Peter, whatever happened with Eric P. Dollard?

[Raui]

I spoke to Tom B, and heard he donated 500 to Eric, (who should just open source BTW) i heard Eric is still an angry SOB thats okay we tolerate bothers with an attitude but only for so long hope he works with the open source community soon


Ash

Ash,
Concerning the open source, isn't that what Eric is doing? It may not be at any fast pace or anything but Eric is saying that our understanding of electricity is faulty and that with his understanding we can go to the next step of building a working device. Right now he is trying to give us a new understanding. Eric is more concerned that we know how electricity works than having us know how to blindly build a device and to be honest anyone serious about their researches into this area should feel the same way. Like the saying goes, you need to learn to walk before you can run.

Lamare,
I have some spare hosting space, if your interested PM me
Raui

[t-rex]

[ashtweth]

HI Raui, lamare , T-rex at all.

thanks for that guys, I also got told about a lot of potential there so yes my friend your right they are going to need this crowd to "pick up the batten", i heard about all sorts of pancake cols and amazing stuff that was demonstrated over the years, i am glad you guys "have not missed the boat" with what they were demonstrating, this was the concern i was told about that people are off in other areas whilst letting Eric's stuff subside.

Ash

[Redisnoc]

Bump...

"one for the gipper"

[Raui]

Eric,
I came to the same conclusion by taking the time derivative of C = i times t, divided by v
to get the resultant being a conductance so that when the capacitance is increased you get a positive conductance and when the capacitance is decreased you get a negative conductance. Same applies to a change of inductance except the equation used was L = v times t, divided by i which gave a result of resistance (v divided by i). I've attached a pdf with my working as I know you don't like text math. So really my question is - since a negative change in inductance leads to a increase in current and a negative change in capacitance leads to an increased voltage, are these increases caused by a negative resistance or conductance or is this just a mathematical coincidence?

@All,
Also I have looked high and low for a chapter in Electric Discharges, Waves and Impulses and cannot find a section on Velocity Measure but I can swear I have seen it before. I have looked in a lot of his other works and cannot find it either. Does anybody have a copy of the edition with this chapter in it?

Raui

[Web000x]

Eric,
I came to the same conclusion by taking the time derivative of C = i times t, divided by v
to get the resultant being a conductance so that when the capacitance is increased you get a positive conductance and when the capacitance is decreased you get a negative conductance. Same applies to a change of inductance except the equation used was L = v times t, divided by i which gave a result of resistance (v divided by i). I've attached a pdf with my working as I know you don't like text math. So really my question is - since a negative change in inductance leads to a increase in current and a negative change in capacitance leads to an increased voltage, are these increases caused by a negative resistance or conductance or is this just a mathematical coincidence?

@All,
Also I have looked high and low for a chapter in Electric Discharges, Waves and Impulses and cannot find a section on Velocity Measure but I can swear I have seen it before. I have looked in a lot of his other works and cannot find it either. Does anybody have a copy of the edition with this chapter in it?

Raui

Raui,

A couple of quotes by Eric Dollard:

This is a result of the variation of capacitance (C in Farrads) with respect to time (T in seconds) which results in a negative conductance G (in Siemens).

This is another example of synchronous parameter variation. In this case inductance (L in Henrys) time (T in seconds) gave rise to positive resistance (R in Ohms)

I'm sure you were already aware of these statements. Good eye.

Dave

[t-rex]

[KurtNalty]

Eric's DC Transmission Line

Eric Dollard's DC Transmission Line Exercise

Eric posted a transmission line puzzle. Here is my answer

************* Original Posting ***********************

I have a D.C. transmission line, the conductors are 2 inches in diameter, spacing is 18 feet.
How many ounces of force are developed upon a 600 foot span of this line, for the following;

1. 1000 ampere line current.

2. For 1000 KV line potential?

I am waiting.

****************** My Answer **************************

The magnetic repulsion between the two conductors:

1) Calculate B

B = mu_0 H = mu_0 (I/(2*pi*r)) = (mu_0 I)/(2*pi*r) r = 18 feet = 5.48m, I = 1000A.

= (4.0e-7*pi)(1000)/(2.0*pi*5.48) = 2.0e-4/(3.14*5.48) = 11.6 uT (very small compared to terrestrial magnetism)

Calculate Force/Length

F/l = IxB = 1000A*11.6uT = 11.6 mN/meter

Calculate total for for 600' span.

l = 600' = 182.88 meter
F = 11.6 mN/meter * 182.88 meter = 2.12 N

Convert to ounces force. 1 lb = 4.45 N = 16 oz. 1 N = 3.596 oz.

F (per 600 span) = 7.62 oz. (Magnetic repulsion)


2) Calculate Electrostatic Attraction

I use the principle of virtual work with parallel plate capacitors
approximated by the 2 in diameter conductors separated by 18 feet.
I model the capacitor as a flat ribbon with 18 feet separation. The
curvature of the cylindrical conductor introduces a small error of the order
2in/18ft = 0.9%, so no problem.

E = (1/2) C V^2 = (1/2) ((epsilon A)/(d)) V^2

F = (del E / del d) = -(1/2) ((epsilon A)/(d^2)) V^2
= -(1/2) ((8.854E-12*182.88m*0.0508m)/(5.48m*5.48m)) (1.0E6V)^2

= -1.36955 N = -4.92 oz (Electrostatic attraction)

We see that reducing the current can balance mechanical forces from repulsion and
attraction. There will be a characteristic impedance associated with this balanced
system.

