[Oldmanandhistoy]

Could some one point me to some information about calculating what power I have going into my cutting tool?

So far the only information I have is the motor is 350W and 240Vac I think, two mechanical speed reductions and a variable electronic speed control. It’s the X2 mill and I would like to know what power I have to calculate feeds, speeds and depth of cuts using ME Consultant 2.0.

As always any help or advice is very much appreciated,

John

[MicroMill]

I think 746 watts generally = 1 HP. So you starting off with just a shade under 1/2 hp.

The voltage is really not a factor. Of course you can calculate current drawn by the motor using the formula: I (current) = P (watts) / E (voltage). Nice to know information if you are wiring it into a breaker panel, but useless in your problem solution. Here's a handy link. http://www.the12volt.com/ohm/ohmslaw.asp

[bruno.chaparro]

Hi,

firts what do you want to cut? Steel, Aluminium, foam, wood?
second: dimension of maximum mill tool (10, 20, 40 mm)
thirth: deep of each pass (2, 3, ...10 mm)

Bruno

[gar]

070528-0920 EST USA

Oldmanandhistoy:

Do you want to measure the power going into your cut, or do you want to do a theoretical calculation of what that power is?

To measure the total power input into the work, tool, and chips measure torque applied to the chuck or collet and the spindle RPM. Then calculate the horsepower from the equation HP = T * RPM/5252 where T is in #-FT.

By measuring torque at the chuck you eliminate all losses up to that point. Windage of the chuck and part is the principal loss not eliminated.

To get the distribution of power into each of the three components - part, tool, and chips - is more difficult.

If you can not measure torque at the chuck, then you need to measure power input further back in the electromechanical path, and estimate the efficiency to get the power at the part-cutter interface.

The furthest back point is the AC power input to the motor drive. You need a wattmeter for this. Volt * current is not a real good estimator.

If you want to run experiments on relative cutting efficiency of different tools, and materials, then input wattmeter measurement would be relatively satisfactory.

If you hold speed constant then power is a good measure. If you want to make comparisons at different spindle speeds, then use energy per unit volume of material removed. Energy is the time integral of power.

.

[Oldmanandhistoy]

Hi all and thanks for replying,

I have attached an image from ME Consultant below.

I have as yet not used it in practice just learning the ropes. Looking at the picture where I have circled in red it gives you the power needed at the cutting tool (I think) to do the cut as it is set up in all the other boxes. So if I know what power is getting to the cutter I can make the necessary changes to keep power needed within the range of my spindle and machine. I think I will have to make some changes to the program to fine tune the setting to give more accurate figures but I am hoping they will be a good starting point.

John

[Oldmanandhistoy]

A quote from the help document

“HP
This value can't be changed directly.
It shows the machine power, expressed as horsepower, required to satisfy the current machining parameters.
The calculated HP value is based strictly on the material removal rate for the active material. No adjustment is made for the type of tool being used.”


John

[Oldmanandhistoy]

Sorry about the multiple posts.

First I need to know if a motor is rated 350Watts; what the actual power is coming out of the motor.
Then I can work out from the mechanical speed reduction what will be getting to the cutter.
So the only thing I need to know now is how much power is lost using electronic speed controllers?

John

[Kipper]

Sorry about the multiple posts.

First I need to know if a motor is rated 350Watts; what the actual power is coming out of the motor.
Then I can work out from the mechanical speed reduction what will be getting to the cutter.
So the only thing I need to know now is how much power is lost using electronic speed controllers?

John

Don't forget the mechanical losses and the fact that manufacturers often quote wrong numbers. Personally I don't care about the figures and just use RPM/type of material/depth of cut/dia of tool and let Mach do the hard work then increase the speed till it cuts nice

[gar]

070528-1202 EST USA

Oldmanandhistoy:

You will never get more power out of any system than you put in. The law of conservation of energy. However, in a mechanical system you can change the mechanical advantage from input to output. Thus, you can adjust your gear ratio to get more torque, but not more power. In fact as you increase the torque output by increasing the gear ratio you will get less power output because of increased inefficiency.

If your motor is rated as 350 W of mechanical output power on a continuous basis at your ambient temperature, then that is 350/746 = 0.47 HP out of the motor. Now assume 50% to get to the actual cut and you are at about 1/4 HP.

If you want a deep cut, then gear down and feed slower. But this may not be optimum. Suppose you want to cut Rc 60 with a CBN bit, then you want a lot of heat at the cutting interface.

.

[MicroMill]

Oldmanandhistoy:

Find an AC current meter and measure you input current to the motor. Measure the actual input voltage and perform the power calculation I gave you. 80% is probably a more realistic figure to apply for a power factor. As you see, the program has already factored in the efficiency. The .8 is of course 80%. Finding a "power" meter is easier said than done. You get into peak power vs RMS power vs 3 phase vs single phase, and on and on. For your basic calculation, an AC current meter works fine. As far as calculating all the ancilliary components, the sum of all the losses of power, cannot be more than what you started with.
That program looks pretty slick.:rainfro:

[Oldmanandhistoy]

Thanks for the help

If anyone would like to give ME Consultant a try; you can get it here as freeware http://www.freedownloadscenter.com/B...onsultant.html


I suppose the only information I really needed was; how does an electronic speed controller affect torque?

