Re: servo sizing

--- In CAD_CAM_EDM_DRO@egroups.com, mgrady <mgrady@i...> wrote:

>
>
> thinking about using geckos servo drives to put on
> a small lathe (old hc chucker) or maybe a mill later on
> from what i have read here servos have a different power curve
> than steppers
> what size motors would be used on small lathes and bridgeport
> size mills
> also sense you can use regular brush dc motors
> and these are marked with hp or volts and amps
> is there a conversion from hp/watts to in/oz
>
> hopefully gecko will make a nice ref on servos like they did on

steppers

>

mike

Mike,

Hi, per your request this is another long-winded reply. Any comments
regarding corrections, (be they grammatical or otherwise), omissions,
mathematical errors, confused explainations, etc. would be greatfully
appreciated. I have used as little math as is possible. Here goes..

DC servomotors have a parabolic power vs. torque curve, power being
defined as speed times torque. Power output is zero at no-load and
when stalled. Power output peaks at ½ of stall torque, which is
also ½ of no-load speed.

Another characteristic is motor efficiency cannot exceed 50% at peak
power output. Efficiency nears 100% for very small loads and falls to
zero when stalled.

At peak power, for every watt the motor delivers to the load, it
generates an equal watt of heat. Nearly all of it is in the form
of "I squared R" losses in the armature resistance. The
armature has
a very poor thermal path to the ambient where this heat must be
dissipated. It would rapidly burn up if asked to continuously deliver
peak power.

Consequentially the motor has a rated continuous torque, which when
multiplied by its speed, is its maximum continuous power output. This
varies from motor but is about ½ of the motor's peak power.

The following equation (Eq. 1) calculates the motor power output in
Watts:

W=(1 – T/S)N*T/1351.8 (Eq. 1)

Where:

W = Watts
T = Torque applied (oz-in)
S = Stall torque (oz-in)
N = No-load speed (RPM)

The following equation (Eq. 2) calculates the motor speed in RPM for
any given applied torque:

RPM=(1–T/S)N (Eq. 2)

Where again:

T = Torque applied (oz-in)
S = Stall torque (oz-in)
N = No-load speed (RPM)

For peak power Eq. 1 simplifies into:

W=S*N/4*1351.8 (Eq. 3)

Also you can get the same answer from W=(Rated VDC X Stall
Amps)/4
Or if you know the armature resistance, W=(VDC squared/R)/4

Motor efficiency can be calculated as:

E=1–T/S (Eq. 4)

Where:

E = Efficiency in percent
T = Torque applied (oz-in)
S = Stall torque (oz-in)

Current draw for the motor can be calculated as:

I=(Is/S)*T (Eq. 5)

Where:

I = Motor current in Amps
Is = Stall current in Amps
S = Stall torque (oz-in)
T = Torque applied (oz-in)

As a practical example, a typical size 23 DC servomotor has the
following specs:

Rated Voltage 60VDC
Stall Current 30 Amps
Stall Torque 400 oz-in
Continuous torque 55 oz-in
No Load Speed 6,000 RPM

Peak power (Eq. 3): W=400 X 6,000/4 X 1351.8 = 2,400,000/
5407.2 = 444 W

The speed will be 6,000 / 2 or 3,000 RPM of course

For continuous power, use (Eq. 1) and make T equal to the motor's
continuous torque rating of 55 oz-in. Use (Eq. 2) to calculate the
speed at that load.

W=((1–55/400)6,000 X 55)/1351.8 =284,625/1351.8 =211W

RPM=(1–55/400)X 6,000 = 5,175 RPM

Now let's put the motor to work, using the previously calculated
values. Use the following identities to define work:

1 HP = 550 ft-lbs/sec
1 HP = 746 Watts

From these identities one can derive force generated at a given speed
for a known value of power:

F=(531* W)/V (Eq. 6)

Where:

F = Force or thrust in pounds
W = Power in Watts
V = Velocity in IPM (inches per minute)


Let's assume a 600 IPM (inches per minute) speed is required.

What force in lbs can this motor generate continuously? Using Eq. 6
yields:

F=(531 X 211)/600 = 112,041/600 = 187 lbs of thrust at 600 IPM

The motor would draw 4.125 Amps from a 60VDC power supply and
dissipate 26.5 Watts as heat.

What force would the motor generate at peak power? What would be the
current draw and motor dissipation?

F=(531 X 444)/600 = 235,764/600 = 393 lbs of thrust at 600 IPM

The motor would draw 15 Amps from a 60VDC power supply and dissipate
456 Watts as heat!! (This is 17 times more heat than the motor can
continuously dissipate; use this power level sparingly).

At stall this motor would deliver 786 lbs of thrust, draw 30 Amps
from a 60VDC supply and dissipate 1800 Watts.

Finally, gearing has to be considered. This is very important because
the motor used in the example delivers the above-calculated power at
a specific speed, i.e. 211 Watts at 5,175 RPM. The motor must be
matched to the machine via an optimal gear ratio to deliver this
power to the load.

Let's assume a 5-pitch leadscrew is used. For it to move
something at
600 IPM it must turn 3,000 RPM (600 X 5). The motor meanwhile
delivers its rated power at 5,175 RPM. From this one gets a 1.725:1
(5,175/3,000) motor to leadscrew gear ratio to get the desired
performance.

All of the above neglects to take into account second-order effects,
the primary one being leadscrew and gearing inefficiencies. As they
say in commercials, "Your results may vary"

Mariss