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doctordoctor
10-08-2014, 03:42 PM
So I have a 10" Atlas

I also have a Baldor 1/2 horsepower DC motor, 1750rpm, and a Sabina DC motor drive.

So combining the two, I could have a variable speed DC motor running the Atlas instead of its current AC motor and pulley system.

I would think this would be a lot smoother and alot more convenient for spindle RPM control.

Anyone done anything like this? Is this too easy to be true?

http://i242.photobucket.com/albums/ff197/acannell/20141008_123831_zps7f3a9jhf.jpg (http://s242.photobucket.com/user/acannell/media/20141008_123831_zps7f3a9jhf.jpg.html)

http://i242.photobucket.com/albums/ff197/acannell/20141008_123837_zpsafo2ary9.jpg (http://s242.photobucket.com/user/acannell/media/20141008_123837_zpsafo2ary9.jpg.html)

http://i242.photobucket.com/albums/ff197/acannell/20141008_123858_zpsuquikqxv.jpg (http://s242.photobucket.com/user/acannell/media/20141008_123858_zpsuquikqxv.jpg.html)

http://i242.photobucket.com/albums/ff197/acannell/20141008_123906_zpsrs80tdot.jpg (http://s242.photobucket.com/user/acannell/media/20141008_123906_zpsrs80tdot.jpg.html)

doctordoctor
10-08-2014, 03:42 PM
http://i242.photobucket.com/albums/ff197/acannell/20141007_112836_zpsxic6dngm.jpg (http://s242.photobucket.com/user/acannell/media/20141007_112836_zpsxic6dngm.jpg.html)

Gary Paine
10-08-2014, 04:24 PM
Yes and Yes. Several of my lathes sport just such a motor and I am very happy with them. You will still need the pulley system, though, to provide the necessary power at low RPM as when you are turning large diameter workpieces. The motor horsepower will drop with speed as in the formula: Horsepower = Torque*RPM/5252. Smaller diameter work can often be accomplished with the belts in high range and just adjustment of speed control.
No affiliation or dealer preference, but I bought one of these: http://www.ebay.com/itm/Digital-Laser-Photo-Tachometer-Non-Contact-RPM-Tach-Meter-Motor-Speed-Gauge-USA-/331075626354?pt=LH_DefaultDomain_0&hash=item4d15a52572 and I find it quick and easy to know just what speed I am going.

MaxHeadRoom
10-08-2014, 04:45 PM
If you want stop start, (blank space on the drive) KB/Baldor have a special switch that goes from FWD to REV but stops at the centre to stop you from going from one to the other in one motion, it also has a braking resistor attached.
If that one is not PWM, but simple SCR control, you may want to pick one up at a later date for superior operation, especially at low RPM.
As to the Tach off ebay, I suggest you remove the batteries between operations if not using it for a while as the battery runs down fairly fast even thought it is 'Off'.
BTW, most motors, DC included are rated maximum torque at zero rpm.
Max.

doctordoctor
10-08-2014, 04:45 PM
Yes and Yes. Several of my lathes sport just such a motor and I am very happy with them. You will still need the pulley system, though, to provide the necessary power at low RPM as when you are turning large diameter workpieces. The motor horsepower will drop with speed as in the formula: Horsepower = Torque*RPM/5252. Smaller diameter work can often be accomplished with the belts in high range and just adjustment of speed control.
No affiliation or dealer preference, but I bought one of these: http://www.ebay.com/itm/Digital-Laser-Photo-Tachometer-Non-Contact-RPM-Tach-Meter-Motor-Speed-Gauge-USA-/331075626354?pt=LH_DefaultDomain_0&hash=item4d15a52572 and I find it quick and easy to know just what speed I am going.

I like that tach. Although I have so much electronics junk if I dont whip up a tack then basically I'm just a hoarder lol

Plus I think I want something that is always on and powered by and attached to the lathe so I can always observe speed.

Do you have a picture of your motor setups?

To be clear, no transmission can increase the power or maintain the power level available from the motor. All it can do is reduce it slightly via friction.

The existing pulley system doesn't increase the power at lower rpms, it just makes the lower rpms possible, and provides you with an increase in torque proportional to the drop in rpm. The product and therefore the power remains the same, ignoring friction losses.

Take two situations, one 5" dia workpiece, and one 1" dia. Same material, same SFM, same cut/depth of cut. You change rpm to keep the SFM the same since the surface of the larger diameter moves faster for a given rpm than a smaller one, since its circumference is 5 times larger.

So the cutting force on the tool remains the same. Lets say its 30lbs.

At 5" 30lbs is 12.5 foot lbs.

At 1" 30lbs is 2.5 foot lbs.

And you increased the rpm by a factor of 5 from 5" to 1" to maintain a constant SFM.

For discussions sake, lets say you get your target SFM at 5" at 500rpm.

So using your forumula:

For the 5": 12.5 * ( 500/5252) = 1.19 hp

For the 1": 2.5 * (2500/5252) = 1.19 hp

Power remains the same. So your transmission just makes it so you CAN have a variable rpm, since the motor rpm is fixed on these AC style motors. You end up with the same available power whatever you set the pulleys to.

I'm not sure what my DC motors torque vs rpm curve is..I may still need a countershaft system if it has a drop in torque at either end, or depending on how the speed control affects torque.

JRouche
10-08-2014, 04:54 PM
I'm not sure what my DC motors torque vs rpm curve is..I may still need a countershaft system if it has a drop in torque at either end, or depending on how the speed control affects torque.

Prolly be dropping tq at the higher rpm... JR

Gary Paine
10-08-2014, 05:39 PM
I like that tach. Although I have so much electronics junk if I dont whip up a tack then basically I'm just a hoarder lol

Plus I think I want something that is always on and powered by and attached to the lathe so I can always observe speed.

Do you have a picture of your motor setups?

To be clear, no transmission can increase the power or maintain the power level available from the motor. All it can do is reduce it slightly via friction.

The existing pulley system doesn't increase the power at lower rpms, it just makes the lower rpms possible, and provides you with an increase in torque proportional to the drop in rpm. The product and therefore the power remains the same, ignoring friction losses.

Take two situations, one 5" dia workpiece, and one 1" dia. Same material, same SFM, same cut/depth of cut. You change rpm to keep the SFM the same since the surface of the larger diameter moves faster for a given rpm than a smaller one, since its circumference is 5 times larger.

So the cutting force on the tool remains the same. Lets say its 30lbs.

At 5" 30lbs is 12.5 foot lbs.

At 1" 30lbs is 2.5 foot lbs.

And you increased the rpm by a factor of 5 from 5" to 1" to maintain a constant SFM.

For discussions sake, lets say you get your target SFM at 5" at 500rpm.

So using your forumula:

For the 5": 12.5 * ( 500/5252) = 1.19 hp

For the 1": 2.5 * (2500/5252) = 1.19 hp

Power remains the same. So your transmission just makes it so you CAN have a variable rpm, since the motor rpm is fixed on these AC style motors. You end up with the same available power whatever you set the pulleys to.

I'm not sure what my DC motors torque vs rpm curve is..I may still need a countershaft system if it has a drop in torque at either end, or depending on how the speed control affects torque.

Well, as I read this I thought you had it, but I guess not. A DC motor will produce rated power at the rated RPM only. As the speed is reduced, even with the modern controllers, horsepower will decrease LINEARLY with RPM. At zero rpm, even with the higher stall torque, there is zero horsepower. At half rpm, there is half horsepower.
You will definitely need the mechanical gear reduction. I maintained my entire countershaft assembly. That way, just as with an AC motor, you can run the motor at higher rpm, and therefore more power, with the spindle going slow enough for the job at hand.

PStechPaul
10-08-2014, 05:41 PM
It appears that your motor might be a series wound type, as the nameplate seems to show only a voltage specification for the armature. This will make it difficult to control the speed, and impossible to reverse electrically. A series DC motor is capable of lots of torque and thus is often used for forklifts and EVs where the operator uses an accelerator to provide torque as needed to attain and hold a desired speed. But it is the wrong motor for a machine tool.

A shunt wound motor, with a separate field as seems to be the design of the controller, can provide better speed control and electrical reversing capability and dynamic braking. As with most brushed motors with field and armature coils, very high torque can be obtained, up to 5x or 10x for very short times, based on I^2*R heating and current capacity of the brushes and commutator.

A permanent magnet brushed DC motor can be controlled by means of PWM and a shaft position/speed transducer can be used by a controller to hold and change speed quickly and accurately. But they may be more expensive, and generally must not be heavily overloaded or overheated because of possible demagnetization.

