Hello!! And Help!!!

[Ed84]

As the other Ed just pointed out, the power equation P=IV shows the current increase directly. I should have grabbed that one. Been out of the classroom too long.

A variac is essentially an adjustable transformer, so I think Ohms law would have to be written differently to account for there being two separate current loops (primary and secondary), with different voltages, different coil resistances, and different currents. Don't the same work with half the voltage takes twice the current.





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[dusty]

Wrong equation. Use one for power, P=V*A so if you want the same power you need to do what if the voltage drops? ie if P=1000 and V=100 then A= (P/A)=10. Now lets lower the voltage to 50, (solving for new current) A=P/V = 20, or a current increase. This of course doesn't quite work this way but does show that if the voltage is lowered the current will have to go up to keep the same power level.

Ed

But if the voltage goes down, why would you think the power remains the same. The need for power may remain the same but the power delivered does not. That is the issue. It goes down with a decrease in voltage ie: P = IE. The voltage goes down and the motor (attempting to do the same level of work) bogs down for lack of power.

[reible]

"Commutated electric motors, such as universal motors, will run at reduced speed or reduced torque. Depending on the motor design, no harm may occur. However, under load, the motor may draw more current due to the reduced back-EMF developed at the lower armature speed. Unless the motor has ample cooling capacity, it may eventually overheat and burn out."

However when you get to induction motors:

"An induction motor will draw more current to compensate for the decreased voltage, which may lead to overheating and burnout."

Since the discussion here is on motors like the ones the older shopsmiths use this applies.

It is just like when your equipment can run on either 120 or 220 volts, the 220 takes have the current then the 120 does. We have talked about many times without having to go into details but it is the same power equation that applies.

Ed

Wrong equation. Use one for power, P=V*A so if you want the same power you need to do what if the voltage drops? ie if P=1000 and V=100 then A= (P/A)=10. Now lets lower the voltage to 50, (solving for new current) A=P/V = 20, or a current increase. This of course doesn't quite work this way but does show that if the voltage is lowered the current will have to go up to keep the same power level.

Ed

But if the voltage goes down, why would you think the power remains the same. The need for power may remain the same but the power delivered does not. That is the issue. It goes down with a decrease in voltage ie: P = IE. The voltage goes down and the motor (attempting to do the same level of work) bogs down for lack of power.

[InspireVeterans]

I fully understand. Do you know what the operating rpm of a shopsmith motor is? I tend to run into motors here and there. Possibly I could swap the original for a lower rpm motor. Again my main useage for this will be turning so having an overall reduced speed will not be a bad thing for me.

[BuckeyeDennis]

Wrong equation. Use one for power, P=V*A so if you want the same power you need to do what if the voltage drops? ie if P=1000 and V=100 then A= (P/A)=10. Now lets lower the voltage to 50, (solving for new current) A=P/V = 20, or a current increase. This of course doesn't quite work this way but does show that if the voltage is lowered the current will have to go up to keep the same power level.

Ed

But if the voltage goes down, why would you think the power remains the same. The need for power may remain the same but the power delivered does not. That is the issue. It goes down with a decrease in voltage ie: P = IE. The voltage goes down and the motor (attempting to do the same level of work) bogs down for lack of power.

Dusty, the load demand (i.e the motor output power) is indeed the primary determiner of motor input (i.e. electrical) power. Practical motors are designed to be quite "stiff", meaning that their speed doesn't drop a lot as the torque demand increases. They do indeed draw more current as required to deliver higher power demands. But if the power demand is excessive, and the induction motor speed drops too far, it will stall. And a huge winding current will be drawn as a result.

Ohm's law, as you wrote it, applies only to resistive circuits (although it can be generalized for use with reactive circuits by using complex variables in place of real numbers). We're talking about reactive circuit in this case. For a "simplified" equivalent circuit of an induction motor, check out this article: http://myelectrical.com/notes/entryid/2 ... nt-circuit. And that one is for a 3-phase motor -- single-phase motors are more complex yet. Suffice it to say that modeling a spinning induction motor as a fixed resistor is nowhere close to reality.