I will go and do some searching for this particular information; I would guess it’s out there some where.

John

[Al_The_Man]

John, This may answer some of your questions as I assume you are using a Universal motor with Triac or SCR control which is tyical for these motors.
http://www.faqs.org/faqs/woodworking/motors/
Al.

[Oldmanandhistoy]

Thanks for the link Al; very interesting read.

John

[gar]

070528-1436 EST USA

Oldmanandhistoy:

Electronic speed control and torque. You need to define your motor type and something about the speed control.

Fundamentally the primary motor limitation for output power is temperature rise in the motor, and in turn the peak absolute temperature of the maximum hot spot in the motor windings.

In a DC motor most of the heat loss is from current in the armature and the DC resistance of the armature. Torque is proportional to armature current in a permanent magnet DC motor, or for fixed excitation in a wound field shunt motor. Thus, on the surface it appears that torque can be a constant for any speed. This is true if you use an external fan to force air thru the motor. However, for an internally cooled motor the maximum steady state torque has to be reduced at lower speeds because there is less air circulation from the internal fan.

The AC motor is a more complex problem to describe.

Note: the power input to a motor is losses plus mechanical load. If no mechanical load, then the only input power is to supply loss. At full mechanical load the input power is the mechanical power output plus losses which are different than the no-load losses.

Electronic speed controllers can have built in torque limiting. Depending upon the design this limiting can have almost any reasonable shape vs speed.

It is not generally the speed controller that determines torque it is the mechanical load you put on the motor if this is below the system ratings for the point at which you are operating.

You may still want to more clearly define your question.

.

[Oldmanandhistoy]

Some more information

On the motor plate

TYPE 95D-1
350w
DC 230V
2.2A
6000 RPM

SPINDLE SPEED
LOW 0~1100 RPM
HIGH 0~2500

VOLTAGE 220~240V AC
FREQUENCY 50Hz/60Hz

As for the type of electronic speed controller; I don’t know but will try to find out.

John

[gar]

Oldmanandhistoy:

Nameplate helps.

If this is a PM (permanent magnet) motor, then the two different spindle speed ranges have to come from gearing or belts.

If this is a PM motor the 2.2 A rating makes the rated input power 230 * 2.2 = 506 W. Rated power output is 350 W or 0.47 HP. This means the motor is about 69% efficient at full load. For a PM motor a simple DC ammeter in series with the motor armature will give you a reading proportional to motor shaft load torque. The two different speed ranges will be provided by mechanical means and the spindle torque will depend on the ratio. So the meter has to have two scales. One for each of the two different speed ranges.

If the motor is a wound shunt motor, then by changing the excitation to the field you can change the speed range. The higher speed range is obtained by lowering the excitation, but this lowers the torque for a given armature current. Roughly Torque = Constant * Current * Field excitation, while roughly Speed = Different constant * Armature Voltage / Field excitation.

.

[Oldmanandhistoy]

Oldmanandhistoy:

Nameplate helps.

If this is a PM (permanent magnet) motor, then the two different spindle speed ranges have to come from gearing or belts.

If this is a PM motor the 2.2 A rating makes the rated input power 230 * 2.2 = 506 W. Rated power output is 350 W or 0.47 HP. This means the motor is about 69% efficient at full load. For a PM motor a simple DC ammeter in series with the motor armature will give you a reading proportional to motor shaft load torque. The two different speed ranges will be provided by mechanical means and the spindle torque will depend on the ratio. So the meter has to have two scales. One for each of the two different speed ranges.

If the motor is a wound shunt motor, then by changing the excitation to the field you can change the speed range. The higher speed range is obtained by lowering the excitation, but this lowers the torque for a given armature current. Roughly Torque = Constant * Current * Field excitation, while roughly Speed = Different constant * Armature Voltage / Field excitation.

.

Thanks Gar, now we are getting there.

The two ranges are indeed gearing.

I now think I have all the information I need to do the calculations.

Thank you for your time and help it is very much appreciated

John

[nh_eng]

Measuring motor power using a wattmeter is not so simple, as the X2 has a DC motor with loss between the AC supply side and the actual power into the motor. An even bigger factor is that the current into the motor is a function of load. Until you have the motor under maximum load, you will not draw max current. In other words you can only measure max motor power when motor is at max load.

It is not clear if the 350W rating of the X2 is power into the motor (electrical watts) or power out of the power (mechanical watts). From my experience, I would guess it is power in. Fractional horsepower motors such as this are generally in the 75-85% efficiency range. Losses in the gear train are difficult to measure, but I would estimate about 65%-70% efficiency. This would give a net output of about 50%, or 175 watts at the cutter. JMHO.

[gar]

070530-0658 EST USA

nh_eng:

In the initial post, #1, the motor was not identified as a DC brush type.

A wattmeter at the input of the motor control will include the losses of the controller as well as the motor load as you point out. Move the wattmeter to the motor input and it will measure the motor input. Put a torque tansducer on the motor shaft and an RPM indicator and you can measure the power output.

In post #15 we got the nameplate information on the motor. From this it became clear that the rated motor power output was 350 W. In post #16 I calculated the motor efficiency at about 69% from the nameplate information.

.

[smabhyan]

Dear John,

I have a pdf file about formulae for machining power calculations for various materials & processes.
If you send me your e-mail ID, I will send them to you.

Sanjay