields manifest for a proton at rest versus in motion for these examples.

This shows this equation to be in working order and accurate to reality.

This equation is actually more complex than the long equation as it gives a single state for everything rather than a large number of multiple Planck States like this one.

If you wanted more detail of the Quarks within the Proton you could graph the equation with the same set of coordinates including the quarks with the same result.

For the Rest proton with quarks in finer detail.

9.9023511969154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Now the charges varies given the details of the quarks within the proton which as now the charges vary you will have two varing graphs one for the Strong Nuclear Force or Color and one for the Electromagnetic Force or Charge being the (+2/3/(dx + dy +dz),R/(dx + dy +dz)) + (+2/3 /(dx + dy +dz), B/(dx + dy +dz)) + (-1/3/(dx + dy +dz), G/(dx + dy +dz)) = (+1/(dx + dy +dz),RGB/(dx + dy +dz))

The equation can be used to whatever detail you would like it to be this being a more exact map of the proton next would be to add gluons if you wanted or even more protons and neutrons to construct an atom, but it is always exact to the planck length, no matter what detail is used.

Overlapped Charge and Color Map, (+2/3/(dx + dy +dz),R/(dx + dy +dz)) + (+2/3/(dx + dy +dz), B/(dx + dy +dz)) + (-1/3/(dx + dy +dz), G/(dx + dy +dz)) = (+1/(dx + dy +dz),RGB/(dx + dy +dz))

Which solves perfectly making the graphing equation even physically correct next we will try something more challenging like a Feynman diagram using this equation, it should be able to graph anything in the universe to the planck length is the test.

The Feynman Diagram we are going to test this on is Beta Decay of Carbon 14 into Nitrogen 14 to start off with the calculations need to be done for the planck state of an Electron and Neutron as beta decay is P+ > N + e- + Ve , so we willl start with mapping the quarks within the proton which a proton's state is 9.9023511969154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Then the neutron can be described as a Planck State too which is

9.9023511949154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

The electron has a smaller state 1.1998578848809383445875560276978 * 10^51 (-1/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .511/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

The Neutrino has State of 28722.600151171579743008314436886(0/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .2/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) being much smaller than all of them

This Completes the Feynmann Diagram for Beta minus decay and satisfies P+ > N + e- + Ve

9.9023511969154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

=

9.9023511949154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

+

1.1998578848809383445875560276978 * 10^51 (-1/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .511/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

+

28722.600151171579743008314436886(0/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .2/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))\

All properties have been conserved.

This shows the volume of the neutron to be slightly smaller in size to the proton by .0000002%.

This calculator can also be used to find the effects of Dark Energy on the particle in question for a proton you could solve the amount of Dark Energy on the particle on Nucleon, we can find that Dark Energy has a velocity currently of 54 meters per second using a simple equation E = (1/2)MV^2 , V = 54 m/s . The Mass of the Dark Energy Particles are unknown so I will use a mass of electron or mass of proton. Giving each section of space a energy of 1.458 Kev outward with the push of Dark Energy if mass of electron or mass of proton it would be 1.313 Mev , now we can write the proton effected by Dark Energy.

9.9023511969154288921026543960449 * 10^59 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)- 938.28/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)- 938.28/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)- 938.28 /(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Now the Proton is displaying the expansion of Dark Energy upon the Proton.

It has been shown that this graphing tool can be used to graph anything that is contained with the universe using the information about its dimensions, so this test has been concluded about the graphing equation as successful, but I wanted to note that (dx^2 + dy^2 +dz^2) = (Planck State)^2 being R^2 in Planck lengths which is why the dimensions are divided by (dx + dy +dz) and that the Planck state( C ) data is used being the dimensions that the field is over being the Complex Manifold. The manifold of space (Euclidean Space) is being used as (dx + dy +dz) which can also be (dx' + dy' +dz') if you wanted to directly start to use special relativity (Makowski space) on it where as the Field dimensions are from Quantum field theory to be put over the manifold which is a type of quantum gravity.

Next will be a proof of the big equation which will take longer to test which will give a ds^2 value based on a complex system which can be used with the graphing equation to graph the actual state of the entire universe exactly without error based on a complex set of 18 variables or kept in its natural state for a ds^2 value which is a Grand Unified Field equation that takes in account the Strong Nuclear Force, Weak Nuclear Force, Gravity and Electromagnetism all in one equation yielding E8 Killing Vectors. This Metric takes in account General Relativity, Special Relativity, Quantum Mechanics, and Quantum Field Theory to arrive at the solution in Killing Vectors which are then placed in Minkowski space.

∇'(x,y,z,t,ω_{s},ω_{p,}M,I,k,φ,**S,X,Z,μ,Y,q,a,**β**) = (d ^{2}/**

(d^2/∇') - (Ctp)^2 = ds^2

One solved solution for this equation already is for ∇' being d2/dx'2 + d2/dy'2 + d2/dz'2 , The original solution for the equation was LGhost QE Which states that Quantum Entanglement is the same as creating a wormhole between two spaces or universes, and that theoretically if you did quantum entanglement on matter between universes you can transmit matter just like is often done across space during standard Quantum Entanglement experiments.

I am changing the (dx,dy,dz) parameters to display a special relativistic 4 current, now including the evolution of the state over time and not just in a static point.

Luniverse = (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy) , ∇' (x,y,z)= d'(x,y,z)∇ , d(x,y,z)' = d(x,y,z) (1-(V(x,y,z)^2 /C^2))^1/2 , E(x,y,z) = (1/2)MV(x,y,z)^2

https://www.wolframalpha.com/input/?i=(∇'Q%2C+∇'u%2C+∇'F+%2C∇'g++-+∇'D)

This shows the parameters as a function of kinetic energy in a direction or velocity in a direction now, now space properly dilates in the presence of energy at a given time giving a value for L that is special relativistic. The original equation was relativistic however this equation was not.

L'universe = (∇'Charge,∇'Color,∇'flavour,∇'gravity - ∇'Dark Energy)

https://www.wolframalpha.com/input/?i=∇+x(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+y(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+z(1-(V^2+%2FC^2))^(1%2F2)

The time coordinate can be ignored but I am still doing the A.I. analysis of it anyways, which shows that our analysis of Dark Energy and Gravity are Valid with (C+V, C-V)

https://www.wolframalpha.com/input/?i=∇+t+(1-(V^2+%2FC^2))^(1%2F2)

This is the proof that the space contraction equation does not interfere with the "Higher Dimensions" such as gravity or dark energy or charge.

https://www.wolframalpha.com/input/?i=∇+Q+(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+g-d+(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+q+u+f+(g-d)+(1-(V^2+%2FC^2))^(1%2F2)

However this does prove that it takes the dilation effect upon dimension x upon Q or space upon charge that the space is changing however not charge.

https://www.wolframalpha.com/input/?i=∇+Q+x+(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+Q+y+(1-(V^2+%2FC^2))^(1%2F2)

https://www.wolframalpha.com/input/?i=∇+Q+z+(1-(V^2+%2FC^2))^(1%2F2)

]]>