Pete C. wrote:
Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.


I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

Well, maybe. I have very underpowered 1/8 Hp continuous motors on my
Series I Bridgeport.
Yes, it rounds off corners a bit when trying to make a sharp corner at
higher speed. I really don't mind that, it rarely can even be detected.
I calculated that neglecting losses, I can deliver about 750 Lbs of
linear force to the table. It is able to snap off a 3/8" end mill
without the servo amps faulting or getting a following error. (That's
why I now do setup at lower speeds - the machine is faster than my eyes
can detect interference.)

Iggy mentioned 500 - 600 Lbs force, that might be iffy on a Series II.

Jon

On 2010-06-29, Jon Elson wrote:
Pete C. wrote:
Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.


I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

Well, maybe. I have very underpowered 1/8 Hp continuous motors on my
Series I Bridgeport.
Yes, it rounds off corners a bit when trying to make a sharp corner at
higher speed. I really don't mind that, it rarely can even be detected.
I calculated that neglecting losses, I can deliver about 750 Lbs of
linear force to the table. It is able to snap off a 3/8" end mill
without the servo amps faulting or getting a following error. (That's
why I now do setup at lower speeds - the machine is faster than my eyes
can detect interference.)

Iggy mentioned 500 - 600 Lbs force, that might be iffy on a Series II.

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

Now, I contend that 300 lbs is a lot of force when it comes to milling.
It is far more than I am able to exert on my manual Bridgeport.

I may have an error in my calculation. I just did an experiment. I ran
my X table at 30 volts. It moved at appx. 1 inch per second at 1.45
amps.

I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.

i

On Jun 29, 1:14*am, Ignoramus28517 ignoramus28...@NOSPAM.
28517.invalid wrote:
....
I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.
i

At the maximum handle force I would apply to a 1/2" end mill, a spring
scale between the spindle and table reads 60 lbs.

jsw

On 2010-06-29, Jim Wilkins wrote:
On Jun 29, 1:14?am, Ignoramus28517 ignoramus28...@NOSPAM.
28517.invalid wrote:
....
I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.
i

At the maximum handle force I would apply to a 1/2" end mill, a spring
scale between the spindle and table reads 60 lbs.

Makes perfect sense to me.

i

On Jun 29, 9:18*am, Ignoramus8895
wrote:
On 2010-06-29, Jim Wilkins wrote:
...
At the maximum handle force I would apply to a 1/2" end mill, a spring
scale between the spindle and table reads 60 lbs.

Makes perfect sense to me.

i

This is the spring scale, much capacity for little money:
http://www.harborfreight.com/big-game-scale-65613.html

You can multiply the range with rope and pulleys at the cost of less
accuracy from pulley friction.

jsw

Ignoramus28517 wrote:

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

These calculations based on slope of the screw thread make my head hurt.
A leadscrew can be made equvalent to a drum with a rope around it.
For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This
assumes the rope is infinitely thin.) Then you can convert torque to
linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =
40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the
leadscrew. 80/0.0318 = 2513 linear pounds force on the table.
I **think** I have done this calculation correctly, I have been using
this scheme for a decade.

Jon

On 2010-06-29, Jon Elson wrote:
Ignoramus28517 wrote:

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

These calculations based on slope of the screw thread make my head hurt.
A leadscrew can be made equvalent to a drum with a rope around it.
For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This
assumes the rope is infinitely thin.) Then you can convert torque to
linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =
40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the
leadscrew. 80/0.0318 = 2513 linear pounds force on the table.
I **think** I have done this calculation correctly, I have been using
this scheme for a decade.

Jon, your calculation makes some intuitive sense. Yesterday at night,
I realized that I forgot 2*pi in my calculation, which would then
yield a number similar to yours.

i