A BLDC is possibly even better because it does not use brushes and is essentially a stepper motor which is driven by pulses that accurately determine the position and speed of the shaft. But they tend to be very expensive and fragile.

A three phase AC induction motor is the least expensive for the power rating, and is very rugged and nearly maintenance-free. A good VFD can provide good control of torque and speed over a wide range, and 2-3 times the rated torque can be obtained. It can also be overclocked to at least twice the rated speed, with a corresponding reduction of torque, or with a higher voltage source, torque can remain constant and power can be more than doubled. A well-designed three phase motor and VFD can exhibit very low torque ripple compared to single phase AC motors.

doctordoctor
10-08-2014, 05:52 PM
It appears that your motor might be a series wound type, as the nameplate seems to show only a voltage specification for the armature. This will make it difficult to control the speed, and impossible to reverse electrically. A series DC motor is capable of lots of torque and thus is often used for forklifts and EVs where the operator uses an accelerator to provide torque as needed to attain and hold a desired speed. But it is the wrong motor for a machine tool.

A shunt wound motor, with a separate field as seems to be the design of the controller, can provide better speed control and electrical reversing capability and dynamic braking. As with most brushed motors with field and armature coils, very high torque can be obtained, up to 5x or 10x for very short times, based on I^2*R heating and current capacity of the brushes and commutator.

A permanent magnet brushed DC motor can be controlled by means of PWM and a shaft position/speed transducer can be used by a controller to hold and change speed quickly and accurately. But they may be more expensive, and generally must not be heavily overloaded or overheated because of possible demagnetization.

A BLDC is possibly even better because it does not use brushes and is essentially a stepper motor which is driven by pulses that accurately determine the position and speed of the shaft. But they tend to be very expensive and fragile.

A three phase AC induction motor is the least expensive for the power rating, and is very rugged and nearly maintenance-free. A good VFD can provide good control of torque and speed over a wide range, and 2-3 times the rated torque can be obtained. It can also be overclocked to at least twice the rated speed, with a corresponding reduction of torque, or with a higher voltage source, torque can remain constant and power can be more than doubled. A well-designed three phase motor and VFD can exhibit very low torque ripple compared to single phase AC motors.

Thanks for the info. Are you sure? It has a label on it that shows the wires can be reversed to reverse rotation. Is there some way I can confirm what type it is?

A bigger problem might be that its "recessed shaft" whatever that means..I need to pull that hydraulic pump head off and see whats what. Maybe I'd have to make an adapter.

It would, of course, be awesome to have a BLDC or VFD type setup but I cant justify spending money like that for it.

http://i242.photobucket.com/albums/ff197/acannell/20141008_144825_zpsjdwjuda6.jpg (http://s242.photobucket.com/user/acannell/media/20141008_144825_zpsjdwjuda6.jpg.html)

Tony Ennis
10-08-2014, 05:53 PM
I have a 1/2 hp 3 phase and a VFD on mine. I like it. A 1 hp VFD will set you back $100. The motor could be free off Craiglist. Mine was :-D But the electrician was not. I recommend you get your lathe working first, then worry about details.

MaxHeadRoom
10-08-2014, 05:53 PM
I doubt it is a shunt wound field motor, No1 it would usually have four connections and it is easy to tell, just short the two motor/armature connections and try to spin the motor, if it resists a great deal more that without the connections it is a P.M. motor.
Max.

Forrest Addy
10-08-2014, 06:24 PM
The motor is a PM or permanany magnet motor. It needs no field supply as would a shunt wound DC motor. The tip is the motor ius reversible by exchanging the armature polarity. Se series would motor require access to the series foreld and the armature to reverse.

I'm sure you will be happy with the 1/2 DC drive once you learn the few tricks needed to properly exploit its advantages ... yes there is a "but": will 1/2 HP be enough? When I was a kind running my dad's 3/4 HP South Bend 10 it seemed I was always having to baby the motor when I had stock to remove. After I started my apprenticeship I scrounged up a 3/4 HP single phase motor and that doubled the (duhh!) cutting capacity.

Gary Paine
10-08-2014, 07:23 PM
Do you have a picture of your motor setups?


Well, yes. But they aren't very interesting. This is my 12x54. This one shows the countershaft assembly with the covers open:
http://i1117.photobucket.com/albums/k595/gppaine/4336_zps84e3a25d.jpg (http://s1117.photobucket.com/user/gppaine/media/4336_zps84e3a25d.jpg.html)

This shows the GE 1/2 horsepower shunt wound DC motor that I'm very happy with:
http://i1117.photobucket.com/albums/k595/gppaine/4337_zps92af4c04.jpg (http://s1117.photobucket.com/user/gppaine/media/4337_zps92af4c04.jpg.html)

I used a KBIC controller board with a big stop button and a start button with a light in it that tells if the main power is on. I made provision for an E-stop, but am not using it. I mounted the speed control rheostat in the headstock where the old on off switch used to be.
http://i1117.photobucket.com/albums/k595/gppaine/4326_zpse32e28fd.jpg (http://s1117.photobucket.com/user/gppaine/media/4326_zpse32e28fd.jpg.html)

NOW, if you want an interesting photo, how about one that compares the Atlas 12 inch to the sweet little Atlas 6 inch?
http://i1117.photobucket.com/albums/k595/gppaine/4338_zps5f54a038.jpg (http://s1117.photobucket.com/user/gppaine/media/4338_zps5f54a038.jpg.html)

doctordoctor
10-08-2014, 07:30 PM
I used a KBIC controller board with a big stop button and a start button with a light in it that tells if the main power is on. I made provision for an E-stop, but am not using it. I mounted the speed control rheostat in the headstock where the old on off switch used to be.


LOL so its basically just like the sherline! They have a KBIC controller in there too.

And thats how I want mine to be too..DC shunt motor with a speed control.

So how does your torque vary with rpm? You and I have the same type of motor I take it?

Gary Paine
10-08-2014, 07:43 PM
LOL so its basically just like the sherline! They have a KBIC controller in there too.

And thats how I want mine to be too..DC shunt motor with a speed control.

So how does your torque vary with rpm? You and I have the same type of motor I take it?

Gosh, Doc. You need to reread my previous posts. Horsepower varies linearly with speed. Torque is relatively constant at the motor through the range (Ok, stall current...). Torque available to the cutting bit is a direct function of the gearing.

doctordoctor
10-08-2014, 08:07 PM
Gosh, Doc. You need to reread my previous posts. Horsepower varies linearly with speed. Torque is relatively constant at the motor through the range (Ok, stall current...). Torque available to the cutting bit is a direct function of the gearing.

Lets say you have a motor, and it can maintain a constant torque regardless of RPM. Thats pretty close to the real world DC brushed motor we are talking about.

Now lets say you want to use that motor on a lathe, and you have two situations, one is turning a 5" diameter piece, the other, a 1" diameter piece.

The materials, cutters, depth of cut, feed rate, and desired SFM are the same between them.

Therefore the cutting force on the tool is the same.

But the torque on the motor is different, because that cutting force is projected out at 5" in one case and 1" in the other. So its 5 times as much torque for the 5" piece.

But the rpm will be 5 times slower to maintain a constant SFM, in the 5" piece.

So power remains the same between the 5" piece and the 1" piece. Because power is the product between the rpm and the torque.

So with a motor that has a constant torque throughout its rpm range, no pulley system is required.

There is no advantage to having a variable ratio pulley system unless the motor has higher power at a specific rpm, or its rpm cant be controlled. But for a constant torque vs rpm motor like a DC shunt wound motor, thats not the case. Not significantly anyway.

Did you see my previous post on this?

http://bbs.homeshopmachinist.net/threads/64623-Atlas-10-quot-DC-motor-speed-control-upgrade?p=941302#post941302

http://www.reliance.com/mtr/images/mtdfig19a.gif

velocette
10-08-2014, 09:33 PM
Hi
Lots of info on this posting most of it is spot on some a little off you can find the correct information for your Baldor motor at

http://www.baldor.com/products/product.asp?1=1&product=DC+Motors&family=All+Families|vw_DCMotors

I am not familiar with the controller you have. IF there is provision for a Tachometer - Generator then check what the specs are and use it if possible

this will give better speed control under load.

One tip I keep harping on about is to fit a 20 - 0 - 20 Amp Meter to the motor leads on the DC motor to check loading when cutting.
Retain the variable ratio drive to retain flexible choice for the torque required.
Set the maximum amps for the control at not more than 1.5 times the motor rated amps (Done with the rotor locked and amp meter on the motor leads)
Be very aware that this is NOT to be attempted unless you FULLY Understand what you are doing BE SAFE.