But back to the original question, running a single-phase induction motor at a substantially reduced voltage poses yet another hazard. Depending on the design, the start relay may never disconnect the start windings. This can release the "magic smoke" PDQ.

[BuckeyeDennis]

I fully understand. Do you know what the operating rpm of a shopsmith motor is? I tend to run into motors here and there. Possibly I could swap the original for a lower rpm motor. Again my main useage for this will be turning so having an overall reduced speed will not be a bad thing for me.

You'd have a lot of options along that line if you had bought an older Shopsmith 10ER. But I don't believe that standard-frame motors can be fit into the Mark 5 headstock, as a practical matter.

But I do seem to recall someone rigging up an outboard belt-drive reducer. Inside the headstock, there's a fixed-ratio belt drive between the lower "auxilary" shaft and the spindle. If you're ambitious, should be possible to remove the belt from that fixed internal drive, and then install a reducing drive outboard of the headstock. Some of the guys here are deep experts on the headstock, and they can tell you if I'm crazy or not.

[everettdavis]

Apart from how to build the watch discussion above, you asked if you could swap the motor for a lower RPM. The Shopsmith drive mechanism uses a special extended length shaft and a mounting not present on other motors.

There is a somewhat expensive upgrade to the motor / drive system that would give you the slow speeds, and the torque through all the speed ranges but likely not in your budget, though an amazing upgrade.

556177 Do-It-Yourself PowerPro Upgrade Kit ... $1,689.00

Look at the the Shopsmith Speed Reducer which is more approachable and seems in line with what you want to do in turning.

555428 Shopsmith Speed Reducer $372.53

Occasionally, they come up on eBay but condition is unknown and warranty likely expired.

Everett

Shopsmith Speed Reducer.png (421.05 KiB) Viewed 7560 times

PP Upgrade.png (344.2 KiB) Viewed 7560 times

[JPG]

Ohm's law applies to resistive circuits.

An induction motor is a different animal.

FWIW, a dc/universal motor does not either.

[dusty]

Ohm's law applies to resistive circuits.

An induction motor is a different animal.

FWIW, a dc/universal motor does not either.

Ohm's law applies to both ac and dc circuits. If we really want to get into this and I don't think we do, I'll dig out the lessons plans I used when I taught "The Principles of Reactive Circuits" in technical school.

I = E/Z where Z equals the effective resistance to current flow in an RCL circuit. The major difference, during theoretical discussions, is that in DC circuits the voltage and current are in phase. In an AC circuit they are not and this phase differential makes for a significantly different understanding. Does the voltage lead or lag the current. ELI the ICE man. Voltage leads the current in a purely inductive circuit (ELI) and voltage lags the current (ICE) in a purely capacitive circuit. Phase relationships in an RCL circuit are a study in themselves but ohm's law still applies. One must simply keep in mind that in AC circuits everything is a matter of "an instant in time" where all things constantly change.

Thus the term "Effective Power".

After the discussion on "Behavior of an RCL Circuit", we will have to delve into "AC Motor Theory"; a class of its own where all of the previously learned facts are altered by reality. All matters are different in the "Real World".

[BuckeyeDennis]

I fully understand. Do you know what the operating rpm of a shopsmith motor is? I tend to run into motors here and there. Possibly I could swap the original for a lower rpm motor. Again my main useage for this will be turning so having an overall reduced speed will not be a bad thing for me.

You'd have a lot of options along that line if you had bought an older Shopsmith 10ER. But I don't believe that standard-frame motors can be fit into the Mark 5 headstock, as a practical matter.

But I do seem to recall someone rigging up an outboard belt-drive reducer. Inside the headstock, there's a fixed-ratio belt drive between the lower "auxilary" shaft and the spindle. If you're ambitious, should be possible to remove the belt from that fixed internal drive, and then install a reducing drive outboard of the headstock. Some of the guys here are deep experts on the headstock, and they can tell you if I'm crazy or not.

Check out the following link for an example of a similar home-brew external speed reducer, and be sure to watch the video linked to in the first post. It’s not quite the same, but you’ll get the idea.

http://www.shopsmith.com/ss_forum/woodw ... 17779.html