How much spindle HP do I need to do this?

[cnczoner]

I'm machining aluminum, mostly 6061, and mostly cutting holes/details in 1/4" to 1/2" plate.

I've been doing this with a 1.5hp motor at 1500 rpm, at ~20ipm feedrate, up to 0.1" D.O.C. with a 3/8" or 1/2" endmill. And I'm happy with this cut rate. But I need more spindle speed (as recommended by every tool-bit source for a better finish) so I'm considering a motor swap. (110V, 1.5hp VFD's are not easy to find, but a 3600 rpm motor will get me 3000 spindle rpm).

My mill is larger than an X3, which has something like 3/4 hp or 1 hp and runs at 2000rpm. But having more table travel should have little to do with spindle HP. So going back to the basics of what I'm trying to do and the rate at which I need to do this, how would I back-calculate the motor hp required? I'm sure I've seen formulae for this in the past, but can't find them now.

But why would I step back the HP? Because on spindle startup I occassionally trip the breaker in my garage (20A outlet/breaker from the garage-door opener), and I've just moved to an apartment with a garage but don't have access to the breaker panel. The first time I trip this breaker, I'll be hearing from the apartment management, and probably get banned from doing this anymore, just due to their lack of understanding of what a mill is. 1 hp would significantly ease the current requirements. And on a related note, the apartment rules explicitly forbid me from drilling for oil or refining oil on the premises!

Cheers,
-Neil.

[Al_The_Man]

Originally Posted by cnczoner I so I'm considering a motor swap. (110V, 1.5hp VFD's are not easy to find, but a 3600 rpm motor will get me 3000 spindle rpm). But why would I step back the HP? Because on spindle startup I occassionally trip the breaker in my garage (20A outlet/breaker from the garage-door opener),.

1ph VFD's are virtually non-existent for good reason.
If you go with a VFD, 1ph 240v in, 240 3phase out, you do not necessarily need to cut back on HP as the accel rate can be programmed so you do not get the normal high inrush of a direct connected motor, just need to be 3 phase motor.
Also with a 4 pole motor you can double the motor rpm by setting the VFD to 120Hz max.
Al.

[cnczoner]

Originally Posted by Al_The_Man 1ph VFD's are virtually non-existent for good reason. If you go with a VFD, 1ph 240v in, 240 3phase out, you do not necessarily need to cut back on HP as the accel rate can be programmed so you do not get the normal high inrush of a direct connected motor, just need to be 3 phase motor. Also with a 4 pole motor you can double the motor rpm by setting the VFD to 120Hz max. Al.

True, and I was looking for exactly that, but now that I moved, I don't have 220V in the garage. I've heard that I can get a 1ph 110Vin, 3ph 220Vout VFD, but the current requirements would still be the same at the input, or worse due to another device in the system (the VFD) which would definitely be less than 100% efficient.

[toastydeath]

max radius of cutter = .25 in

Area of a circle -
pi * .25^2 = .196 in^2

volume per minute -
.196 in^2 * 20 ipm = 3.92 in^3/min

.3 hp per in^3/min in aluminum -
3.92 in^3/min * .3 hp = 1.176 hp (est. sharp tool)
3.92 in^3/min * .4 hp = 1.568 hp (est. dull tool)

So, 1.176 HP is required, minimum, for this cut. As the tool dulls, the power requirement will increase and the spindle will stall. If you use a 1 hp motor, you will have to back off the feed rate.

1 hp / .4 = 2.5 in^3 of aluminum per minute, max
1 hp / .3 = 3.33

2.5 in^3/min / .192 in^2 (plunging area of 1/2" endmill) = 13.02 in/min feedrate for a 1hp cut with a dull cutter.
3.33 / .192 = 17.35 in/min sharp cutter

So there you go. Rough values for what you'll need to do. You should probably aim for a 15 ipm-ish value.

[neilw20]

Toastydeath

Aluminum .3HP per in^3/minute.
Thats nice and easy.
What figures do you use for other materials?
What will give me blue chips in K1040 without coolant?
Using coolant cause instant heat treatment of chips and HP goes up by about 3 to 5 times as soon as you cool it!. Air is great.

Thanks in advance.

[toastydeath]

neilw20 -

You'll have to find an HP chart that combines both various steels and hardnesses to find that answer, I just know the aluminum numbers off the top of my head. Steels are usually in the 1.5 - 3.0 in^3/hp range, annealed. I believe machinery's handbook has such a chart, but my copy is at work.

I'm sure you know this, but for the sake of others reading: Blue chips are a function of the spindle RPM/surface speed, not the feed rate. Upping a cut from 10 ipm to 20 ipm does little to the heat generated, even though you're now removing twice as much metal. So thus, outside of high speed machining, feed rate and depth are the real two optimization parameters, not spindle speed (which you should pretty much set at whatever the cutter can handle and leave it there). The power required per volume actually drops when you increase feed rate without increasing the speed of the spindle.