Dedicated DC Nut

Eric

doctordoctor
10-08-2014, 09:42 PM
Hi
Lots of info on this posting most of it is spot on some a little off you can find the correct information for your Baldor motor at

http://www.baldor.com/products/product.asp?1=1&product=DC+Motors&family=All+Families|vw_DCMotors

I am not familiar with the controller you have. IF there is provision for a Tachometer - Generator then check what the specs are and use it if possible

this will give better speed control under load.

One tip I keep harping on about is to fit a 20 - 0 - 20 Amp Meter to the motor leads on the DC motor to check loading when cutting.
Retain the variable ratio drive to retain flexible choice for the torque required.
Set the maximum amps for the control at not more than the motor rated amps (Done with the rotor locked and amp meter on the motor leads)
Be very aware that this is NOT to be attempted unless you FULLY Understand what you are doing BE SAFE.

Dedicated DC Nut

Eric

OOH I like that ammeter idea..definitely going to do that...even if I keep the AC motor that would be a great way to measure cutting forces

flylo
10-08-2014, 10:21 PM
I've used ConSew industrial sewing machine motors on Atlas & other small lathes. They are DC 3/4 hp VS reversible with brakes for about $100. I didn't look for the cheapest but here's a sample eBay item number:131312271588

flylo
10-08-2014, 10:33 PM
eBay item number:121449324516 Here's one I haven't tried with digital readout, hmm?

Gary Paine
10-08-2014, 10:45 PM
I'll insert my responses next to your statements:
Lets say you have a motor, and it can maintain a constant torque regardless of RPM. Thats pretty close to the real world DC brushed motor we are talking about. OK

Now lets say you want to use that motor on a lathe, and you have two situations, one is turning a 5" diameter piece, the other, a 1" diameter piece. OK

The materials, cutters, depth of cut, feed rate, and desired SFM are the same between them.

Therefore the cutting force on the tool is the same. OK

But the torque on the motor is different, because that cutting force is projected out at 5" in one case and 1" in the other. So its 5 times as much torque for the 5" piece. OK, BUT ONLY WITHIN THE RANGE OF AVAILABLE TORQUE FROM THE MOTOR

But the rpm will be 5 times slower to maintain a constant SFM, in the 5" piece. OK

So power remains the same between the 5" piece and the 1" piece. Because power is the product between the rpm and the torque. NO, YOU JUST SAID RPM IS SLOWER TO MAINTAIN DESIRED SFM AND THE TORQUE SUPPLIED BY THE MOTOR IS 5 TIMES AS MUCH. THE REQUIRED HORSEPOWER IS 5 TIMES LARGER ON THE 5" PIECE. YOUR 1/2 HP MOTOR CAN SUPPLY 16500 FT-LB/MINUTE. THE MAXIMUM TORQUE SHOULD BE ABOUT 125 FT-LB. AT FULL MOTOR SPEED

So with a motor that has a constant torque throughout its rpm range, no pulley system is required. NICE TRY, BUT YOU ARE CONFUSING AVAILABLE MOTOR TORQUE WITH CUTTING TORQUE REQUIRED. TRY THINKING OF IT IN TERMS OF FORCE. YOUR MOTOR PULLEY HAS A FIXED RADIUS AND CAN PROVIDE A PULL ON THE V-BELT IN POUNDS EQUAL TO 125 DIVIDED BY THE MOTOR PULLEY RADIUS IN FEET. IF THERE ARE NO OTHER PULLEYS THAN THE DRIVEN IN THE SYSTEM, THE AVAILABLE SPINDLE TORQUE IS THE SAME, IGNORING LOSSES. THE BELT TENSION CANNOT EXCEED THE 125/R. NOW IMAGINE DOUBLING THE DIAMETER OF THE DRIVEN PULLEY. WITH THE SAME BELT TENSION (FORCE), THE TORQUE IS DOUBLED BECAUSE THE RADIUS AT WHICH IT IS APPLYING THE FORCE IS DOUBLED.

There is no advantage to having a variable ratio pulley system unless the motor has higher power at a specific rpm, or its rpm cant be controlled. But for a constant torque vs rpm motor like a DC shunt wound motor, thats not the case. Not significantly anyway. DISAGREE

J Tiers
10-08-2014, 10:59 PM
Lets say you have a motor, and it can maintain a constant torque regardless of RPM. Thats pretty close to the real world DC brushed motor we are talking about.

Now lets say you want to use that motor on a lathe, and you have two situations, one is turning a 5" diameter piece, the other, a 1" diameter piece.

The materials, cutters, depth of cut, feed rate, and desired SFM are the same between them.

Therefore the cutting force on the tool is the same.

But the torque on the motor is different, because that cutting force is projected out at 5" in one case and 1" in the other. So its 5 times as much torque for the 5" piece.

But the rpm will be 5 times slower to maintain a constant SFM, in the 5" piece.

So power remains the same between the 5" piece and the 1" piece. Because power is the product between the rpm and the torque.

So with a motor that has a constant torque throughout its rpm range, no pulley system is required.

There is no advantage to having a variable ratio pulley system unless the motor has higher power at a specific rpm, or its rpm cant be controlled. But for a constant torque vs rpm motor like a DC shunt wound motor, thats not the case. Not significantly anyway.


Don't get too fond of that idea.....

Material removed per unit time tends to go as power..... not torque.

Power goes as torque x rpm (rpm being distance per unit time, and torque being force).

If your motor can produce a given torque, and has a given base speed, it produces half power at half speed. So it produces 1/5 power at 1/5 speed.

Comparing that motor straight, with that same motor through pulley reduction, at a speed of 1/5 of base speed, you get 1/5 power with the straight coupling. But with a 5:1 pulley reduction, you get 5x torque at the same output shaft speed. So with the pulley you can still get full power, but with the straight coupling you are limited to 1/5 power.

That can be the difference between being forced to whittle off tiny threads of material, vs being able to take a reasonable cut. I have a 1/4 HP 1725 rpm motor on my mill. At the 35 RPM back gear speed, it happily runs a 3" slabbing cutter and takes a good cut. No way could I do that if I had only a direct coupled electronically variable speed, such as a DC motor and controller, or VFD and AC motor. It's a 50:1 torque increase......

PStechPaul
10-08-2014, 11:01 PM
I have thought about getting a small three phase sewing machine motor, perhaps for a motorized bike. Here is a 1/2 HP version:
http://www.ebay.com/itm/SEWLINE-NEW-3450-RPM-220-VOLT-3-PHASE-MOTOR-FOR-INDUSTRIAL-SEWING-MACHINE-/221565007340

http://i.ebayimg.com/03/!Bjcl2OQBGk~$(KGrHqMH-C0Es-ZkrfpLBLTlmyEmK!~~_35.JPG

I'm satisfied with the original Chinese 3/4 HP 4 pole AC motor on my lathe, but I also have three 1 to 2 HP three phase motors and a 5 HP and also a 2 HP DC treadmill motor I could use. And I have a couple of working VFDs. All of these were under $100 each.

flylo
10-08-2014, 11:18 PM
eBay item number:301325318086 is the exact motor I've used. Worked well for me but I just replaces the 1/3 hp motor & ran the stock set of pulleys on the slow setting but the rpms on the new motor were up to double the old & the HP was double also, plus variable speed, reverse & brake which I didn't use. The reverse & variable speed switched can be moved to a more convenient place also. I payed $100/shipped looks like they've gone up but it's been a couple years. I should have 2 left in the shop. Also flex shaft sewing lights are a lot cheaper than machining lights as you can see here eBay item number:250901728976

mikem
10-08-2014, 11:23 PM
Most of my lathe work is about 200-300 rpm. Your motor is only going to generate about 20% rated Horse power at that speed using an electronic speed controller. By using a 1 to 5 pulley reduction, your motor will run at full RPM for full horsepower output and your torque will be increased by a factor of 5 at 350 rpm. I think that you should try it, to prove it to yourself, rather than trying to convince those of us that have tried it that we are wrong! :)

JRouche
10-08-2014, 11:47 PM
For some reason I like DC motors. Always seems to be the most controllable. But I dont know all the types out there.

I do have a DC motor on my lathe though and its a dream. Silky smooth speed control even down to a snails pace (not that I spin it that slow much). But at a crawl the spindle is unstoppable. I do have a back gear and use that instead of taxing the electronic speed control.

Oh? My lathe is a Monarch 10EE with some happy thyratron tubes glowing a nice blue/purple :) JR

Mike Amick
10-09-2014, 12:13 AM
I think you have a perfect setup.
My last lathe I used the exact motor .. and .. it was a beast.

