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Re: Impedance, Mechanical, Electrical



Original poster: Jim <branley1@xxxxxxxxxxx>

Tesla list wrote:
Original poster: "Jared Dwarshuis" <jdwarshuis@xxxxxxxxx>


Impedance, Mechanical, Electrical



I wrote this to help the novice coiler understand a rather difficult topic. There is nothing familiar or intuitive about the mathematical approach to understanding impedance. Brains are not "wired from the factory" for grasping second order differential equations and imaginary numbers. Such stuff takes practice, and faith! Fortunately there are tangible,

" hands on" physical objects that can be related to. This will help us to resolve the mystery.



So why should I care about this topic, where is the impetus ?



Always a valid question!



Hmmm, well??.



A Tesla coil should have a power factor correction for the inductive "load"

A Tesla coil might need radio frequency suppression, a "choke"

Optimal transfer of energy to the load will require careful impedance matching.



You need to be armed with some theoretical weaponry, or none of this will ever make any sense.

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When we have a mass tied to the end of a spring we have an object that behaves mathematically very much like a series inductor and capacitor.



Capacitors are like springs. Theoretically you could store charge forever just like you could keep a spring under tension forever. Both cases concern potential energy.



Inductors are very much like a mass. Lenz law and Newton second law of motion are essentially equivalent. Both cases concern forms of kinetic energy. (energy of motion!)



Since the lumped model of LCR and dampened mass/spring share the exact same energy equations, they also share the same equations for impedance. The amplitude equations which describe how much mass and spring bob up and down along the X axis are very old, I don't know who first wrote them. Here I will demonstrate that the familiar impedance formula can actually be derived from mass spring amplitude equations.



M dv/dt + CV +KX = F e(iwt) (where C is Kg/s)



L di/dt + RI + Q/C = V e(iwt) (where this C is capacitance)



Let:



X(t) = A e i(wt-phi)



Q(t) = Ae i(wt-phi)



Then:



M d2/dt2 A e i(wt-phi) + C d/dt A e i(wt-phi) + K A e i(wt-phi) = F e iwt



Where: M = Kg C = Kg/s K = Kg/ss F = Newton



L d2/dt2 A e i(wt-phi) + R d/dt A e i(wt-phi) + 1/C A e i(wt-phi) = V e iwt



Where: L = Kg (m/coul)sqrd R = Kg/s (m/coul)sqrd

1/C = Kg/ss (m/coul)sqrd V = N (m/coul)



Equating the real and imaginary parts of the previous equations, squaring both sides, using the trigonometric identity: (sine phi) sqrd + (cos phi)sqrd = 1



We get:



X = F / ((K- M(w)sqrd)sqrd + Csqrd(w)sqrd))^1/2



Q = V / ((1/C- L(w)sqrd)sqrd + Rsqrd(w)sqrd))^1/2



(w) can be pulled from the denominator.



Then:



X(w) = Velocity = F/ [ ( K/(w) ­ M(w) )sqrd + Csqrd ]^1/2



Q(w) = Current = V/ [ ((w)L ­ 1/(w)C )sqrd + Rsqrd ]^1/2



We recognize impedance in the denominator for both systems



On the surface it may seem a bit odd that X corresponds to Q, but in reality both are just describing displacements from an equilibrium value. In the case of a dampened mass/spring, the equilibrium is simply where the device comes to rest.



So far a lot of math. But what does it all mean??



Lets look at the easiest part first, namely resistance. (named C in the mechanical world.)



C in the mechanical world has units of kg/s. When we push a box across a uniform floor at a constant velocity we can describe this with a simple force equation. Force = velocity x C The units are: Kg m/ ss = m/s x kg/s (C describes how slippery the floor is)



R in the electrical world is analogous. Voltage takes the roll of force. Current becomes velocity (moving charge) and R takes the roll of C.



*C and R are both linear in these models, not always true in either the electric or mechanical worlds.*



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Now an informal look at reactance!



Let us imagine a smooth block attached to a spring which is at the other end anchored to a wall such that our smooth block can slide on a slightly rough floor.



We can always wiggle and jiggle the contraption to find exactly when to push, so as to build up pushes for maximum jiggle.



Why is that?



Well suppose the block is really massive but attached to a bitty joke of a spring.



Question?



Since it is attached to a bitty spring is it ok to give it a good kick?



Answer!



No! You will break your foot, nearly all of the force is going into moving the large mass



Ok! Now push the contraption slowly. Did you will find that nearly all the force went into the collapsing the spring?



This is the essence of reactance!



When we found our maximum jiggle, (likely without realizing it) we had found the perfect compromise point between pushing slowly and giving it a good kick. And of course since we are talking about different amounts of time between pushes, we are actually talking about frequency. (And so it is!, reactance is frequency dependant.)



We have not exhausted the topic, there is also the matter of phase angle. Also there are several other forms of impedance that we have not even talked about. But enough for now.


End.

Jared Dwarshuis w/07




Jiggle vs Kick, both methods use the same amount of force I presume. One applies the force gradually, the other applies the force rapidly. The jiggle conserves energy?
Jim