Shopsmith Forums

There's been considerable discussion on the forum, on and off, about the current draw of Mark V's. Most recently today, over at this thread. As luck would have it, I had at home today the perfect equipment to add some solid data to these discussions. Specifically, a digital oscilloscope and a 100A current probe. Hopefully, the scope images will aid in the understanding of the no-load current draw, the associated power consumption, and the start-up current draw of of a conventional Mark V induction motor.

First up, we have a scope trace of my 2005 Mark V model 520 running at "SANDING DISC" speed, with a sanding disc installed but not being used. The yellow trace shows the line voltage, at 114V RMS measured. The current probe (blue trace) shows 7.48V RMS. So the power is 114V * 7.48A = 853W (1.14 hp), right?
Nope, that's wrong. The voltage and current waveforms are out of phase (by 63.4 degrees, according to the scope). You can multiply the instantaneous voltage by the instantaneous current, and indeed you will get the instantaneous power. But that same math simply does not work out when using RMS values -- those are time averages.

Next up, we have the same voltage and current waveforms, but the scope is set up to calculate the instantaneous power waveform as well. The vertical scale for power (red waveform) is 1000 W per division.
You can tell by eye that the average power is no more than 500W. In fact, the waveform is not sinusoidal, and the scope reports an accurate average power of 328.5W (0.44 hp). This number determines how much you pay the power company. Interestingly, you can see that the instantaneous power goes negative twice per AC cycle. This is due to the energy stored in the motor inductance, literally being pumped back out onto the power grid.

So where is that 0.44 hp of actual power going? Well, some of it is being turned into heat in the motor. But induction motors are quite efficient, so my educated guess is that the majority of it is going into mechanical transmission losses in the variable-speed belt transmission (where it is also turned into heat).

As an aside, if my Mark V is representative, this would point to a dramatic difference in useable power with a 1-1/8 hp motor, as opposed to the 3/4 hp motors on the old machines. On my machine, about half the power of a 3/4 hp motor would be used up just to spin the spindle under no load!

Sorry guys, I do not yet own a PowerPro, so I can't show you those waveforms. But they would be very similar to the CNC equipment that I design, and trust me, the waveforms look very different. However, the efficiency improvement over an induction motor will be largely due to it not having to drag around that variable-speed transmission.

While I had the scope and current probe set up, I took a few start-up current draw measurements as well. My machine is on a dedicated 20A branch circuit, about 50 ft from the main 200A load center. I also had about 6 feet of 14-gauge extension cord plugged in series, as part of my test harness.

To capture entire start-up waveforms, I changed the scope time scale to 200 ms per division. You can no longer see the details of any particular AC cycle, but what we're interested in now is the "envelope" of the waveforms.

The first shot shows the voltage and current waveforms when powering up the Mark V with a steel sanding disc installed, and the speed control set to "DISC SAND". You can see that when I turn on the power switch, the current immediately jumps from zero to 76A peak (54A RMS). This is the "locked rotor" current of the motor. It remains near that peak level for about 800 milliseconds, and then declines rapidly as the motor spins up to speed.
You can also see that when supplying that much current, my line voltage (at the machine) drops by about 20V peak (14V RMS).

Next, I repeated the same test, but with the speed control set all the way to "SLOW". This didn't affect the start-up current magnitude, but it did cut the peak-current duration to less than 400 milliseconds.
Then, I removed the sanding disk, and took another shot while still on "SLOW". The peak-current duration was reduced yet again, to about 200 milliseconds.
As a final test, still with no tooling on the spindle, I ran the test with the speed control set to "FAST". Now, the start-up current stayed above 70A peak (50A RMS) for almost 1.4 seconds.
So what I learned from this set of experiments is that the magnitude of the start-up current is always the same -- it's a function of the motor itself. But either adding inertia to the spindle, or cranking up the speed control, increases the duration of the peak start-up current.

Thank you very much for these very revealing screen shots and for the time and effort expended to provide them. These pretty much put an end to the debate about start up currents. The data shows vividly why the breaker might trip when starting up with a heavy mechanical load.

I would like to see my own data collected with this scope.

Nice experiments. I don't know much about circuit breakers but I would like to know why your 20A breaker didn't trip on that last experiment (50A for almost 1.4 seconds).

"Time"! If that load had remained for a bit longer, it would have tripped. Some breakers are 'more or less' instantaneous while others have an intentional time delay.

Here is one manufacturers detailed explanation.

https://www.carlingtech.com/sites/defau ... Delays.pdf

mountainbreeze wrote:Nice experiments. I don't know much about circuit breakers but I would like to know why your 20A breaker didn't trip on that last experiment (50A for almost 1.4 seconds).

Dusty got it right -- most household circuit breakers are of the "inverse time" variety, which means that the time to trip varies inversely with the magnitude of the over-current event. The primary purpose of the breaker is to prevent the branch wiring from overheating. Since it takes a while to heat up all that copper wiring, it's OK to allow large surges for short periods of time -- this prevents "nuisance tripping". You can do the same thing with fuses, by installing time-delay (a.k.a. "slow-blow") fuses. These are generally needed for motor circuits.

To answer your question specifically, I looked up the trip curve for the breaker that is installed on my Shopsmith circuit. It's on page 22 of this document (which also has a good operating overview on page 19). A 50A startup current is 2.5x the rated current of the breaker. So on the trip curve, you first find 2.5 on the horizontal axis ("Multiples of Rated Current"). Then follow that line up vertically, and you'll see that the trip time can be anywhere between about 4.5 seconds and 18 seconds. So with a 1.4 second start-up current, I actually had a safety factor of at least three, as far as tripping the breaker goes.

I didn't try spinning up a heavy inertia with the speed control on FAST, because that 1.4s surge was already pushing my own comfort zone. Just because it doesn't trip the breaker doesn't mean that such abuse wouldn't damage the Mark V motor or switch. Protecting an arbitrary device that is plugged into the circuit is not the function of a branch-circuit breaker.

This is a test to see if I can convert this to a URL.

dusty wrote:This is a test to see if I can http://www.shopsmith.com/ss_forum/woodw ... ml#p189439 to a URL.

http://www.shopsmith.com/ss_forum/viewt ... 45#p189445


http://www.shopsmith.com/ss_forum/viewt ... 39#p189439

?????

Obviously, it did not work. I also thought that I had deleted it very soon after I posted but that too did not work. Sorry for the unnecessary and non-productive interruption.

Holey Cow, Buckeye! I now know (- ahem - can look up)- more information about start up amps, power, circuit breakers etc. that I ever thought about before.

The last ocilloscope I used was in the Army back in 1954. Only green screens. Color is nifty.