I have a similar motor on my current mill .. and also .. right of the motor control box
I have a box with digital RPM, Voltage, and analog current.

http://www.mikeamick.com/misc/mill.jpg

Mike A

doctordoctor
10-09-2014, 12:29 AM
I'll insert my responses next to your statements:
Lets say you have a motor, and it can maintain a constant torque regardless of RPM. Thats pretty close to the real world DC brushed motor we are talking about. OK

Now lets say you want to use that motor on a lathe, and you have two situations, one is turning a 5" diameter piece, the other, a 1" diameter piece. OK

The materials, cutters, depth of cut, feed rate, and desired SFM are the same between them.

Therefore the cutting force on the tool is the same. OK

But the torque on the motor is different, because that cutting force is projected out at 5" in one case and 1" in the other. So its 5 times as much torque for the 5" piece. OK, BUT ONLY WITHIN THE RANGE OF AVAILABLE TORQUE FROM THE MOTOR

But the rpm will be 5 times slower to maintain a constant SFM, in the 5" piece. OK

So power remains the same between the 5" piece and the 1" piece. Because power is the product between the rpm and the torque. NO, YOU JUST SAID RPM IS SLOWER TO MAINTAIN DESIRED SFM AND THE TORQUE SUPPLIED BY THE MOTOR IS 5 TIMES AS MUCH. THE REQUIRED HORSEPOWER IS 5 TIMES LARGER ON THE 5" PIECE. YOUR 1/2 HP MOTOR CAN SUPPLY 16500 FT-LB/MINUTE. THE MAXIMUM TORQUE SHOULD BE ABOUT 125 FT-LB. AT FULL MOTOR SPEED

So with a motor that has a constant torque throughout its rpm range, no pulley system is required. NICE TRY, BUT YOU ARE CONFUSING AVAILABLE MOTOR TORQUE WITH CUTTING TORQUE REQUIRED. TRY THINKING OF IT IN TERMS OF FORCE. YOUR MOTOR PULLEY HAS A FIXED RADIUS AND CAN PROVIDE A PULL ON THE V-BELT IN POUNDS EQUAL TO 125 DIVIDED BY THE MOTOR PULLEY RADIUS IN FEET. IF THERE ARE NO OTHER PULLEYS THAN THE DRIVEN IN THE SYSTEM, THE AVAILABLE SPINDLE TORQUE IS THE SAME, IGNORING LOSSES. THE BELT TENSION CANNOT EXCEED THE 125/R. NOW IMAGINE DOUBLING THE DIAMETER OF THE DRIVEN PULLEY. WITH THE SAME BELT TENSION (FORCE), THE TORQUE IS DOUBLED BECAUSE THE RADIUS AT WHICH IT IS APPLYING THE FORCE IS DOUBLED.

There is no advantage to having a variable ratio pulley system unless the motor has higher power at a specific rpm, or its rpm cant be controlled. But for a constant torque vs rpm motor like a DC shunt wound motor, thats not the case. Not significantly anyway. DISAGREE

That is incorrect. lets try again.

Power does not go up in any drivetrain, no matter what. Agree? If you believe this to be untrue then you believe in perpetual motion machines.

Belt tension is irrelevant. The belt is irrelevant. The only thing important about the drive train is the overall ratio, ignoring friction losses.

It does not require 5 times as much power to turn a 5" workpiece compared to a 1" workpiece, if its the same depth of cut and the same SFM.

Look at it this way, the cutter "sees" the same amount of material go past it in both cases, because the SFM is the same. Same material removal rate, same power.

Power is the product of torque and rpm. If the torque goes up 5 times, and the rpm goes down 5 times, the product is the same. The power is the same.

You are not removing material faster by increasing the diameter of the workpiece if you are maintaining the same SFM and cutting parameters. Maintaining the same SFM means lowering the rpm as diameter goes up.

Show mean your calculation to establish your logic. You wont be able to because its impossible.

I already showed you mine, but here it is again:



Take two situations, one 5" dia workpiece, and one 1" dia. Same material, same SFM, same cut/depth of cut. You change rpm to keep the SFM the same since the surface of the larger diameter moves faster for a given rpm than a smaller one, since its circumference is 5 times larger.

So the cutting force on the tool remains the same. Lets say its 30lbs.

At 5" 30lbs is 12.5 foot lbs.

At 1" 30lbs is 2.5 foot lbs.

And you increased the rpm by a factor of 5 from 5" to 1" to maintain a constant SFM.

For discussions sake, lets say you get your target SFM at 5" at 500rpm.

So using your forumula:

For the 5": 12.5 * ( 500/5252) = 1.19 hp

For the 1": 2.5 * (2500/5252) = 1.19 hp

doctordoctor
10-09-2014, 12:38 AM
Don't get too fond of that idea.....

Material removed per unit time tends to go as power..... not torque.



Correct. But material removal rate is not going up as workpiece diameter increases if SFM is maintained constant and the cutting parameters remain the same.




Power goes as torque x rpm (rpm being distance per unit time, and torque being force).



Incorrect. RPM is not distance per unit time. Distance would require the diameter of the part to be incorporated into the calculation.



If your motor can produce a given torque, and has a given base speed, it produces half power at half speed. So it produces 1/5 power at 1/5 speed.



Correct. If the torque of the motor remains the same, and only RPM changes, the power is less as RPM goes down. This idea does not directly correlate to turning various workpiece diameters at different rpms and different torques.




Comparing that motor straight, with that same motor through pulley reduction, at a speed of 1/5 of base speed, you get 1/5 power with the straight coupling. But with a 5:1 pulley reduction, you get 5x torque at the same output shaft speed. So with the pulley you can still get full power, but with the straight coupling you are limited to 1/5 power.



You dont "get" any specific power by setting the transmission to a certain ratio. Just like the engine in your car is not outputting a certain power based on what you have the transmission set to. The final output torque and rpm are all that matter when determining the power output of the system.



That can be the difference between being forced to whittle off tiny threads of material, vs being able to take a reasonable cut. I have a 1/4 HP 1725 rpm motor on my mill. At the 35 RPM back gear speed, it happily runs a 3" slabbing cutter and takes a good cut. No way could I do that if I had only a direct coupled electronically variable speed, such as a DC motor and controller, or VFD and AC motor. It's a 50:1 torque increase......

No transmission increases power. A given cut depth in a given material with a given cutter at a given SFM requires a specific amount of cutting force. That force remains the same whether the workpiece is 1 inch or 1 trillion miles in diameter. If SFM is maintained constant as workpiece diameter changes, RPM must go up (smaller diameter) or down (bigger diameter), and torque will go up (bigger diameter) or down (smaller diameter). The product remains the same. I.e. the power remains the same.

doctordoctor
10-09-2014, 12:40 AM
Most of my lathe work is about 200-300 rpm. Your motor is only going to generate about 20% rated Horse power at that speed using an electronic speed controller. By using a 1 to 5 pulley reduction, your motor will run at full RPM for full horsepower output and your torque will be increased by a factor of 5 at 350 rpm. I think that you should try it, to prove it to yourself, rather than trying to convince those of us that have tried it that we are wrong! :)

You are wrong if you believe the logic as stated.

Now whether or not the speed control and motor actually have a constant torque vs rpm relationship that works within the SFMs and MRRs desired is a different question, and its totally possible a transmission may still be needed.

But if you think power goes up when you add a transmission, you have a major problem with your understanding of basic physics and how lathes work.

Use any method you like to calculate the power required to make a cut on a lathe. If you keep the depth of cut, material, and SFM the same, and vary only the workpiece diameter, you will end up with the same power regardless of workpiece diameter.

doctordoctor
10-09-2014, 12:55 AM
To be absolutely clear, the material removal rate in the scenarios I am describing remains the same. Im not saying you can remove more material per second and have the power say the same. So you would have to vary the axial feedrate so that the MRR stays constant, as well as the SFM.

But thats secondary to the points I'm making. A transmission of any kind is not increasing the power available to the cut, assuming the motor has a constant torque over its rpm range, and the lathe is operating within that rpm range.

Now, if the motor has a fixed rpm, then you need a transmission. If it has significantly lower power output at everywhere but a certain rpm, then you need a transmission so you can let it run at that rpm and change the transmission to change speeds.

But if it has a constant torque over its rpm range, and that rpm range overlaps the rpms you need for your machining, then there is no reason for a transmission.

