This link has some pics of the Presto 8D cutter. This cutter is missing the motor. Besides that I think it would run ok. I am also missing the belt. If I got a motor I would definitly be using that lathe instead of the Rek-o-kut V. Sorry about the picture quality.
http://heidelberg.en.craigslist.org/wan/1007017044.html
This is the Rek-okut that I am experimenting with now.http://honolulu.craigslist.org/oah/wan/1007024315.html
You can also see the two main heads I have. The first Presto works but the Presto 1-D is fryed. I took it all apart and I will try to rewind the bottom coil. The top coil comes up on the meter as 1000 ohms which would make sense if they were in parallel (head marked at 500 ohms), but the bottom coil has a short and registers 40 ohms on the meter. The thin wires going to the terminal post were charred.
This is the amp I am using. It is a Westinghouse Program Amp. Like an idiot I didn't write down the specs on it; I will post them next week.http://hickory.craigslist.org/wan/1007046992.html
Ok so this is what I am working on
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Ok so this is what I am working on
Last edited by VRCM on Sun Jan 25, 2009 2:04 pm, edited 1 time in total.
Tim
Wierd. What about this link. Someone else got to this one finehttp://hartford.craigslist.org/wan/1001003061.html
Tim
Presto 8D etc.
If you need any info on the Presto 8D check the Presto History page. Just google Presto History.
If you have some specific questions, perhaps I can help. PM me.
Alan.
If you have some specific questions, perhaps I can help. PM me.
Alan.
I did have a question about the 1-D head. Mine is blow. When I took it apart I could see that the wires going the coils seemed charred and were disconnected. Now I tested the resistance of the two coils and the top one was 1000 ohms and the other 40 ohms. They should both be 1000 right ---to make up the 500 ohms? Even so, I was wondering if it would be a good idea to rewind them to make the head an 8 ohm head. I heard some other people talking about it but I don't know. I didn't see a lot of specifics on how it is actually done. I plan to use a slightly thicker wire, approx. 31 gauge, instead of what appears to be 36 or so gauge. I looked up the length of wire I would need for such a resistance and figured upon 128 feet or so per coil to make each coil about 16 ohms. --together in parallel they should drop it to 8. I am not very good with electronics so I am looking for second opinions on this. I have the feeling I made a mistake in my thinking somewhere.
Tim
Just keep in mind that resistance (which you can measure with multimeter) is not the same as impedance! ...although both are expressed in ohms.
Difference is because you use sinusoidal voltage and in this case the impedance is relation between effective voltage and effective current.
... hmmm ... my english is very weak, I know
The best thing to do is ask someone who has proper working head to measure the resistance which you can compare. BUT the impedance of the head is the one which must match with the amp's.
Just to know... I'm not expert in this but I think I understand some basics.
Here is copy/paste from wikipedia:
According to Ohm’s law the impedance of a device can be calculated by complex division of the voltage and current. The impedance of the device can be calculated by applying a sinusoidal voltage to the device in series with a resistor, and measuring the voltage across the resistor and across the device. Performing this measurement by sweeping the frequencies of the applied signal provides the impedance phase and magnitude.
Difference is because you use sinusoidal voltage and in this case the impedance is relation between effective voltage and effective current.
... hmmm ... my english is very weak, I know
The best thing to do is ask someone who has proper working head to measure the resistance which you can compare. BUT the impedance of the head is the one which must match with the amp's.
Just to know... I'm not expert in this but I think I understand some basics.
Here is copy/paste from wikipedia:
According to Ohm’s law the impedance of a device can be calculated by complex division of the voltage and current. The impedance of the device can be calculated by applying a sinusoidal voltage to the device in series with a resistor, and measuring the voltage across the resistor and across the device. Performing this measurement by sweeping the frequencies of the applied signal provides the impedance phase and magnitude.
Ok--because I went on wikipedia and found the math a little confusing---I did an experiment. I tested the resistance of an 8 ohm speaker (what I want the head to be and similar in construction to a cutter head) and it came up as 8 ohms. Maybe it is a coincident or something or maybe they actually are related. Also the speaker crackled when I did it (from battery in meter). So anyways, what I got out what bancho and wikipedia said is that impedence is related more to the voltage and current---but if the meter is testing the resistsnce with voltage then maybe that is the impedence? I don't know. I guess it would be best to get the resistance of the 1-d from someone else. But even if I rewound the coils to the actual resistance of someone elses head would that necessarily mean that the impedence would match?
Last edited by VRCM on Mon Jan 26, 2009 11:57 pm, edited 1 time in total.
Tim
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cuttercollector
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I posted this somwhere else before but now I can't find it, so here we go again.
A normal digital meter measures resistance with DC voltage.
Impedance is the resistance AT some AC frequency and in fact will vary with frequency to some extent and not be the same as at "0" Hz. or DC.
The change in the value depends on both the inductance and capacitance of the device being measured. A carbon resistor would probably be about the same impedance from DC to a fairly high frequency - well above the audio band. But a speaker or cutter has a coil of wire, which has a general characteristic of higher and higher resistance as freqency increases.
The general expectation is that the DC resistance would be somewhat lower than the AC impedance at, say, 1KHz. but you still should be able to get a rough idea if there is a dead short "0 ohms" or if it is open (infinite ohms)
As an aside, Voltage, Current and Resistance are bound together by ohm's law, whether AC or DC.
Voltage = Current X Resistance
Resistance = Voltage / Current
Current = Voltage / Resistance
and power in Watts is always Current X Voltage
A normal digital meter measures resistance with DC voltage.
Impedance is the resistance AT some AC frequency and in fact will vary with frequency to some extent and not be the same as at "0" Hz. or DC.
The change in the value depends on both the inductance and capacitance of the device being measured. A carbon resistor would probably be about the same impedance from DC to a fairly high frequency - well above the audio band. But a speaker or cutter has a coil of wire, which has a general characteristic of higher and higher resistance as freqency increases.
The general expectation is that the DC resistance would be somewhat lower than the AC impedance at, say, 1KHz. but you still should be able to get a rough idea if there is a dead short "0 ohms" or if it is open (infinite ohms)
As an aside, Voltage, Current and Resistance are bound together by ohm's law, whether AC or DC.
Voltage = Current X Resistance
Resistance = Voltage / Current
Current = Voltage / Resistance
and power in Watts is always Current X Voltage