Balance magnitudes of attraction and repulsion

((mu I^2)/(2 pi)) (L/d) = (1/2) ((epsilon L*WireDiameter)/(d^2)) V^2

V^2/I^2 = (mu/epsilon) ((d)/(pi*WireDiameter)) = Z^2

Z = 377 sqrt(d/(pi*WireDiameter) Ohms

Enjoy

Kurt Nalty

[Web000x]

Eric Dollard's DC Transmission Line Exercise

Eric posted a transmission line puzzle. Here is my answer

************* Original Posting ***********************

I have a D.C. transmission line, the conductors are 2 inches in diameter, spacing is 18 feet.
How many ounces of force are developed upon a 600 foot span of this line, for the following;

1. 1000 ampere line current.

2. For 1000 KV line potential?

I am waiting.

****************** My Answer **************************

The magnetic repulsion between the two conductors:

1) Calculate B

B = mu_0 H = mu_0 (I/(2*pi*r)) = (mu_0 I)/(2*pi*r) r = 18 feet = 5.48m, I = 1000A.

= (4.0e-7*pi)(1000)/(2.0*pi*5.48) = 2.0e-4/(3.14*5.48) = 11.6 uT (very small compared to terrestrial magnetism)

Calculate Force/Length

F/l = IxB = 1000A*11.6uT = 11.6 mN/meter

Calculate total for for 600' span.

l = 600' = 182.88 meter
F = 11.6 mN/meter * 182.88 meter = 2.12 N

Convert to ounces force. 1 lb = 4.45 N = 16 oz. 1 N = 3.596 oz.

F (per 600 span) = 7.62 oz. (Magnetic repulsion)


2) Calculate Electrostatic Attraction

I use the principle of virtual work with parallel plate capacitors
approximated by the 2 in diameter conductors separated by 18 feet.
I model the capacitor as a flat ribbon with 18 feet separation. The
curvature of the cylindrical conductor introduces a small error of the order
2in/18ft = 0.9%, so no problem.

E = (1/2) C V^2 = (1/2) ((epsilon A)/(d)) V^2

F = (del E / del d) = -(1/2) ((epsilon A)/(d^2)) V^2
= -(1/2) ((8.854E-12*182.88m*0.0508m)/(5.48m*5.48m)) (1.0E6V)^2

= -1.36955 N = -4.92 oz (Electrostatic attraction)

We see that reducing the current can balance mechanical forces from repulsion and
attraction. There will be a characteristic impedance associated with this balanced
system.

Balance magnitudes of attraction and repulsion

((mu I^2)/(2 pi)) (L/d) = (1/2) ((epsilon L*WireDiameter)/(d^2)) V^2

V^2/I^2 = (mu/epsilon) ((d)/(pi*WireDiameter)) = Z^2

Z = 377 sqrt(d/(pi*WireDiameter) Ohms

Enjoy

Kurt Nalty

Hey Kurt,

I've looked at most of you logic and I see something that is worth a second look. Having talked to Eric personally about the subject of "internet math", I am going to suggest that you find a way to make it more "text-book-ish" or he might have trouble deciphering it.

He told me that he would have looked this up himself but doesn't have any books left. He only wants this info (I think) so that he can give suggestions on designing a machine that has the attractive(dielectric) and repulsive forces(magnetic) cancel while generating usable power.

Dave

[Raui]

Kurt,
Great to see someone else helping on the equations! Dave mentioned making them more 'textbookish' as so Eric can read them, when I send Eric equations I use this LaTeX generator and he's understood every equation I've ever sent him; Online LaTeX Equation Editor - create, integrate and download

Eric,
Hope your doing well.

Raui

[KurtNalty]

Online Equation Typesetter

Thanks for the link to the online equation typsetter.

In the meanwhile, I've posted a PDF for the math at

http://www.kurtnalty.com/ForceBalancedDCTransmissionLine.pdf

[t-rex]

[Aether]

@All,
Also I have looked high and low for a chapter in Electric Discharges, Waves and Impulses and cannot find a section on Velocity Measure but I can swear I have seen it before. I have looked in a lot of his other works and cannot find it either. Does anybody have a copy of the edition with this chapter in it?

Raui

After doing a quick search on google I found this:

Lectures on Electrical Engineering - Charles Proteus Steinmetz - Google Books

Which is in the Chapter "Line Oscillations" in Electric Discharges, Waves and Impulses on pg 83.

[hoggel]

has the discussion moved elsewhere? or why is it so quiet in here?....just wondering.

[Web000x]

has the discussion moved elsewhere? or why is it so quiet in here?....just wondering.

I am reading everything that is posted and would love to spend my time discussing. But I'm busy building a synchronous belt driven power generation setup. I will have two alternators with pulleys of different ratios driving them so that I may investigate Eric's suggestion of using magamps for energy synthesis. I have pulleys for the alternators at a 2/1 and 3/1 ratio so that I may look into modulated an Alexanderson magamp at 2nd and 3rd harmonic. Eric told me that modulation at the 2nd harmonic represents power and at the 3rd harmonic represents energy. I shall investigate and report my findings.


Dave