So for blue chips, turn your spindle up!

[neilw20]

Thanks.
Did you know there is an electronic version of Machinery's Handbook.
It is a bunch of PDF's - about 33 Meg, and coolant doesn't damge it.(LOL)

[skmetal7]

Originally Posted by Al_The_Man 1ph VFD's are virtually non-existent for good reason. If you go with a VFD, 1ph 240v in, 240 3phase out, you do not necessarily need to cut back on HP as the accel rate can be programmed so you do not get the normal high inrush of a direct connected motor, just need to be 3 phase motor. Also with a 4 pole motor you can double the motor rpm by setting the VFD to 120Hz max. Al.

what happens to the torque/ HP when you go above 60 hz?

[cnczoner]

Originally Posted by toastydeath max radius of cutter = .25 in Area of a circle - pi * .25^2 = .196 in^2 volume per minute - .196 in^2 * 20 ipm = 3.92 in^3/min .3 hp per in^3/min in aluminum - 3.92 in^3/min * .3 hp = 1.176 hp (est. sharp tool) 3.92 in^3/min * .4 hp = 1.568 hp (est. dull tool) So, 1.176 HP is required, minimum, for this cut. As the tool dulls, the power requirement will increase and the spindle will stall. If you use a 1 hp motor, you will have to back off the feed rate. 1 hp / .4 = 2.5 in^3 of aluminum per minute, max 1 hp / .3 = 3.33 2.5 in^3/min / .192 in^2 (plunging area of 1/2" endmill) = 13.02 in/min feedrate for a 1hp cut with a dull cutter. 3.33 / .192 = 17.35 in/min sharp cutter So there you go. Rough values for what you'll need to do. You should probably aim for a 15 ipm-ish value.

I like the simple math, BUT ... shouldn't the depth of cut be factored in above for the volume calcs? The actual volume would be pi * d^2 * DOC = 0.0196. So volume per minute at 20ipm would be 0.392 cu in / min. And using .4 for a dull cutter, I get 0.157 hp. Or did you factor this in somewhere else in the constants etc?

Cheers,
-Neil.

[cnczoner]

Originally Posted by skmetal7 what happens to the torque/ HP when you go above 60 hz?

From a theoretical standpoint, hp should be similar, but torque would drop. I'm no motor expert so others correct me if this is incorrect.

BTW, prior to posting this, I spoke with various motor sources who told me I could use a "reduced-voltage controller" to provide a soft start, but I'd still need a 3-phase motor to work with that, and both together was in the order of $900. That's why a 1hp, 110V, 1ph motor for just over $100 is very appealing, if it can do the job. Also, they were against pushing most of these motors beyond 90Hz.

Cheers,
-Neil.

[toastydeath]

Originally Posted by cnczoner I like the simple math, BUT ... shouldn't the depth of cut be factored in above for the volume calcs? The actual volume would be pi * d^2 * DOC = 0.0196. So volume per minute at 20ipm would be 0.392 cu in / min. And using .4 for a dull cutter, I get 0.157 hp. Or did you factor this in somewhere else in the constants etc? Cheers, -Neil.

I read that to be you were plunging, not doing straight line milling. If you are milling, you do not include the area of the face in any way as that doesn't affect the volume. If you are plunging, it doesn't matter how far you are going in depth total, just that you are going there at a certain rate.

The volume for full width slotting is the radial depth of cut and the axial depth of cut. So, for a full width slot at .1" deep on a .5" cutter:

.1 * .5 * 20 = 1 in^3/min
1 * .3 = .3 hp

If you did area of the cutter face (in^2) * depth of cut (in^1) * inches a minute (in^1/min), you'd be into inches^4/min, and that's not what we're looking for.

[LeeWay]

Have you considered DC motors and controllers? Some of these are quite adjustable with speeds, torque, acceleration etc? I am working on one with a fairly large treadmill motor and once I get the 220 VAC hooked in, should produce about 2 HP @ 180 VDC. The the drive I am using now is too small. It only runs on 110 @ 130 VDC. Not getting near the max on the motors rating. Probably half or less.
I have the new drive and will be hooking that up today.
I know though, that living in an apt. 220 hookup will likely be pretty difficult to lay your hands on. If you have a clothes dryer connection or an electric range, then you could build a nice extension cord.
You could do this type setup for about $200 with cord.
I'll let you know how well the 220 works.

I have another question regarding the math symbols in the formula's above.
I found this link that has some math symbols. I do not see what this is a symbol for. ^
Thanks.