There is some room for error. But I wouldn't worry too much.
Please add a closeup photo of the motor label too. Often the model number designates the size. I'm fairly confident we can figure it out.
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Variable speed press drill
- Thread starter Marc Moreau
- Start date
There is some room for error. But I wouldn't worry too much.
Please add a closeup photo of the motor label too. Often the model number designates the size. I'm fairly confident we can figure it out.
Hey @Marc Moreau , I was reading back through this thread and noticed a link that @Janger provided to the King Industrial Nova Voyager DVR Drill Press.
The information in that link suggests that you do have a 2HP motor in your machine.
The information in that link suggests that you do have a 2HP motor in your machine.
Yes. We used to make jokes about that. There is HP and then there is Sears HP. You can guess which one was bigger. Forgetting about losses the reality is that simple (watts / 745) formula. Your Sears Shopvac will not have 4.5HP on a 115VAC circuit for anything more than a millisecond or two as the motor starts up.
So if the motor draws 13A at 115VAC then we're talking about 1500W or about 2 HP. The motor turning at the rated RPM won't be drawing 13A. And it won't be 100% efficient at that power so the reality is a much lower HP.
But still we don't really care that much. As @Susquatch stated so eloquently, the torque is the important part and the HP is how fast you can apply that torque. We might find that a 2HP drill press with a VFD is terrible at drilling 5/8" holes at 300 RPM while does a great job with the motor turning 1750 RPM with a 5.8:1 step down pulley. That's simply because the motor rated torque has been multiplied by 5.8.
In both cases the drill bit is turning 300 RPM but one will stall and the other will make chips. So if a 1/3 HP motor is rated on a drill press with 5.8:1 pulley reduction down to 300 RPM for drilling 5/8" holes, then you will likely require a 2HP motor with a VFD turning it at 300 RPM to have the same torque with a 1:1 pulley setup.
That's why even though I have an AC servo (encoder feedback on the motor driver) that at full speed matches the old single phase motor I did make a two step pulley in case doing something at low RPM stalls the motor.
Actually it would cause a motor fault and that would stop everything. I have the AC servo drives on my motors showing current consumption rather than RPM. So far everything I've done hasn't come close to the current rating of the motor.
This is true if the diameter (and feed per rev) remained the same, i.e 1/16 drill at 2500 and then at 250, 1/10 the removal rate of course requires 1/10 the HP which you get if torque remained constant. However he's trying to lower the speed because the diameter went up (as is almost always the case). If he's got 1/2HP at 2500 and slows it 250 via VFD, he's got 1/20 of HP. I think you would agree you are not going to drill a 1/2 hole with 1/20 HP.
You want to think in terms of torque, ok. Torque is a function of force at a radius. What's happened to the radius of the cutting in going from 1/16 to 1/2? That changes teh torque required to turn it.
I believe the following are facts. Any argument with them? They are the extent of my point/claim
- HP = RPM x Torque
- Removal rate (cubic in inches/min say) is a function of HP
- VFD's don't increase torque over their range, when less than line frequency.
- The formula dictates, that if you don't increase torque as speed goes down, you get less HP.
- If that less HP is still good enough for what you need it do to for whatever reason, peace, but it doesn't change the physics of what is going on
- No one is drilling a 1/2 hole with 1/20 HP, which is about the power you could expect taking 1/2 HP @ 2500 and reducing it to 250 via VFD
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Right! and that is using a mechanical transmission to reduce speed and multiply torque so are still getting 2HP @ 300 rpm, which the VFD cannot do.....which is my whole point
Marc Moreau
Marc Moreau
Marc Moreau
Marc Moreau
Marc Moreau
Marc Moreau
Marc Moreau
Marc Moreau
This is exactly the same point I made and the reason the whole thing gets confusing.
Actually, I only sort of agree. You boil it down to how much hp is needed to do this job. I boil it down to how much torque is needed.
