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Capacitor Impedance for Tormek 1200?

Started by dellis356, July 04, 2025, 11:30:01 AM

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dellis356

What is the micro Farad required for this machine?

Ken S

Welcome to theforum, Derek.

I suggest you email support, (support@tormek.se).

Ken

HaioPaio

#2
What you are looking for is called capacitance, not impedance.
Tormek support will help.

tgbto

Impedance covering capacitance+inductance+resistance (Z=R+jLw+1/jCw), OP was correct...

Ken S

A number of our members have an electrical/engineering background much more advanced than mine. I appreciate that. My electrical background is generally limited to installing and troubleshooting telephone voice and data circuits and self study in home wiring electricity.

I can read charts like Tormek's and understand why a 50 Hertz, 240 Volt motor in not compatable with 60 Hertz 120 Volt US commercial power. I also understand why a motor drawing 200 Watts requires more power than one drawing 120 Watts.

I can not understand the working of the posted formula, or why it is important. I would like to understand this. I believe I am not alone in this. I would humbly request replies going beyond just mouthing formulas. Please share your expertise by educating the rest of us and explaining the conceps involved and whythey are important

I once watched a marvelous VHS tape explaining Einstein's E=MC2. It gave the example of when a baseball is thrown at the speed of light squared, the baseball (matter) becomes Energy. for me, that was very informative. I would like to better understand the Tormek motors.

Ken

tgbto

Ken, the idea of the formula was to substantiate a reply to an objection regarding the choice of words of the OP.

I am no Richard Feynman, so I will not be able to be as clear as a professional teacher. Still, to make a long story short :

The concept of impedance is a generalization of the concept of resistance, i.e. the proportional relationship between current and tension. With a resistance, everything is simple : tension U is related to current I by the U=R.I relationship.

For a capacitor, U is related to the integral of the current over a period of time. But using mathematics in the complex world, integrating is "simply" dividing by j.w (assuming a sinusoidal current, and where j²=-1 and w or "omega"  is the pulsation of the current). So one can write U=Z.I with Z=1/jCw.

Conversely, for a solenoid, U=Z.I with Z=jLw.

Therefore, in electronics, impedance is a generic term covering at once R,L and C. And asking for the impedance of a capacitor is another way of asking for its capacitance. Pretty much in the same way that asking someone for their address will give you the name of their hometown, and a bit more.

As for E=m.c², if I may, there might be a slight twist in your VHS or your recollection of it. To apply E=mc² to your baseball, you would have to thoroughly disintegrate said baseball, and the energy released (in the form of massless photons) would be m.c², m being the rest mass of said baseball. If you could accelerate the ball to lightspeed, its mass would change and things get pretty messy from here. But e=mc² is always true, even for a baseball at rest. And you cannot accelerate it to c², which is not a speed.

Hope this helps,

Nick.

HaioPaio

I was only able to find a capacitor for the T4 and T8 at that website https://www.sharpeningsupplies.com/collections/tormek-replacement-parts.

The capacitance of those capacitors is specified in microfarads, while there is no mention of impedance.
For the T1200, I would ask support.

Herman Trivilino

#7
Quote from: tgbto on July 07, 2025, 08:21:32 AMImpedance covering capacitance+inductance+resistance (Z=R+jLw+1/jCw), OP was correct...

Impedance is measured in ohms, capacitance is measure in farads. OP wanted to know the number of farads, so he was looking for capacitance, not impedance.
"Knowledge isn't free, you have to pay attention." R.P. Feynman

Herman Trivilino

Sorry, Ken, to have to use a formula. The impedance, in ohms, of a capacitor is given by 1/(2*pi*f*C), where f is the frequency in hertz, and C is the capacitance in farads. So, for the T8 the capacitance is 25 microfarads, or 0.000 025 farads. If you're in North America the frequency is 60 hertz (in Europe it would be 50 hertz). That works out to about 210 ohms.

So, loosely speaking, the capacitor offers a resistance to the flow of electricity of about 210 ohms.
"Knowledge isn't free, you have to pay attention." R.P. Feynman

Herman Trivilino

Quote from: Ken S on July 07, 2025, 12:25:46 PMI once watched a marvelous VHS tape explaining Einstein's E=MC2. It gave the example of when a baseball is thrown at the speed of light squared, the baseball (matter) becomes Energy. for me, that was very informative. I would like to better understand the Tormek motors.

Yeah, you gotta be careful of those popular science publications and videos. What the formula really tells us is that the baseball has a rest energy of mc2. So when at rest, it has an energy of about (0.145 kg)(300 000 000 m/s)2, or about 13 000 000 000 000 000 joules. For comparison, if you heated 0.145 kg of water from room temperature to the boiling point (at sea level) it would take about 50 joules of energy. You could convert the energy of the baseball into, say, radiation by annihilating it, but unfortunately that would be very dangerous. Such things have to be done very carefully, using the correct materials (not baseballs) under carefully controlled conditions. We call such devices nuclear reactors.