PStechPaul
10-09-2014, 01:06 AM
Similar discussions have been going on at www.DIYelectricCar.com for a long time, and are central to the question of retaining a transmission or connecting an electric motor directly to the driveshaft. The power, torque, thrust, speed, RPM, and gear ratios are determined by the performance required, which may be equivalently expressed by acceleration or ability to climb a certain percent grade, or both. With no transmission, it requires a much larger motor than with one. Some smaller motors, particularly series wound, can briefly provide enough torque for adequate performance for the short bursts of acceleration desired, but AC induction motors with 2-3x maximum torque "boost" need to be about three times larger.

With a single fixed drive ratio on a lathe, and a typical AC or DC motor, the continuous torque available at the spindle will also be fixed, and will be essentially constant from zero to maximum RPM. Thus the power will also be proportional to speed. Since an AC motor (and some DC motors) can be "overclocked" or driven to higher speeds with higher than rated voltage, it is possible to get 2x or more power due to the higher RPM, and by using a higher gear or pulley/belt ratio, higher torque can be obtained.

You may be able to utilize the full power of a motor for metal cutting by running at or near maximum RPM and adjusting the DOC and feed accordingly, but it may generate more heat and less efficient cutting which might be improved by using carbide and/or coolant.

Gary Paine
10-09-2014, 01:08 AM
That is incorrect. lets try again.

Power does not go up in any drivetrain, no matter what. Agree? If you believe this to be untrue then you believe in perpetual motion machines.

Belt tension is irrelevant. The belt is irrelevant. The only thing important about the drive train is the overall ratio, ignoring friction losses.

It does not require 5 times as much power to turn a 5" workpiece compared to a 1" workpiece, if its the same depth of cut and the same SFM.


Look, I'm trying to help you as are others, but you adamantly insist your thinking is right. You are not, I'm afraid.
My masters degree in mechanical engineering does not seem to help me straighten you out. What's up with that?
By gearing a machine down 5:1 you will increase the torque available at the spindle five fold. Torque/workpiece radius will be the same in both your cited instances but the motor can apply 5 times the cutting force to the smaller radius workpiece than it can to the larger one.
I do, however believe in perpetual motion. Laws are made to be broken. I just haven't invented the way to make it happen - yet.

doctordoctor
10-09-2014, 01:10 AM
Similar discussions have been going on at www.DIYelectricCar.com for a long time, and are central to the question of retaining a transmission or connecting an electric motor directly to the driveshaft. The power, torque, thrust, speed, RPM, and gear ratios are determined by the performance required, which may be equivalently expressed by acceleration or ability to climb a certain percent grade, or both. With no transmission, it requires a much larger motor than with one. Some smaller motors, particularly series wound, can briefly provide enough torque for adequate performance for the short bursts of acceleration desired, but AC induction motors with 2-3x maximum torque "boost" need to be about three times larger.

With a single fixed drive ratio on a lathe, and a typical AC or DC motor, the continuous torque available at the spindle will also be fixed, and will be essentially constant from zero to maximum RPM. Thus the power will also be proportional to speed. Since an AC motor (and some DC motors) can be "overclocked" or driven to higher speeds with higher than rated voltage, it is possible to get 2x or more power due to the higher RPM, and by using a higher gear or pulley/belt ratio, higher torque can be obtained.

You may be able to utilize the full power of a motor for metal cutting by running at or near maximum RPM and adjusting the DOC and feed accordingly, but it may generate more heat and less efficient cutting which might be improved by using carbide and/or coolant.


I am not talking about cuts which have different power requirements.

In a scenario where diameter goes up, I'm talking about modifying the feedrates and rpm so that the material removal rate and SFM remain the same as a smaller workpiece.

doctordoctor
10-09-2014, 01:12 AM
Look, I'm trying to help you as are others, but you adamantly insist your thinking is right. You are not, I'm afraid.
My masters degree in mechanical engineering does not seem to help me straighten you out. What's up with that?
By gearing a machine down 5:1 you will increase the torque available at the spindle five fold. Torque/workpiece radius will be the same in both your cited instances but the motor can apply 5 times the cutting force to the smaller radius workpiece than it can to the larger one.
I do, however believe in perpetual motion. Laws are made to be broken. I just haven't invented the way to make it happen - yet.

Ive been REALLY clear about explaining my thinking and your masters degree cant seem to produce an equation describing what you are talking about. Whats up with that?

Whats hilarious, is that in true internet forum tradition, you assume that your masters degree somehow trumps whatever degree I do have, when you have no idea what education I may or may not have. LOL

How can you say workpiece radius will remain the same in both instances I cited, when one is 1" and the other is 5"?

Ive been making it clear that Im talking about equal cuts and equal MRR's on varying workpiece diameters and constant SFM. I'm not talking about the "possible" available torque. And in any case, the overall power DOES NOT INCREASE.

Gary Paine
10-09-2014, 01:26 AM
Ive been REALLY clear about explaining my thinking and your masters degree cant seem to produce an equation describing what you are talking about. Whats up with that?

Whats hilarious, is that in true internet forum tradition, you assume that your masters degree somehow trumps whatever degree I do have, when you have no idea what education I may or may not have. LOL
You've spent a lot of effort trying to describe what you think, but are ignoring the equations I have given you. The relationship between torque and horsepower, the effect of gearing to increase torque, the torque of your half horse motor.....
And to the best of my knowledge, everything I have tried to explain to you is accurate. I have explained that torque can and does increase with "gearing". Force, or belt tension if you will, certainly does define the issue. The motor applies its torque through the belt in the form of a pulling force on the top of the driven pulley. A driven pulley twice as big will produce twice the torque for the same pulling force in the same way that a workpiece 1 inch in diameter requires 1/5 the torque to make the same cut that a 5 inch piece will require.

doctordoctor
10-09-2014, 01:39 AM
You've spent a lot of effort trying to describe what you think, but are ignoring the equations I have given you. The relationship between torque and horsepower, the effect of gearing to increase torque, the torque of your half horse motor.....
And to the best of my knowledge, everything I have tried to explain to you is accurate. I have explained that torque can and does increase with "gearing". Force, or belt tension if you will, certainly does define the issue. The motor applies its torque through the belt in the form of a pulling force on the top of the driven pulley. A driven pulley twice as big will produce twice the torque for the same pulling force in the same way that a workpiece 1 inch in diameter requires 1/5 the torque to make the same cut that a 5 inch piece will require.

Okay lets start over. Because sometimes the truth gets lost in misunderstandings.

Heres what Im saying in the simplest possible terms. I am not IMPLYING anything else. I am saying only these following things. You may be saying entirely different things, and making points unrelated. In other words, we may be discussing totally different things.

No transmission increases power.

If material removal rate remains the the same, power required for that cut remains the same, even if workpiece diameter changes.

Material removal rate, and therefore power, can be kept constant as workpiece diameter changes by varying axial feedrate and RPM.

The thing you said that started this discussion was this:



You will still need the pulley system, though, to provide the necessary power at low RPM as when you are turning large diameter workpieces.


That is not true, because the transmission will not increase the power, because thats impossible.

If you just swap the word "power" for "torque" in what you said then its true.

Gary Paine
10-09-2014, 01:57 AM
Okay lets start over. Because sometimes the truth gets lost in misunderstandings.
..........

No transmission increases power.



Wrong, and I should have quit a long time ago, but I'm a stubborn old man. No transmission increases power, but a transmission does DEFINITELY increase TORQUE. Remember that available horsepower at a given rpm is a constant. That is at the motor. Now gear it down with pulleys. HP= TN/5252. Therefore if the speed has been geared down by half, the torque is up by a factor of 2 and the horsepower is the same.
The same is happening in your car. That engine has a variable speed too, but try to get it moving from a standstill at low engine speed in hi gear. In the lower gears the transmission is taking available engine torque and increasing it immensely through transmission gearing at the expense of vehicle speed in that gear. That torque is applied to the road through the rolling radius of the tires. You can accelerate much more rapidly through the low gear than you could have in hi gear due to the force on the road applied by the tire. The horsepower didn't increase, the torque did.
Think about it some more.

doctordoctor
10-09-2014, 02:14 AM
Wrong, and I should have quit a long time ago, but I'm a stubborn old man. No transmission increases power, but a transmission does DEFINITELY increase TORQUE. Remember that available horsepower at a given rpm is a constant. That is at the motor. Now gear it down with pulleys. HP= TN/5252. Therefore if the speed has been geared down by half, the torque is up by a factor of 2 and the horsepower is the same.
The same is happening in your car. That engine has a variable speed too, but try to get it moving from a standstill at low engine speed in hi gear. In the lower gears the transmission is taking available engine torque and increasing it immensely through transmission gearing at the expense of vehicle speed in that gear. That torque is applied to the road through the rolling radius of the tires. You can accelerate much more rapidly through the low gear than you could have in hi gear due to the force on the road applied by the tire. The horsepower didn't increase, the torque did.
Think about it some more.