As long as the rigidity of your machine and its feed rate is controllable, it is totally possible to drill a 1/2 hole at 250 RPM IF YOU REALLY DO HAVE CONSTANT TORQUE. You just have to reduce the feed rate. That's hard to do on a hobby machine because the drill would prolly grab due to lack of smooth feed control and rigidity in the feed mechanism and the rotating spindle. But that's a different problem than the force or power required to do it.
A guy doing it for a living would throw up in disgust at the drill rate because it would take forever. A pure hobbiest might not care.
The torque is required to cut the steel and the hp required is a function of how fast the feed the feed rate is.
I agree with everything else you say totally except for the very last point which is the same as above.
Yes you can. PROVIDED YOU REALLY DO HAVE THE SAME TORQUE AND PROVIDED THAT YOU REDUCE THE FEED RATE.
Right! and that is using a mechanical transmission to reduce speed and multiply torque so are still getting 2HP @ 300 rpm, which the VFD cannot do.....which is my whole point
No argument with your point here. That's the ideal outcome. No need to reduce feed rate cuz you have much higher torque through gear multiplication to keep right on cutting bigger chips.
You say HP, I say Torque! Such a fun mostly theoretical debate!
Wish we lived closer. I'd drive you totally bonkers!
What a beautiful machine!
Hey Marc, are you serving up a photo of the motor and a closeup of the data plate on the motor shortly?
Based on this photo alone, I'd say it really could be a 2hp motor. I'd need a better photo to say with more certainty.
Regardless, you have the bones there for an awesome drill press. If you replace that motor with a VFD Rated motor of the same size, add a VFD, and keep all the pulleys for those times you need them, you will be happy happy happy!
You can go with a regular 3phase motor too which costs less, but you won't get the full range of speeds you could get with a VFD Rated motor. It's your wallet.
Your post is great. Please don't take my comments as arguing with you. I'm not. I'm just interested in what you said because you experience with slowing the quill with your hands has been used before as support for why you need more hp.
What is the method by which you change your mills speed? Is it a VFD or a variable speed motor or a sheeve system?
This is important. In my experience, you would never be able to stop or slow a VFD or a sheeve system that way. It would take your wrist off too - albeit not nearly as smartly as a torque multiplied system like your lathe has. However, a regular variable speed motor does NOT have constant torque. The torque and hp BOTH go down with the speed.
On the other hand, if your speed control is via a VFD then I'm gunna have to eat a lot of crow and/or go completely nuts trying to figure out why my vfd system is not like yours. There is no way I could ever slow mine down with my hand like that.
To be clear, I'm mostly interested so I can understand the basis for the experiences that some machinists have on this subject.
Agreed, however I'm speaking in practical terms to give the OP info on how to or not proceed. The 1/20 HP is at the drill (the spindle if you will). That comes from the formula, it torque is constant and you dropped rpm from 2500 to 250. I think you are in agreement with that. What feed rate do you propose to drill a 1/2" hole with 1/20HP? 1/10,000 per rev? The edge any mortal could put on the drill would simply skip over surface, i.e. would not be sharp enough to set up a shear plane for so small a chip.
If you could put an edge on that would take that small a chip, to go through a 1/2" piece of steel @250 rpm would take half an close to half an hour, actual several hours because the drill would need several fine sharpenings. I stand by my statement, no one is going be drilling a 1/2" hole with 1/20th of a HP.
Whatever the feedrate and time, wouldn't you agree that 1/20HP at the drill and ages to drill a hole is suboptimal compared to 1/2 HP (ignoring losses) via a mechanical transmission at the drill and minutes to make the hole? Thats what matters to the OP's inquiry and was my point.
I'll disagree with that, twice
. Torque, HP and rpm are inexorably linked and I think this confusion happens because people get focused on one rather how they change together and in the context of a desired result such as not take hours to drill a hole...or stated another way, see your removal rate (volume /time) go down as you reduce your speed. I also think it is practical not theoretical as the OP was looking for advice on how to achieve a 10x speed reduction and would a VFD be a good way to do so.
