Shopsmith Forums

RFGuy wrote: ↑Sun Dec 13, 2020 8:02 am

reible wrote: ↑Sat Dec 12, 2020 7:29 pm The blade listed is OK if it is still in like new condition. If it has been used a lot then I would worry. It needs to be sharp and clean.

A shopsmith has the advantage of being able to lower the speed and up the torque which is not common on most saws so while the HP seems lower it is really not that under powered. Just set to the lower range for ripping.

It is also critical that the machine have a nice clean waxed table for the wood to slide on, it must be aligned. You will also need some supports as both in feed and out feed.

Ed

Don't forget though that this "automatic transmission" created by the sheaves is just that, an automatic transmission so there is power loss associated with it compared to a direct drive. Does anyone know how efficient the Shopsmith design is, out of curiosity? Not exactly the same, but I think CVT transmissions in cars are only like 80% efficient if I recall correctly. So, this has to be taken into consideration as well when considering Shopsmith versus standard tablesaw horsepower.

I once measured my Mark V 520 drawing 0.44 hp of true electrical input power while spinning a sanding disk under no load. I have little doubt that the majority of that is transmission loss.

You can see the scope waveforms at the link below. Note that image ordering got scrambled during some server port or another, so you’ll need to manually match the images to their description in the text.

viewtopic.php?f=9&t=16474

BuckeyeDennis wrote: ↑Sun Dec 13, 2020 6:54 pm I once measured my Mark V 520 drawing 0.44 hp of true electrical input power while spinning a sanding disk under no load. I have little doubt that the majority of that is transmission loss.

You can see the scope waveforms at the link below. Note that image ordering got scrambled during some server port or another, so you’ll need to manually match the images to their description in the text.

viewtopic.php?f=9&t=16474

Dennis,

Thanks. Good work on measuring this in the past. That is an interesting thread. A little surprised that much power is being lost to the transmission. I have no experience with a 3HP of 5HP commercial tablesaw, but boy they must cut through hardwood like cutting through butter by comparison. Don't get me wrong, I have no complaints with my Mark V (non PowerPro). It cuts fine through 8/4 cherry with a sharp, clean blade. Just surprised it can do it if that much power is being lost in translation before reaching the blade...

RFGuy wrote: ↑Sun Dec 13, 2020 7:40 pm [It cuts fine through 8/4 cherry with a sharp, clean blade. Just surprised it can do it if that much power is being lost in translation before reaching the blade...

Well, you and a handsaw can probably cut through 8/4 cherry, too, and the average human can only generate about 0.1 hp continuously.

BuckeyeDennis wrote: ↑Sun Dec 13, 2020 6:54 pm I once measured my Mark V 520 drawing 0.44 hp of true electrical input power while spinning a sanding disk under no load. I have little doubt that the majority of that is transmission loss.

You can see the scope waveforms at the link below. Note that image ordering got scrambled during some server port or another, so you’ll need to manually match the images to their description in the text.

viewtopic.php?f=9&t=16474

I think I recall learning this about AC Amps Vs Power in my childhood but I'm struggling and will look for a refresher. What I can't remember at all is the math formula your scope is applying in the captures. I frequently use a clamp-on AC/DC Ammeter. So, for example, if at some speed with no load the conventional Headstock is about 8.2A and the PowerPro is 2.8A that comparison is not meaningful in terms of power being consumed. Is that correct? Does that phase relationship shown in the other thread (between voltage and amperage) change with motor load and/or RPM on a given motor?

- David

DLB wrote: ↑Mon Dec 14, 2020 9:26 am

BuckeyeDennis wrote: ↑Sun Dec 13, 2020 6:54 pm I once measured my Mark V 520 drawing 0.44 hp of true electrical input power while spinning a sanding disk under no load. I have little doubt that the majority of that is transmission loss.

You can see the scope waveforms at the link below. Note that image ordering got scrambled during some server port or another, so you’ll need to manually match the images to their description in the text.

viewtopic.php?f=9&t=16474

I think I recall learning this about AC Amps Vs Power in my childhood but I'm struggling and will look for a refresher. What I can't remember at all is the math formula your scope is applying in the captures. I frequently use a clamp-on AC/DC Ammeter. So, for example, if at some speed with no load the conventional Headstock is about 8.2A and the PowerPro is 2.8A that comparison is not meaningful in terms of power being consumed. Is that correct? Does that phase relationship shown in the other thread (between voltage and amperage) change with motor load and/or RPM on a given motor?

- David

That phase relationship exists with any reactive(non resistive) load. It results in a variable called power factor.

Thus P NOT equal to IE, but is equal to IE x (power factor). A resistive load has a power factor of 1.0.

I doubt that was included in thy childhood exposure to electricity.

P = power
I = amperes
E = voltage
power factor = cos(phase shift)
phase shift can be + or minus(current leads or current lags)
Capacitive loads current leads voltage
Inductive loads current lags voltage

Power factor ranges between 0 and 1

P.S. throw in a switching power supply and all this goes out the window. Both the above and that clamp on ammeter assumes a non varying load. A 'switching' power supply ain't non varying!!!

JPG wrote: ↑Mon Dec 14, 2020 9:40 am I doubt that was included in thy childhood exposure to electricity.

