Questions and Answers, CTT, December 2015

[webenda]

In Classic Toy Trains December 2015 issue on page 22 a question is asked about a Lionel postwar No. 282 crane's electromagnet, "Is there a way to increase the magnetic power of the crane without burning out the motor?" The author of the question states they are using an old starter set power pack.

 

CTT answers (in part),"…when it comes to an electromagnet (an inductor), there is a sensitivity to fast rise times…"

 

What a strange thing to say. An inductor's impedance is proportional to frequency, but this 282 electromagnet is not a sensitive little thing that will get upset about the 60 Hz waveform shape from a modern transformer. 

 

CTT explains about this sensitivity, "This (fast rise time) causes heat…"

 

That is a false statement. There is less current flow in an inductor during the fast rise time portion of a chopped sine wave than there is at the same time during a pure sine wave. Less current = less heat.

 

CTT goes on to state, "…which causes resistance; so it draws more current and gets even hotter."

 

The statement is so wrong. More resistance would reduce current so it would not get as hot.

It is not resistance either, but reactance or impedance that is a function of rise time. The equaton is X_L = 2pfL .

I am considering nominating this 282 answer for the Ig Nobel award.

[lionelsoni]

Wayne, I had a reaction similar to yours; but you beat me to the punch.  I wrote up these comments yesterday, but hadn't posted them yet:

There are errors in the answer to a question in the December CTT about trying to increase the load that a Lionel 282 crane will pick up.  This same question popped up recently on the forum.

Here's how it does work:  The current through the electromagnet increases as the voltage applied to it increases.  This is true whether the voltage is a pure sinusoid or is phase controlled.  In either case you can get however much current you want by varying the voltage.

The instantaneous force produced by the magnetic field and the heat power produced are both proportional to the square of the instantaneous current.  So, whatever waveform the current has and whichever voltage waveform you use to produce that current waveform, the same average current corresponds to the same average heating and to the same average lifting force.  Therefore, it doesn't matter which kind of power source you use:  At some voltage setting, you get the same result from either.

The article comes to the opposite conclusion, that you should use a 1950s power supply with the 282 and not a modern one.  The explanation includes the incorrect statements that inductors have "a sensitivity to fast rise times", and that increased magnet resistance due to heat causes increased current "and gets even hotter".  The current in an inductor in fact responds sluggishly to sudden voltage changes; and current through a resistor decreases with increased resistance rather than increases, not that these things have anything to do with the question.

The oddest part of the answer is the suggested exercise of "coloring in a single wave on both waveforms".  I don't see how this can help anyone "to better understand what is happening", and wouldn't even if the waveform illustrations were accurately drawn.

[webenda]

Thank you Bob, for adding more information on this subject.

[webenda]

lionelsoni

Therefore, it doesn't matter which kind of power source you use: 

Maybe it does Bob.

Waveform from CW-80 set to 12 Vrms:

 

 

There is not much magnetic force from the electromagnet during 8.7 milliseconds of zero current each half cyle. The electromagnet is going to drop the object during this time period. How far will it fall?

 

Distance d travelled by an object falling during time t is d = 0.5(gt²)

where g = 32.2 ft / sec²

and t = 0.0087 seconds

d = 0.5(32.2 x 0.0087²) = 0.001218609 feet

 

Converting feet to inches gives 14.6 thousandth of an inch. Is 14.6 thousandths too far for the magnet to pull the falling object back from? I don't know.

---

Let's examine the advantage of using DC. We will add a full wave rectifier and capacitor filter to the line powering the electromagnet.

 

 

Advantage 1: Rectified and filtered DC voltage is nearly equal to peak value of the transformer's waveform.

 

Advantage 2: No dead time in the electromagnet.

 

Advantage 3: No reactance with DC current so same DC voltage as AC rms produces more current and magnetic strength.

 

There are disadvantages also:

 

Disadvantage 1: Not as simple as just being satisfied with picking up lighter loads.