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This online calculator relates power required for drilling mode using the inputs shown. So related to ones own machine you have to kind of self monitor min & max rpm range which relates to the input SFM input parameter. Feed (FN) is pulling on the handle. So one could do iterations by adjusting FN to maintain resultant calculated power < installed power. That's probably where the 'you will be there all day on a big hole' comes from, something has to give way. SFM & other parameters I suspect are typical machining correlations, deviating much from those infers you have some specific tooling/knowledge outside the norm.
Here are a few snapshots just varying diameter, drilling aluminum. I believe hole saws & annular cutters do not belong in this calculation because they have a different removal volume vs a 'cylinder' hole volume of a typical drill.
www.machiningdoctor.com
Here are a few snapshots just varying diameter, drilling aluminum. I believe hole saws & annular cutters do not belong in this calculation because they have a different removal volume vs a 'cylinder' hole volume of a typical drill.
Machining Power Calculator and Formulas - Machining Doctor
Not just an Online Machining Power Calculator! Separate calculations for Milling, Turning, Grooving, and Drilling, plus all the relevant formulas.
www.machiningdoctor.com
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Marc Moreau
Marc Moreau
No problem I know you try to help me I have just need to understand. Sometime if I don't see I have a hard time to understand. If I buy a 2 HP 3 phase and keep all my pulley . If I set the pulley for 1500 RPM with the VFD starting at 250 to1500 rpm I should have good torque ? my vision is to don't have to play with pulley all the time. exemple start 1/8 drill step to 1/4 up to3/8 and finally 1/2 do you think this will be good ? if not I have to change 4 time the belt I will be happy.
Let me throw another monkey wrench into the works here.
Torque = Amps x Turns.
What follows is another one of my long winded explanations.
One of the reasons the VFD manuals often say add external fan cooling to the motor if it runs for long periods at low speed under load is because in fact the VFD is designed, like stepper and servo motor drivers to limit the current at the motor specified current. And at low speeds the motor isn't turning fast enough to have the fan push enough air over the motor.
So. A motor like the one in the drill press has a 12A rating on it. Likely about 1.5HP with an 85% efficiency. So under full load at the standard 1725 RPM (I think that's the speed for 60Hz) 12A flows through the windings.
Now add the micro-processor to the mix along with a 1.5HP 3 phase motor. The VFD takes the 220VAC in, converts it to DC of about 310 VDC. The driver transistors to the 3 phase windings are energized with what is called Pulse Width Modulation to create three voltages that simulate 60 Hz signals that are 120 degrees out of phase relative to each other.
To do this they turn on the driver for a particular phase until the voltage reaches where it should be based on the sine wave it's trying to create. Say it's 100 slots in that 16.667 millisecond 60 hz wave. It won't turn on the driver for the full period in each slot: shorter at the lower part of the AC voltage; longer at the peak AC voltage. Perhaps fully on during the whole slot period at the peak voltage. So at 60 Hz and 1725 RPM at the peak of the AC wave form the voltage is also 310V. (I'll leave out the since that the current lags the voltage because we're dealing with an inductor since that will just confuse the point I'm trying to make).
The key thing is the pressure on the windings (voltage) at the peak of the AC waveform is 310 VDC which I get from 1.41414 x AC waveform and this forces the maximum 12A current into the windings.
OK. Let's change the frequency to 10 Hz which is generally the lowest output you'd get from a VFD. Again at the peak of the AC 3 phase waveform, the applied voltage will be 310 VDC. But likely a shorter time since the motor speed and the frequency is lower. We might still have 100 slots but the time the voltage is ON during those slots is shorter.
So why is that? Back to AMPs x Turns = Torque. Apply a voltage to a coil of wire and it takes a certain amount of time to start current flowing because of the inductance of the windings which itself is a result of the number of turns in the coil of wire.
And here's the key point! If the frequency is lower but the time to create 12A current in the winding remains the same then the time each driver in each phase slot is on is just shorter. But the 12A current can still be achieved.