I never said transmissions don't increase torque.

What you write above contradicts what you wrote before, which is what I am taking issue with:


You will still need the pulley system, though, to provide the necessary power at low RPM as when you are turning large diameter workpieces.

Because no transmission can increase power. Or "provide the necessary power". I think you meant to say torque but you said power, using it colloquially even though its technically incorrect.

Gary Paine
10-09-2014, 02:36 AM
I never said transmissions don't increase torque.

What you write above contradicts what you wrote before, which is what I am taking issue with:



Because no transmission can increase power. Or "provide the necessary power". I think you meant to say torque but you said power, using it colloquially even though its technically incorrect.

No, I meant power. What seem to be the problem is an issue of whether you are looking at the motor or the spindle. When you have to lower the rpm of the dc motor to get down to the correct SFM, you lower the horsepower that motor can deliver. The torque, with no gearing, is the same but RPM is down. Now if I add gearing to the machine, I can run at a higher motor speed and still have the lower spindle speed. With constant torque and higher speed, I deliver more power to the cut.

RichR
10-09-2014, 02:40 AM
OK, I'll take a shot at this. Your motor and spindle both have 1" pulleys and you have a 1" piece of material in the chuck. The torque required for your
cut is equal to X and your SFM is equal to Y. When you replace the 1" piece with a 5" piece your SFM increases to 5Y. You slow down your motor
to bring the SFM back down to Y so the torque required at the cutting edge is once again equal to X. The torque now required at the pulley will be 5X.
Think of it like a lever. One end is the edge of the pulley, the spindle is the pivot, and the other end is the edge of your workpiece.

doctordoctor
10-09-2014, 02:53 AM
No, I meant power. What seem to be the problem is an issue of whether you are looking at the motor or the spindle. When you have to lower the rpm of the dc motor to get down to the correct SFM, you lower the horsepower that motor can deliver. The torque, with no gearing, is the same but RPM is down. Now if I add gearing to the machine, I can run at a higher motor speed and still have the lower spindle speed. With constant torque and higher speed, I deliver more power to the cut.

I agree with what you describe here, but it doesnt seem consistent with what you say here:



I'll insert my responses next to your statements:
Lets say you have a motor, and it can maintain a constant torque regardless of RPM. Thats pretty close to the real world DC brushed motor we are talking about. OK

Now lets say you want to use that motor on a lathe, and you have two situations, one is turning a 5" diameter piece, the other, a 1" diameter piece. OK

The materials, cutters, depth of cut, feed rate, and desired SFM are the same between them.

Therefore the cutting force on the tool is the same. OK

But the torque on the motor is different, because that cutting force is projected out at 5" in one case and 1" in the other. So its 5 times as much torque for the 5" piece. OK, BUT ONLY WITHIN THE RANGE OF AVAILABLE TORQUE FROM THE MOTOR

But the rpm will be 5 times slower to maintain a constant SFM, in the 5" piece. OK



So far so good..but then..



So power remains the same between the 5" piece and the 1" piece. Because power is the product between the rpm and the torque. NO, YOU JUST SAID RPM IS SLOWER TO MAINTAIN DESIRED SFM AND THE TORQUE SUPPLIED BY THE MOTOR IS 5 TIMES AS MUCH. THE REQUIRED HORSEPOWER IS 5 TIMES LARGER ON THE 5" PIECE. YOUR 1/2 HP MOTOR CAN SUPPLY 16500 FT-LB/MINUTE. THE MAXIMUM TORQUE SHOULD BE ABOUT 125 FT-LB. AT FULL MOTOR SPEED

You acknowledge that, compared to the 1" workpiece, the 5" piece will be run at a 5 times slower RPM, with a 5 times higher cutting torque, but then in the next sentence state that the power required will be 5 times as much. How does that make sense? The product of the RPM and torque is the power. And if you divide one by 5 and multiply the other by 5, the product remains the same.

doctordoctor
10-09-2014, 03:05 AM
OK, I'll take a shot at this. Your motor and spindle both have 1" pulleys and you have a 1" piece of material in the chuck.



Ok.



The torque required for your cut is equal to X and your SFM is equal to Y.


The cutter doesnt see torque, it sees cutting force. Thats an important distinction and I'm not trying to nitpick semantics. The spindle sees torque. Read on to see the relevance:




When you replace the 1" piece with a 5" piece your SFM increases to 5Y.


Agreed. (assuming RPM doesnt change of course)



You slow down your motor to bring the SFM back down to Y.


Agreed.



so the torque required at the cutting edge is once again equal to X.


Just to be clear, a given SFM does not equate to a given torque. Logic as thus: the cutting force for equal cuts at equal SFM exerts different torques on the spindle depending on workpiece diameter. So in other words, matching the SFM to the 1" part by slowing the 5" part RPM down does not make the torque the same. I dont think you are actually intending to be incorrect here..you just need to replace the word "torque" with "force".



The torque now required at the pulley will be 5X.
Think of it like a lever. One end is the edge of the pulley, the spindle is the pivot, and the other end is the edge of your workpiece.

Agreed.

But what conclusion are you coming to based on all this? Because I agree with everything you are saying, besides the mixup of the terms torque and force.

Gary Paine
10-09-2014, 04:26 AM
You acknowledge that, compared to the 1" workpiece, the 5" piece will be run at a 5 times slower RPM, with a 5 times higher cutting torque, but then in the next sentence state that the power required will be 5 times as much. How does that make sense? The product of the RPM and torque is the power. And if you divide one by 5 and multiply the other by 5, the product remains the same.

Let' go back to basics. Quoting the Machinery's Handbook:
Force is any cause tending to produce or modify motion. The units by which a force is usually measured are pounds or tons.
Work is the product of force by distance and is expressed by a combination of units of force and distance, as in foot pounds.
Power is the product of force by distance divided by time, or the performance of a given amount of work in a given time, and is expressed as inch pounds per minute, foot -pounds per minute, etc.
Horsepower is the unit of power adopted for engineering work. One horsepower is equal to 33,000 foot-pounds per minute.

So, you cannot ignore time (or distance if you will). The one inch diameter in your example requires 1/5 the torque needed by the 5 inch diameter to achieve the cutting force and runs 5 times faster at the same SFM. Assume in both examples, the cut is ongoing for several minutes. The horsepower required for the 1 inch piece is Y foot-pounds per minute. The larger diameter part will require 5 times Y foot pounds per minute at the same cutting force because of the larger diameter at the same SFM, or 5 times the horsepower is required for that same amount of material removal.

J Tiers
10-09-2014, 09:00 AM
Correct. But material removal rate is not going up as workpiece diameter increases if SFM is maintained constant and the cutting parameters remain the same.

Sorry to break it to you but you have the wrong idea there..... You are "starting from the wrong end". Your material removal rate will be a LOT SMALLER than it is at higher speeds with same torque.... You are comparing totally wrong things.

If you have the power to remove a certain size chip at a certain SFM....you have the TORQUE to remove that chip (force), and the POWER to do it at a certain SFM (distance per unit time). When you slow the motor down 5x, you can remove that same chip because the TORQUE is the same, but you do NOT have the POWER to do it at the same SFM. Force is same, distance cannot be the same.




Incorrect. RPM is not distance per unit time. Distance would require the diameter of the part to be incorporated into the calculation.

Sorry, you are wrong to dismiss that.... both pi and the diameter are constants, so they do not affect the proportionality.... double the rpm and the distance is doubled.....That is true at ANY diameter. If you want the specific number, yes, you need the diameter, but the proportionality is unaffected.




Correct. If the torque of the motor remains the same, and only RPM changes, the power is less as RPM goes down. This idea does not directly correlate to turning various workpiece diameters at different rpms and different torques.

You had better show your work, there, I think you will quickly understand if you do the math.


You dont "get" any specific power by setting the transmission to a certain ratio. Just like the engine in your car is not outputting a certain power based on what you have the transmission set to. The final output torque and rpm are all that matter when determining the power output of the system.

No transmission increases power. A given cut depth in a given material with a given cutter at a given SFM requires a specific amount of cutting force. That force remains the same whether the workpiece is 1 inch or 1 trillion miles in diameter. If SFM is maintained constant as workpiece diameter changes, RPM must go up (smaller diameter) or down (bigger diameter), and torque will go up (bigger diameter) or down (smaller diameter). The product remains the same. I.e. the power remains the same.

