# AC THEORY

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
• jaggedben

#### Joethemechanic

##### Senior Member
• LarryFine

#### jaggedben

##### Senior Member
Music theory has more to do with division of the major scale and oscillation... And harmony.
If you asked mozart about harmonics you might get that look... If you ask jim root he would smile. There is alot of history behind dividing a octave into 11 half steps.You're not telling me anything I don't know.

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#### jaggedben

##### Senior Member
I know. (It was (supposed to be) a joke.) I happen to be a piano player, too.
I knew you knew. Just providing more content for the spectators.

• LarryFine

#### rambojoe

##### Senior Member

I knew everything you just said before getting involved with electrical work, so yeah, it was much easier to glom onto harmonics in electricity when I first heard about it.
Why are words i didnt write in my post?!
Anyway, you knew me when i did band instrument repair? Bach didnt use harmonics either, maybe stravinsky.

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#### jaggedben

##### Senior Member
That places where the lines cross are the notes
Only some of them are close to our conventional 12-tone equal tempered tuning.

#### ActionDave

##### Chief Moderator
Staff member
Some basic structural too. Know what moment, shear, compression, torsion, tension are, And be able to recognize some basic structures like a cantilever, or a bridge or a column. You don't know how many times I see stuff falling down because the guy that installed it didn't know any basic structural concepts
I'll never understand calculus, but I understand this.

#### Joethemechanic

##### Senior Member
I'll never understand calculus, but I understand this.

You are probably already doing some calculus without knowing it

• Carultch

##### Senior Member
Only some of them are close to our conventional 12-tone equal tempered tuning.
Would that be why they are called perfect? #### Besoeker3

##### Senior Member
A Mathematician named Hall
His pecker plus eight
Was a fifth of eighth
Was nothing at all

Boom boom !!

#### drcampbell

##### Senior Member
I'll never understand calculus, but I understand this.
Never say "never".

Get this book: Calculus Made Easy by Silvanus P. Thompson, 1910
Despite the difficult-to-fathom title, that's exactly what this book does. Functional and compact, only 250 pages.

I have no idea why anything other textbook is used for first calculus classes. As a freshman, I struggled with a verbose and academically-rigorous textbook written by a math professor for math professors, which I was able to comprehend only after mastering calculus. </rant>

• Carultch

#### drcampbell

##### Senior Member Had he learned & used mathematics, he might not be on his deathbed at age fifty.

• • mayanees, jaggedben and retirede

#### drcampbell

##### Senior Member
It's called the Fourier Series. ...
The √2 relationship is something entirely different.

• jaggedben

#### __dan

##### Senior Member
I'll never understand calculus, but I understand this.
Spring force F = k * x, which is the equation of a line, y = m * x + b. m , k are constants, the slope of the line. x is the displacement, dimensions of length. b, the offset, b = 0 for the spring. It's linear, freshman year HS.

But you don't want to know the force F, it doesn't tell you what you want to know. That they do teach it this way proves my point, the book is wrong.

The high order skill is to look at the system and write the equation for it. From there the computer will solve it or you can Google for the solution to the equation.

First step in the book method is to disassociate the problem from the model so the math becomes unrelatable or unrecognizable, producing hordes of people who get lost at that point (thrown from the cliff more like it).

Forget about the force balance equation. Look at it and set it up as the Conservation of Energy Law. Total Energy = Potential Energy + Kinetic energy (+ losses = 0). The potential energy of the spring is the integral of force over distance (calc term), which is area under the curve, for that case area is a triangle. So it's the area of the triangle 1/2 k * x^2. Which the same result from the power law of calculus, integral of spring force f k * x, >> 1/2 k * x^2.

Kinetic energy KE = 1/2 m * v^2 (the book will say recall you know this from the last chapter).

Total energy TE is a constant, which makes your life easier. From the conservation of energy law, look at it and write TE = KE + PE (conservation of energy. Then knowing TE is a constant, TE = 1/2 m * v^2 (that's KE) - 1/2 k * x^2 (that's PE). The minus sign because KE = 0 at max PE or PE = 0 at max KE or PE = KE = TE when either KE or PE = 0. At least in that form you can look at it and recognize which is due to the spring and which is due to movement momentum.

Knowing the spring will oscillate if you pull on it and let it go, the proper form to write it is TE (a constant) = 1/2 m * v^2 cos(theta) + 1/2 k * x^2 i sin(theta), which is the model in the complex plane. That gets you to Euler TE (a constant) e^ i (theta) (a constant = 1) = KE + PE. They are a vector sum in the complex plane. There is a hidden sine wave which you do not have to deal with.

The i in the above is a vector rotation operator, by 90 deg, so the PE and KE are at right angles to each other and the sum is by the Pythagoras Theorem. The book gets that part wrong. The TE is the hypotenuse of the triangle.

The book does not teach it his way and is so horribly wrong. But everyone who flunks calc and needs a job is down at Jiffy Lube changing the oil on the teacher's car. So is a sense it does work for them.

The book will write the same problem as a "second order differential equation" then weep along with you about how hard it is to solve. Then to Jiffy Lube so you can make a living at it.

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#### Carultch

##### Senior Member
Spring force F = k * x, which is the equation of a line, y = m * x + b. m , k are constants, the slope of the line. x is the displacement, dimensions of length. b, the offset, b = 0 for the spring. It's linear, freshman year HS.

But you don't want to know the force F, it doesn't tell you what you want to know. That they do teach it this way proves my point, the book is wrong.