The speed of the baseball has nothing to do with mc2. In fact, c is a speed, whereas c2 is not. It's just a conversion factor between units of mass and units of energy. Analogous to the conversion factor between inches and feet.

We all know that inches and feet are both measures of the same thing, distance. What Einstein taught us with mc2 is that kilograms and joules are both measures of the same thing. Historically, we thought that mass was conserved, and energy was conserved. Two separate conservation laws. But Einstein taught us that that is just an approximation. There really is nothing special about nuclear reactions. When you light a match you are creating radiation, and that means the mass of the match and the smoke is decreased. It's just that the decrease is too small to measure. Nuclear reactions, though, are much more energetic, so the reduction in mass is quite measurable.

"Knowledge isn't free, you have to pay attention." R.P. Feynman

tgbto

Quote from: Herman Trivilino on July 26, 2025, 04:59:19 PMSo, loosely speaking, the capacitor offers a resistance to the flow of electricity of about 210 ohms.


The capacitor offers no resistance to the flow of electricity at all, it essentially alters its phase by 90 degrees (and multiplies it by a factor that depends on frequency). The resistance is the real part of the impedance and the impedance of a capacitor is purely imaginary : all the energy is conserved in an electrical form, none is lost to Joule effect.

Herman Trivilino

#11
Quote from: tgbto on July 28, 2025, 08:55:44 AM
Quote from: Herman Trivilino on July 26, 2025, 04:59:19 PMSo, loosely speaking, the capacitor offers a resistance to the flow of electricity of about 210 ohms.


The capacitor offers no resistance to the flow of electricity at all, it essentially alters its phase by 90 degrees (and multiplies it by a factor that depends on frequency).
 The resistance is the real part of the impedance and the impedance of a capacitor is purely imaginary : all the energy is conserved in an electrical form, none is lost to Joule effect.

The impedance is a complex number, containing both a real part and an imaginary part. For an ideal capacitor, the impedance is purely imaginary, with no real part.

But for a real capacitor, there is a real part. Quoting from Wikipedia:

"The magnitude of the impedance |Z| acts just like resistance, giving the drop in voltage amplitude across an impedance Z for a given current I."

My calculation gives the magnitude of the impedance as about 210 ohms.

That's why I said "loosely" speaking. I'm ignoring the imaginary part, which is, as you say, responsible for the phase shift.
"Knowledge isn't free, you have to pay attention." R.P. Feynman

tgbto

#12
Quote from: Herman Trivilino on July 28, 2025, 07:08:18 PMThe impedance is a complex number, containing both a real part and an imaginary part. For an ideal capacitor, the impedance is purely imaginary, with no real part.

But for a real capacitor, there is a real part. Quoting from Wikipedia:

"The magnitude of the impedance |Z| acts just like resistance, giving the drop in voltage amplitude across an impedance Z for a given current I."

My calculation gives the magnitude of the impedance as about 210 ohms.

That's why I said "loosely" speaking. I'm ignoring the imaginary part, which is, as you say, responsible for the phase shift.


Agreed, *but* you based your calculation on the capacitance only, meaning you assume an ideal capacitor.

So you are not ignoring the imaginary part, you are dealing only with the imaginary part. The 210 ohms value you computed is a pure reactance, or said otherwise, no resistance at all.

To quote from the same wikipedia : "A pure reactance does not dissipate any power."


Dutchman

Now we know enough about possible misunderstandings regarding capacitors.
Thank you.

Herman Trivilino

#14
Quote from: tgbto on Yesterday at 07:59:17 AM
Quote from: Herman Trivilino on July 28, 2025, 07:08:18 PMThe impedance is a complex number, containing both a real part and an imaginary part. For an ideal capacitor, the impedance is purely imaginary, with no real part.

But for a real capacitor, there is a real part. Quoting from Wikipedia:

"The magnitude of the impedance |Z| acts just like resistance, giving the drop in voltage amplitude across an impedance Z for a given current I."

My calculation gives the magnitude of the impedance as about 210 ohms.

That's why I said "loosely" speaking. I'm ignoring the imaginary part, which is, as you say, responsible for the phase shift.


Agreed, *but* you based your calculation on the capacitance only, meaning you assume an ideal capacitor.

Yes, you are correct about that. So is it correct to say that what I calculated is the magnitude of the ideal capacitor's impedance?

QuoteSo you are not ignoring the imaginary part, you are dealing only with the imaginary part.

Yes. I did get that wrong.

QuoteThe 210 ohms value you computed is a pure reactance, or said otherwise, no resistance at all.

To quote from the same wikipedia : "A pure reactance does not dissipate any power."

But, quoting from the same article:
"In electrical engineering, impedance is the opposition to alternating current presented by the combined effect of resistance and reactance in a circuit."

So it does offer 210 ohms of opposition to the current, plus whatever contribution is made by the (nonideal) resistance.
"Knowledge isn't free, you have to pay attention." R.P. Feynman