Where'd THAT come from? The point of the transmission is to ALLOW THE MOTOR POWER TO BE APPLIED AT THE DESIRED OUTPUT SPEED. The transmission (ignoring frictional losses) reduces speed, but multiplies torque. The product remains the same, so power is unaffected.

THAT IS IN STARK CONTRAST TO REDUCING MOTOR SPEED! When you reduce motor speed, you throw away power, because the torque is the same and speed is reduced.

Therefore, if you use a transmission, you lower SFM when you reduce output speed, but can increase chip size, because the torque (force) is larger. Net result is that you can maintain metal removal rates.

The entire point of the transmission on a lathe is to allow lower speeds for larger diameters, and tougher materials.

At high speeds, you remove a small chiip, but you remove a lot of chip LENGTH At lower speeds, your removal rate would be reduced with speed, since the SFM is reduced for an equal size part.

BUT, if you have a LARGER part, the torque required for that larger diameter is MORE than at a smaller diameter. The reduced motor speed has only the SAME torque as always, so you MUST remove an even smaller chip.

You have an error in how you are looking at the whole problem. If you do as you suggest, you will quickly come to realize your error. The lathe will seem very weak and underpowered, as indeed it will be at anything less than full motor speed.

doctordoctor
10-09-2014, 11:16 AM
Let' go back to basics. Quoting the Machinery's Handbook:
Force is any cause tending to produce or modify motion. The units by which a force is usually measured are pounds or tons.
Work is the product of force by distance and is expressed by a combination of units of force and distance, as in foot pounds.
Power is the product of force by distance divided by time, or the performance of a given amount of work in a given time, and is expressed as inch pounds per minute, foot -pounds per minute, etc.
Horsepower is the unit of power adopted for engineering work. One horsepower is equal to 33,000 foot-pounds per minute.

So, you cannot ignore time (or distance if you will). The one inch diameter in your example requires 1/5 the torque needed by the 5 inch diameter to achieve the cutting force and runs 5 times faster at the same SFM. Assume in both examples, the cut is ongoing for several minutes. The horsepower required for the 1 inch piece is Y foot-pounds per minute. The larger diameter part will require 5 times Y foot pounds per minute at the same cutting force because of the larger diameter at the same SFM, or 5 times the horsepower is required for that same amount of material removal.

You arent addressing the question I asked. If the torque is 5 times greater, and the rpm is 5 times slower, where are you getting the requirement of 5 times more power?

Im talking about cuts with identical material removal rates. Axial feedrate must be reduced so material removal rate remains constant between workpiece diameters. Is that the source of confusion?

doctordoctor
10-09-2014, 11:43 AM
We're going in circles guys...you are starting to say things to correct what I am saying, but the things you are saying are the things IVE BEEN SAYING...so there is obviously a disconnect of some kind.

My points are:



No transmission increases power.

If material removal rate remains the the same, power required for that cut remains the same, even if workpiece diameter changes.

Material removal rate, and therefore power, can be kept constant as workpiece diameter changes by varying axial feedrate and RPM.


And here is a snapshot of a feed and speed calculator showing what I am talking about.

Equal material removal rate, equal horsepower, different diameters.

http://i242.photobucket.com/albums/ff197/acannell/blag_zps9d2a0a5f.jpg (http://s242.photobucket.com/user/acannell/media/blag_zps9d2a0a5f.jpg.html)

Gary Paine
10-09-2014, 12:55 PM
You arent addressing the question I asked. If the torque is 5 times greater, and the rpm is 5 times slower, where are you getting the requirement of 5 times more power?

Im talking about cuts with identical material removal rates. Axial feedrate must be reduced so material removal rate remains constant between workpiece diameters. Is that the source of confusion?

I hereby pledge to not try to answer even these simple questions in the middle of the nite from here on.

In your calculations, you showed that the horsepower required was the same for identical material removal rates on the two diameters, and that is correct. But it makes the assumption that the power was available because the motor was working in it's normal range with "constant" torque and variable RPM.
Let's use that example further, but stretch it to a limit. Let's make that cut on the 1 inch bar sufficiently large to use up the entire 1/2 horsepower of your motor. The torque applied to the 1 inch bar by the cutter is X foot pounds. The force is Y pounds.
Now we go to the 5 inch bar. We want the same cutting force F and the same SFM with this bar, so the cutting torque needs to increase due to the longer lever arm by a factor of 5 to a value of 5X and the RPM needs to lower by a factor of 5 to achieve that SFM. But wait! We do not have 5 times the torque available. The motor was already delivering the full torque to cut the one inch bar. Reducing the motor RPM by 1/5 to achieve the SFM electronically does nothing to increase torque. The solution is to achieve the cut on the 5 inch bar that is requiring the 1/2 horsepower to make through a 5:1 pulley reduction whereby the torque will amplify by a factor of 5. Just what we need, right. Both bars are cut at the same SFM and DOC and consuming the 1/2 HP, but it took gearing to make it happen.
WHEW! Why couldn't I just say that in the middle of the nite?:o

doctordoctor
10-09-2014, 01:11 PM
I hereby pledge to not try to answer even these simple questions in the middle of the nite from here on.

In your calculations, you showed that the horsepower required was the same for identical material removal rates on the two diameters, and that is correct. But it makes the assumption that the power was available because the motor was working in it's normal range with "constant" torque and variable RPM.
Let's use that example further, but stretch it to a limit. Let's make that cut on the 1 inch bar sufficiently large to use up the entire 1/2 horsepower of your motor. The torque applied to the 1 inch bar by the cutter is X foot pounds. The force is Y pounds.
Now we go to the 5 inch bar. We want the same cutting force F and the same SFM with this bar, so the cutting torque needs to increase due to the longer lever arm by a factor of 5 to a value of 5X and the RPM needs to lower by a factor of 5 to achieve that SFM. But wait! We do not have 5 times the torque available. The motor was already delivering the full torque to cut the one inch bar. Reducing the motor RPM by 1/5 to achieve the SFM electronically does nothing to increase torque. The solution is to achieve the cut on the 5 inch bar that is requiring the 1/2 horsepower to make through a 5:1 pulley reduction whereby the torque will amplify by a factor of 5. Just what we need, right. Both bars are cut at the same SFM and DOC and consuming the 1/2 HP, but it took gearing to make it happen.
WHEW! Why couldn't I just say that in the middle of the nite?:o

Ahh okay, I agree with all that, and actually, I've said exactly the same thing. I think the conversation got sidetracked by alot of sub-disagreements/misunderstandings on related topics.

So, as a cherry on top of all this, what sort of transmissions do modern production lathes have? Do they still have a motor attached to a variable ratio transmission, or do they have some kind of motor that can have its torque proportionally increased at low rpm electrically?

And, lets say that I want to make do with my DC motor, but still get rid of the countershaft assembly. I think I have a good reason to do this. My countershaft assembly is not the factory one, its a homemade one out of hardware store pillow blocks and welded square tube with wood inside. Its wobbly and vibrates alot. I might be able to tighten it up, but most likely its a liability as far as surface finish. What about just using the back gear reduction for the torque multiplying and doing away with the pulley system?

Id probably have to make a map of what sort of speeds and cutting torques would be available, since it would then be hybrid of the speed control and backgears, possibly with 4 pulley ratios of the pulleys that are already there. That seems like it would still offer a wide range overall wouldnt it?

Mike Amick
10-09-2014, 01:40 PM
Im sure you dont intend to hook the motor up " directly". I would think that you are thinking
at least 1 pulley to pulley. I would go with the biggest "gear down" you can come up with, these
DC motor/speed control setups really do stall easier at low speeds ( for whatever reason).

On my lathe I built a little jack shaft to double down on the initial speed.

Just shouting in from the circle, that has formed.

Mike A

flylo
10-09-2014, 02:38 PM
My Consew motor changes were much easier, just change the motors, use the same pulleys so when the belt the the motor to the lathe was in the slowest of the 2 pulleys the motor was full speed 3450 rpm putting out the full 3/4 hp but adding variable speed & reverse on the motor while running the lathe at the same speeds it was designed to run verified by a laser tach. I don't see where all this arguing is constructive. Try it like you think it will work & fix it if it doesn't.

J Tiers
10-09-2014, 06:29 PM
It's constructive if the OP eventually learns some physics here....

Judging by his last post, it seems as if the penny has finally dropped, and he is understanding what we have all been saying in various ways.

doctordoctor
10-09-2014, 07:06 PM
It's constructive if the OP eventually learns some physics here....

Judging by his last post, it seems as if the penny has finally dropped, and he is understanding what we have all been saying in various ways.