 

Disadvantage 2: Sine Peak DC voltage is too high. Need to find a way to reduce to 282's maximum of 18 Vrms.

 

Disadbantage 3: Voltage range specification for 282 is for AC. There is no reactance in a coil running on DC so currents are higher. The voltage range specification may have to be reduced to keep current down.

 

 

[lionelsoni]

Wayne,

I have some concerns about your last post:

o  Your picture of a phase-controlled waveform appears to be a voltage waveform.  If there's any significant inductance in the magnet, the current waveform would be rather different.  And it's the current, or actually its square, that determines the magnetic force on the load.  I would expect the current to start to ramp up at the time the voltage half-cycle waveform turns on; and I would expect it to continue past the axis-crossing of the voltage waveform, as the magnetic field collapses.  The low-pass filtering due to the electromagnet's inductance should make the current waveform more like a sinusoid than the voltage waveform is.

o  You seem to have mislabeled the time dimensions of the waveform.  The duration of a half-cycle is 1/120 second, or 8.3 milliseconds, not 16.7.  This makes the vertical motion 3.7 mils, not 14.6.

o  You assume that the load begins to fall as soon as the magnetic force stops.  But if there is to be no net vertical motion of the load, the average of the combined forces on it must be zero.  If there were no magnetic force for the 8.7 (now 4.35) milliseconds, then the magnetic force during the preceding 8 (4) milliseconds must have been greater than gravity and so must have accelerated the load upward during that time.  So it started the 8.7 (4.35) millisecond interval with an upward velocity.

Of course, if the vertical motion of the load was blocked mechanically by the presence of the electromagnet, then the fall would start at zero velocity, with some amount of downward mechanical force having been supplied by the collision and contact with the electromagnet; but then there would be some greater load that the electromagnet could have attracted that would have just kissed the electromagnet before starting down.  It is that greatest load that we are concerned with; so we can assume that the maximum load would only barely touch the electromagnet at the top of its travel.

The least uniform force waveform that seems possible to me is an impulse, great enough to exactly reverse the maximum load's maximum downward velocity.  The load then follows a parabolic ballistic trajectory during almost the entire half-cycle.  I calculate that this would result in a peak-to-peak position displacement of .085 millimeters, or 3.35 mils.  This is so little that I think it safe to assume that there is no significant change to the geometry of the magnetic field and therefore that the magnetic force remains proportional to the square of the current, unaffected by the microscopic motion of the load.

From this, I conclude that it is the average magnetic force, and therefore the average of the square of the current, and not the 60-hertz waveform that matters in the lifting ability of the electromagnet, as long as it does not saturate the magnet of course.

[webenda]

Thank you for your analysis Bob, I appreciate it.

lionelsoni

o  Your picture of a phase-controlled waveform appears to be a voltage waveform.

It is the voltage at CW-80 output terminals. I have a current shunt but no 282 gantry crane to view the current wave form of a 282 electromagnet. (I had a light bulb in the circuit for a load, but only because my CW-80 has a severe oscillation riding on the 60 Hz wave when operated without a load.)

lionelsoni

o  You seem to have mislabeled the time dimensions of the waveform.  The duration of a half-cycle is 1/120 second, or 8.3 milliseconds, not 16.7.

I just looked at the image and the scope gives my error away, it says 2.5 ms per division. I remember thinking 5 ms per division when I labeled the trace. Indeed I mislabeled the time, thank you for catching that. 

Recalculating:

d = 0.5(32.2 x 0.004352) = 0.00030465225 feet

 

Converting feet to inches gives 3.6 thousandth of an inch. (Same as your answer... I chose not to round up to 3.7.)

lionelsoni

o  You assume that the load begins to fall as soon as the magnetic force stops.

Yes, that is what I assume. With no 282 to experiment with I don't know what really happens when the back side of the sine wave reaches zero.

 

I wonder what 282's are going for on eBay...?