So unlike a DC motor where all you do is reduce the voltage to slow it down, which results in less amps into the turns and therefore less torque, the AC PWM based VFDs try their very best to create a 10 Hz 3 phase waveform that matches the motor plate voltage and current.
And 12A at 10Hz is still the same torque as 12A at 60 Hz which is why HP is a value that just isn't that important.
One more thing. AC 3 phase motors will start to slip once the load is too high and the current is limited to the name plate voltage. That's why VFDs have a minimum frequency output. Lower than that and the motor doesn't work.
What makes AC servos work so well (they really are just 3 phase motors) is that the encoder on the back reports to the controller where the motor shaft is. So initially when tuning it, the waveform can be applied open loop without the encoder. That's how a STMBL AC servo drive kit is first run to make sure the correct phases are connected. Then the encoder is added to the feedback loop and now the AC waveform and current through the windings is tracked and more voltage is applied or reduced to maintain the correct position as the motor turns. I can turn my AC servo spindle motor on the mill at 1 RPM and there is no way I can stop it from turning by grabbing onto it.
Likely not.
You forgot one parameter that is critical in this discussion and applies also to other areas.
Torque = Amps x Turns.
As long as we buy the premise that with a VFD the current through the windings is not the same at 250 RPM as it is at 2500 RPM then your statement is valid!
What follows is another one of my long winded explanations.
One of the reasons the VFD manuals often say add external fan cooling to the motor if it runs for long periods at low speed under load is because in fact the VFD is designed, like stepper and servo motor drivers to limit the current at the motor specified current. And at low speeds the motor isn't turning fast enough to have the fan push enough air over the motor.
So. A motor like the one in the drill press has a 12A rating on it. Likely about 1.5HP with an 85% efficiency. So under full load at the standard 1725 RPM (I think that's the speed for 60Hz) 12A flows through the windings.
Now add the micro-processor to the mix along with a 1.5HP 3 phase motor. The VFD takes the 220VAC in, converts it to DC of about 310 VDC. The driver transistors to the 3 phase windings are energized with what is called Pulse Width Modulation to create three voltages that simulate 60 Hz signals that are 120 degrees out of phase relative to each other.
To do this they turn on the driver for a particular phase until the voltage reaches where it should be based on the sine wave it's trying to create. Say it's 100 slots in that 16.667 millisecond 60 hz wave. It won't turn on the driver for the full period in each slot: shorter at the lower part of the AC voltage; longer at the peak AC voltage. Perhaps fully on during the whole slot period at the peak voltage. So at 60 Hz and 1725 RPM at the peak of the AC wave form the voltage is also 310V. (I'll leave out the since that the current lags the voltage because we're dealing with an inductor since that will just confuse the point I'm trying to make).
The key thing is the pressure on the windings (voltage) at the peak of the AC waveform is 310 VDC which I get from 1.41414 x AC waveform and this forces the maximum 12A current into the windings.
OK. Let's change the frequency to 10 Hz which is generally the lowest output you'd get from a VFD. Again at the peak of the AC 3 phase waveform, the applied voltage will be 310 VDC. But likely a shorter time since the motor speed and the frequency is lower. We might still have 100 slots but the time the voltage is ON during those slots is shorter.
So why is that? Back to AMPs x Turns = Torque. Apply a voltage to a coil of wire and it takes a certain amount of time to start current flowing because of the inductance of the windings which itself is a result of the number of turns in the coil of wire.
And here's the key point! If the frequency is lower but the time to create 12A current in the winding remains the same then the time each driver in each phase slot is on is just shorter. But the 12A current can still be achieved.
So unlike a DC motor where all you do is reduce the voltage to slow it down, which results in less amps into the turns and therefore less torque, the AC PWM based VFDs try their very best to create a 10 Hz 3 phase waveform that matches the motor plate voltage and current.
And 12A at 10Hz is still the same torque as 12A at 60 Hz which is why HP is a value that just isn't that important.
One more thing. AC 3 phase motors will start to slip once the load is too high and the current is limited to the name plate voltage. That's why VFDs have a minimum frequency output. Lower than that and the motor doesn't work.