My understanding of the situation hasnt changed at all. Nothing I said was untrue. My original statements are still valid. Yourself and others made many arguments to attempt to explain what you were thinking and several of the statements and logics you put forth were incorrect, and that formed the basis for much of the discussion. You cant expect your conclusion, whatever it is, to be made evident, if your supporting arguments are clearly false.

Regardless, my statement that a transmission is not required for an electric motor with a constant torque versus rpm is still true. Gary Paine finally put together an explanation of what he was saying that made sense, and I agree with it. However, its only one scenario. If the cutting torque does not exceed the motor torque, then a transmission has no advantage. There has been an implied assumption that the 1" cutting torque is equal to the motor maximum torque, and that to achieve the 5" cutting torque, a transmission would be required. That would be true if the 1" cutting torque was indeed equal to the motor maximum torque, but that is just an assumption, and in no way must be true.

Lets say the maximum motor torque is 50 in lbs. And we want to make a cut that requires 5 lbs of force. The 1" diameter cut would then require 5 in lbs of torque. The 5" diameter cut would require 25 in lbs of torque. Both are within the motors torque capability. So all that is required is to change the motor rpm. A transmission would offer no advantage.

David Powell
10-09-2014, 07:51 PM
I run a Standard Modern 9" lathe with a dc drive of unknown hp. The controller Is a Boston Ratiotrol unit. I need a 20 amp fuse to get sensible results. The headstock has two speeds, direct and a 5-1 reduction. A very competent electrician suggested that motor life would likely be extended if the motor was not " lugged" at low revs but instead encouraged to usually run at half speed or over. For metal removal I up speed and feed until something protests and then back off a bit, blue chips, motor slowing, feed gears growling.If it is a large diameter piece, or hard material I use the low gear, if it is smaller than about 2" or brass or aluminium then high gear. I can thread with a tailstock die holder and allow the motor to stall ( Momentarily), providing the speed controller is set to 10% or less. ( I presume that the voltage applied to the armature is then about 8- 10 volts) I know that cannot be good for the motor, but I have done it for 22 yrs without problems so far. Certainly a larger workpiece needs more torque at the cutting edge than a smaller one, but it also needs a lesser rotational speed to keep at the proper surface speed. the 5-1 and 1-1 gearing of the lathe and DC drive seems to work well for me. Quit arguing over pedantics and get making swarf, you will soon see what works and what will not! regards David Powell.

Henro
10-09-2014, 08:12 PM
Regardless, my statement that a transmission is not required for an electric motor with a constant torque versus rpm is still true. Gary Paine finally put together an explanation of what he was saying that made sense, and I agree with it.

Your statement is true in the theoretical world, but only holds true in the real world if the motor on the lathe is large enough to supply whatever "power" is called for by the work being done, at all spindle speeds that may be used.

It is unlikely that your lathe without a transmission will be able to perform the same range of work that a similar lathe with the same size motor AND a transmission will.

Have you considered this?

J Tiers
10-09-2014, 09:59 PM
My understanding of the situation hasnt changed at all. Nothing I said was untrue. My original statements are still valid. Yourself and others made many arguments to attempt to explain what you were thinking and several of the statements and logics you put forth were incorrect, and that formed the basis for much of the discussion. You cant expect your conclusion, whatever it is, to be made evident, if your supporting arguments are clearly false.


No false statements were made..... sorry to disillusion you..... You simply didn't "get it".

Anyway.... first, go back to Gary Paine's info... it's actually ALL THERE, and it clearly shows your error.

Yes. many of your statements are true, they are just connected up wrongly. A person can say only true things, and still be wrong.... The true things must be relevant IN THE SITUATION.

Our point is that JUST AS YOU ADMIT.... it takes 5x as much torque to exert the same force at 5" diameter as it does for a 1". Yep, true as the Gospel, bro.

NOW.... you assume that the 1" torque is all you need. But what we are talking about here is POSSIBILITY, not whatever you happen to be using at the time. Yes, if you take cuts that are never over 1/5 the power you have, then if you turn down to 1/5 speed, you will still have just as much capability as the tiny fraction of total power you were using before. True... but you assume that will always be enough.

Our point is that you have LOST much of the power you had available, and you will NOT be satisfied to do that forever. You just THINK that now, but calm down and listen, you will not think that forever. That thing ain't a Sherline, and once you get that idea ingrained, you will start to try to get things DONE. Tiny "nibbles" being cut are not gonna do it for you then.

If you keep the back gears and step pulleys, you will be able to apply all the power you have to the work. You WILL want to, so don't make it impossible to do that by limiting the motor power you can apply.

Some tough steels need to be worked at slower SFM, AND they need a solid cut taken. Little "nibbles" tend to work harden the surface, to a point where you cannot cut them very well any longer, even with carbides, and especially with the anemic 1/5 of full power you may have at the rpm needed.

Your proposed gear cutters are an example... if you use a hardenable tool steel, that will need a good cut at relatively low SFM.

Tony Ennis
10-09-2014, 10:51 PM
I am not sure what problem we're trying to solve here. It takes about 3 seconds to move the belt to any of the 4 positions in the headstock. Regardless of the motor, you can't beat mechanical advantage. You can augment it, but you can't be better without it. Run the normal motor. Use the belts. Make chips. Save your time and energy for the parts of the lathe that will require your considerable creativity.

Your lathe was designed to have a motor running at 1760 RPMs. Between the 4 pulleys in the headstock and the 6:1 back gears, you'll be able to slow that thing down and have torque out the wazoo.

darryl
10-10-2014, 12:48 AM
In the real world, a turning job will come along where the torque requirement is higher than the motor is rated for. Then you will have to trade off rpm for higher torque using belts and pulleys, or gears, or a variable ratio drive of some kind. While it is true that you can dial the motor rpm down without using a transmission of some kind, it only works if the cutting force is less than the motors rated torque. With a half horse motor at 1750 rpm or whatever, the torque will be greater than from a 3450 rpm motor of the same rated power. You're a little ahead of the game in that case, but there's still no point in trying to go direct drive with only electronic speed control. You will need torque multiplication at some point, sooner than later probably.

You can follow the logic of taking a progressively smaller bite to mitigate this problem, but that will also run out. Too small of a bite just doesn't 'cut it' sometimes, and again I think you'd find that out sooner than later.

I've been considering a direct drive motor for my lathe, but it would be a large diameter, pancake type motor capable of high torque to begin with. And I'd be paying the price by limiting my top rpm capability, so it may not be viable anyway. I'd be paying an additional price- the inertia of the spindle would increase significantly, which is bad news if something jams.

I like the treadmill motor conversion on my lathe, but I still reduce the motor rpm going to a jackshaft, then I have the original three pulley ratios still. I would not eliminate any of that as I know that I'd be able to stall the motor quite easily if I didn't have that roughly 2.5-1 torque multiplication.

PStechPaul
10-10-2014, 03:47 AM
I think the dead horse has been beaten to a pulp, and I think everyone knows the relationships of force, torque, power, and gear/belt ratio. But something else to consider with a variable speed drive is that, at lower speeds, there may be more "cogging" effect, which can affect the finish on the work. Also, if the motor has an internal shaft-driven fan, it will not provide much cooling at low RPM, but the heat produced from I^2R losses will be the same at the same torque, so it might overheat. Furthermore, a motor running close to its rated speed will have a good deal of rotational inertia, which can help smooth out the cutting process as it encounters irregularities in the material.

It is still quite convenient and desirable to have a variable speed motor for reasons stated previously, but if the belts can be changed easily, and/or if there is a back gear available with the flip of a lever, it is better to use those mechanisms to obtain the maximum available torque for a given speed.

macona
10-10-2014, 04:23 AM
A bigger problem might be that its "recessed shaft" whatever that means..I need to pull that hydraulic pump head off and see whats what. Maybe I'd have to make an adapter.



The recessed shaft might be a bit of a problem. Basically, instead of a shaft coming out and interfacing with the pump they use a standard pump with a keyed (sometime spline) shaft that slips into a mating hole in the face of the motor. This is common with pump/motor units. I suppose you could make a shaft to press in or something but I would be worried it would wobble without an outboard bearing.

What I would do is put the motor/pump combo on ebay and sell it to someone who can used it as it is and use the money to buy the right style motor, 3 phase or DC.

J Tiers
10-10-2014, 07:42 AM
I think the dead horse has been beaten to a pulp, and I think everyone knows the relationships of force, torque, power, and gear/belt ratio.

Everyone EXCEPT the OP, who knows the words, but NOT the practical applications.....

Seastar
10-10-2014, 09:48 AM
Well-----
That was certainly entertaining!
Bill