This situation has a lot of moving parts, even though the forces vs x-position is a straight line. The force determines the acceleration, which determines the velocity, which determines the x-position, which determines the force. So you have this feedback loop of terms that are all inter-dependent on each other, and it is hard to see where to start. That's what DiffEQ is all about, is navigating your way through this feedback loop of dependency. There are also many equally-valid solution methods, that all ultimately agree with each other.

If it is my choice, the way I'd introduce this topic, is to show experimental results that are easily recognized as a sine or cosine wave. Then work backwards, knowing the form of the solution is x=A*cos(w*t), and show how it satisfies the original equation of motion. m*x" + k*x = 0. Take derivatives twice (a simple Calc 1 topic) to get x"=-A*w^2*cos(w*t). Then with a little algebra, you can see that it is all satisfied when w=sqrt(k/m), confirming our original guess at the solution.

#### Joethemechanic

##### Senior Member
freshman year HS.

Not really anymore. They graduate college and can't figure out the square footage of a perfectly square building

• Carultch

#### __dan

##### Senior Member
I know they teach the same problem using a 2 x 2 matrice and linear algebra (don't ask me why, I have no clue). Or I would say they need more help at Jiffy Lube.

Looking at it in that form is impenetrable. I cannot look at any part of it and recognize which part is displacement of x (distance), which is dx/dt (velocity), which is d2x/dt2 (accelleration). Even knowing what to look for, the notation is so far from the model it is invisible.

But ... e^i(theta) = cos(theta) + i sin(theta), Euler's formula, is a right triangle with e^i(theta) as the hypotenuse, which = 1 for the unit circle.

The right hand side sums with Pythagoras Theorem, r^2 = a^2 + b^2. Set the problem up with r is a constant, a is kinetic energy, b is potential energy. They are at right angles to each other. Multiplying by i rotates by 90 deg but the units are still Joules.

The textbook, and the entire internet, does a trick substitution with this at theta = pi. Sin pi = 0 dropping out the i on the right hand side, cos pi = -1. So they declare e^i(pi) = -1. You could say this is true in the existential case for theta = pi and false for every other value of theta.

What is true for every case of theta, e^i(theta) = 1, summing the right hand side using Pythagoras Theorem. For the case of theta = pi, it would be (-1)^2 = 1 on the right hand side. They forget to square and sum, sq root, the right hand side. e^i pi = 1, e^i(theta) = 1 for every value of theta.

The textbook and the entire internet is wrong.

But the people who flunk this stand in line in hordes for the open spots at Jiffy Lube.

#### Joethemechanic

##### Senior Member
You teach one of them the 3 4 5 rule to get things square, and they can't comprehend that the units don't matter

#### Omid

##### Member
I think if you are a residential electrician, you may not need to know all of the material, but you will likely need to know the basics of AC theory in order to troubleshoot problems. But if you are a commercial or industrial electrician, you will likely need to know more of the material.

• Carultch

#### Carultch

##### Senior Member
The textbook, and the entire internet, does a trick substitution with this at theta = pi. Sin pi = 0 dropping out the i on the right hand side, cos pi = -1. So they declare e^i(pi) = -1. You could say this is true in the existential case for theta = pi and false for every other value of theta.

What is true for every case of theta, e^i(theta) = 1, summing the right hand side using Pythagoras Theorem. For the case of theta = pi, it would be (-1)^2 = 1 on the right hand side. They forget to square and sum, sq root, the right hand side. e^i pi = 1, e^i(theta) = 1 for every value of theta.

The textbook and the entire internet is wrong.

It's not that everyone is wrong, it's that there are absolute value signs you are forgetting to include to make the point you are trying to make. Called a modulus in the complex number world, which is the Pythagorean theorem combination of both the real and imaginary parts. Imaginary and real parts are like oil and water, and you lose information when you "add" them this way, so it isn't the same number.

It is a special case that e^(i*pi) = -1, because that's a 180 degree rotation (pi radians) around the unit circle from +1, which is -1.
It is a general case that |e^(i*theta)| = +1, as long as theta is a real number. No matter how much you rotate around the unit circle, you are always a distance of +1 from the origin.
e^(i*theta) in general isn't necessarily equal to +1, except at special cases of full rotations, where theta is a multiple of 2*pi.

What's ultimately behind this, is the Taylor series of e^x, having terms in common with the Taylor series for both sine and cosine. That, and the cycle of powers of i, that rotate from +1 to +i, to -1, to -i, and back to +1. It's very fascinating how this formula puts it all together, and shows us what we can do with all the hidden parts of functions when we are limited to the real numbers.

I know they teach the same problem using a 2 x 2 matrice and linear algebra (don't ask me why, I have no clue). Or I would say they need more help at Jiffy Lube.

Looking at it in that form is impenetrable. I cannot look at any part of it and recognize which part is displacement of x (distance), which is dx/dt (velocity), which is d2x/dt2 (accelleration). Even knowing what to look for, the notation is so far from the model it is invisible.
There is a reason why they do this, and it's primarily to use as an application of the concepts from linear algebra.

What's happening is they are turning a 2nd order equation, into a 1st order system of equations, so that linear algebra finds the solution.

Given:
m*x" + k*x = 0

They construct a system of equations by using v=x' and replacing x" with v'. This gives us the system:
x' = v, the definition of velocity
and
v' = k/m * x, the original equation with v' replacing x"

It's just an alternative way of looking at the same problem, to use different techniques of solving it

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