What makes AC servos work so well (they really are just 3 phase motors) is that the encoder on the back reports to the controller where the motor shaft is. So initially when tuning it, the waveform can be applied open loop without the encoder. That's how a STMBL AC servo drive kit is first run to make sure the correct phases are connected. Then the encoder is added to the feedback loop and now the AC waveform and current through the windings is tracked and more voltage is applied or reduced to maintain the correct position as the motor turns. I can turn my AC servo spindle motor on the mill at 1 RPM and there is no way I can stop it from turning by grabbing onto it.
Dabbler
ersatz engineer
One thing is happening here that is different from "what the manufacturers say" and what the theory behind VFDs would predict.
Despite all the curves, and all the approaches to motor control. The torque actually goes *down* as the frequency decreases. Not a lot but noticeably. Or should I say *usable* torque. Due to slip and heat generation in the motor, it seems that the electrical losses are greater at say, 10Hz, than at say, 25 Hz.
So I use as a rule of thumb that you lose about 10% of your usable torque at 30 hz, and about 25% of the usable torque at 10 Hz. So if you are drilling *really* big holes, You need to consider that in your motor selection, or as I said above, use gearing/belt changes to get the usable torque you need.
Note that *all* my experience is with normal 3PH motors not 'inverter duty' motors.
Despite all the curves, and all the approaches to motor control. The torque actually goes *down* as the frequency decreases. Not a lot but noticeably. Or should I say *usable* torque. Due to slip and heat generation in the motor, it seems that the electrical losses are greater at say, 10Hz, than at say, 25 Hz.
So I use as a rule of thumb that you lose about 10% of your usable torque at 30 hz, and about 25% of the usable torque at 10 Hz. So if you are drilling *really* big holes, You need to consider that in your motor selection, or as I said above, use gearing/belt changes to get the usable torque you need.
Note that *all* my experience is with normal 3PH motors not 'inverter duty' motors.
So its the same torque? I did not dispute that, au contraire, I stated it as is fact. However torque must increase, not stay the same, to have the same power @ 10Hz. Not going to happen without a mechanical transmission or servo. Keep in the mind the application, this is not a conveyor where a VFD works because slowing its speed also slows the rate of work. You don't slow down a machine tool because you want to do less work (need less HP) you do so because the diameter has increased.
Optimal performance maintains power by increasing torque as the speed is reduced. A VFD can't and doesn't do that (agreed a servo can, but that's a different story)...or at least I have not heard of it. Show a spec sheet with torque increasing or HP staying constant as frequency moves down from line.
DPittman
Ultra Member
I used a 1.5hp 3ph inverter duty motor and a vfd to replace my 1.5 hp single phase motor and I kept the original pulley setup. I seldom change belts and really like the setup.
No problem I know you try to help me I have just need to understand. Sometime if I don't see I have a hard time to understand. If I buy a 2 HP 3 phase and keep all my pulley . If I set the pulley for 1500 RPM with the VFD starting at 250 to1500 rpm I should have good torque ? my vision is to don't have to play with pulley all the time. exemple start 1/8 drill step to 1/4 up to3/8 and finally 1/2 do you think this will be good ? if not I have to change 4 time the belt I will be happy.
90 VDC motor with chopped DC / PWM control. Not a VFD, but very vaguely analogous to a VFD. DC narrow pulse width low RPM, motor voltage is high but winding impedance greatly restricts motor current. Low power.
My experience with VFDs is only with centrifugal pumps, and they have very low startup load. But, absolutely clearly demonstrate the RPM / HP correlation. My valve test rig uses a 25 HP centrifugal pump. At 60 Hz, the pump produces 300 USGPM @ 40 PSIG in a 4” line. At 10 Hz, the pump produces about 40 USGPM @ 15 PSIG. Not linear, but pump efficiency is greatly impacted by impeller RPM. But if you think about it as a movement of mass, pounds of water per minute is about as clean a way to demonstrate HP as possible. HP = foot pounds per minute, my pump numbers provide all three variables.