You are right, I exaggerated. Military tech school, so I was ~19. The only part I remembered was 'ELI the ICE man.' I didn't ask my question very well, what I was wondering was if the magnitude of the phase angle between voltage and current changed in response to load or RPM. I've learned that it does, Power Factor on a 'typical' induction motor may range from less than 0.1 to more than 0.8, increasing with load from 0 to 100%.

BuckeyeDennis wrote: ↑Sun Dec 13, 2020 6:54 pm

RFGuy wrote: ↑Sun Dec 13, 2020 8:02 am Don't forget though that this "automatic transmission" created by the sheaves is just that, an automatic transmission so there is power loss associated with it compared to a direct drive. Does anyone know how efficient the Shopsmith design is, out of curiosity? Not exactly the same, but I think CVT transmissions in cars are only like 80% efficient if I recall correctly. So, this has to be taken into consideration as well when considering Shopsmith versus standard tablesaw horsepower.

I once measured my Mark V 520 drawing 0.44 hp of true electrical input power while spinning a sanding disk under no load. I have little doubt that the majority of that is transmission loss.

You can see the scope waveforms at the link below. Note that image ordering got scrambled during some server port or another, so you’ll need to manually match the images to their description in the text.

viewtopic.php?f=9&t=16474

I was pretty curious about this because I had seen before that the power consumed in the headstock drive does not change very much until the speed gets up around saw speed. Call this empirical data, I measured amps and am not set up to capture all parameters. I tested with nothing on the spindle. The motor pulled 7.05 Amps with the belt off, 7.12 Amps with the belt on at speed A. Current increase was very gradual, all the way up to R before it hit 8.13 Amps, then increases were progressively faster up to 11.44 Amps at W. I would conclude that there is almost no loss in the CVT function or anywhere else in the headstock at low speed but starting at saw speeds we are seeing some power consumption in belts and bearings. I hate to disagree with Dennis, but it appears to me that the overwhelming majority of the 0.44 HP electrical input power he measured was converted to heat by the motor.

- David

I think that Dennis would agree that the transmission loss was due to friction that resulted in heat as well as heat created by losses in the motor itself.

That energy HAD to go somewhere!!

Run it on too small of awg and its all out the window !! The number of amps running through a circuit describes the quantity of charge that runs through it each second. A couple other factors the amount of energy that this transfers. The voltage of the circuit specifies the amount of energy that each unit of charge carries. The amount of time for which the circuit runs converts this energy transfer rate to a quantity of energy.

Multiply the number of amps that go through the circuit by the voltage across it. If, for example, 12 amps go through a 240 volt circuit: 12 × 240 = 2,880. This is the power going through the circuit, measured in Watts.

Multiply the power rating by the time for which the circuit runs. If, for instance, it runs for 20 seconds: 2,880 × 20 = 57,600. This is the energy that the circuit transfers, measured in joules. Divide this answer by 1,055, which is the number of joules in a British Thermal Unit (BTU): 57,600 ÷ 1,055 = 54.6. This is the number of BTUs that the circuit carries. https://sciencing.com/how-8498688-conve ... -btus.html

DLB wrote: ↑Tue Dec 15, 2020 4:58 pm I hate to disagree with Dennis, but it appears to me that the overwhelming majority of the 0.44 HP electrical input power he measured was converted to heat by the motor.

- David

No worries about disagreeing with me, David! I merely “suspected” that most of the 0.44 hp losses were in the transmission. Sorry for the delay in responding, but it’s been an unusually busy week.

I don’t think that either of our experiments are conclusive regarding where the no-load losses are happening. I had the equipment on hand (at the time) to measure true input power. But I didn’t test the motor unbelted, so I could only measure total losses. You did measure the motor unbelted, but did not have the requisite equipment to measure true input power.

The graph below shows why a current measurement alone doesn’t tell much about no-load motor input power. The electrical power factor is small, meaning that the motor current is roughly 90 degrees out of phase with the applied voltage, and thus the true electrical power is small. As the load increases, modestly, the power increases primarily due to the power factor increasing. The motor current doesn’t change very much, as you verified in your experiments.

Sources of power loss in such a system include resistive losses in the motor windings, magnetic hysteresis losses in the motor iron, motor-bearing losses (minimal), and motor cooling-fan losses (significant) before we even get to the motor output shaft.

So why did I suspect that most of the power losses are in the transmission, rather than in the motor itself? It’s not very scientific, actually. In my experience, an induction motor running with nothing attached to the output shaft doesn’t get very warm. But an unloaded Shopsmith headstock does get a bit warm, even though it has considerably more surface area to dissipate heat.

But now that you’ve made me think about it, that could be largely due to better heat dissipation of the motor cooling fan in free air. And the graph says that the motor efficiency goes all the way down to diddly-squat at no load. So now I don’t think we have enough data to say, one way or the other.

I would probably cut something big like that with my homemade track saw- worm drive skilsaw with a track made from 1/4 luan plywood and a metal straightedge screwed together, c-clamped in place and set the whole thing on the ground or my old folding work table, resting the wood I'm cutting on scrap wood strips so I don't cut into the table or floor. I hate wrestling large flat stock on a table saw of any sort!