the size of the motor requires that you calculate the momentum equation.
the angular acceleration of a rotating assembly is:
dw/dt= T/Jtotal
where:
dw/dt is the angular acceleration in radians/ sec2
Jtotal is the total first moment of inertia of the assembly in kg.m2
T is the torque in N.m
In your circumstance the total first moment of inertia is the sum of smaller factors, the first moment of the ballscrew, the first moment of the armature of the servo and the first moment of the linearly accelerationg axis mass
refered to the rotating ballscrew
Jtotal=Jballscrew + Jservo + Jlinear
Jballscrew= 1/2. rho. r4. PI. l
where:
r=radius of ballscrew in m
rho=density of steel (aprrox 8000kg/m3)
l= length of ballscrew in m
Jservo is published by the servo manufacturer. For example my 750W Delta B2 series servos have a first moment of 1.16 x 10-4kg.m2
Jlinear= 1/40 p2. m
where:
m= axis mass in kg
p is the pitch of the ballscrew in m
I will start by reviewing the calaculation I made for my machine. I have 32mm 5mm pitch C5 ground ballscrews. They have a travel of 350mm but are 700mm long. The double nut reduces the travel somewhat.
These screws would be similar in size to what you would require for a 300mm travel. I have 750W delta B2 servos, as above. I assumed a linear axis mass of 150kg. The cast iron bed is 115 kg, and the balance is the
weight of the vice and the part.
Jballscrew= 0.5 x 3.141 x 0.0164 x .7
=5.76 x 10-4kg.m2
Jservo=1.16 x 10-4kg.m2
Jlinear=1/40 x 0.0052 x 150
=0.94 x 10-4kg.m2
Jtotal= (5.76 + 1.16 + 0.94) =7.86 x 10-4kg.m2
This is an exceptionally important result, and one you must understand, the rotational momentum is dominated(72%) by the ballscrew....not the linear axis mass(12%).
If you choose to use a 32mm diameter screw then it is likely that the momentum will be dominated by the screw, not the 900kg you pile on the top!!! In this regard pippin's comment about a 32mm screw being
overkill warrants consideration. The momentum of a screw is proportional to the fourth power of its diameter. Thus if you were to decrease the size of the ballscrew to 20mm (rather than 32mm) the
ballscrew momentum would diminish by 85%!!!!
The first moment of the linear axis mass is proportional to the second power of pitch...so pitch is an important design parameter also but much less so than the ballscrew diameter.
Lets say you use the same servo and ballcrew as in my machine (32mm diameter, 5mm pitch,700mm long and 750W Delta B2) and recalculate for 200kg axis mass and 900kg axis mass respectively.
Jtotal,200kg axis= (5.76 + 1.16 + 1.25) x 10-4kg.m2 ......so the linear axis is only 15% of the total momentum.
Jtotal,900kg axis=(5.76 + 1.16 + 5.52 ) x 10-4kg.m2........so the linear axis mass is now 44% of the total momentum.
Despite a heavy axis mass the rotating parts (the ballscrew and servo) still dominate the momentum. Is this what you were expecting?
All you have to decide now is what you want your acceleration to be and that will determine the torque required and therefore the servo required. Lets imagine you chose the Delta B2 as I have....what
would the acceleration be with the 900kg axis, the worst case?
dw/dt= 2.4 / 12.44 x 10-4.....where the rated servo torque is 2.4Nm
=1930 radians/sec2
This corresponds to a linear acceleration (determined by the pitch) of:
accel=1.53 m/sec2 or .15g
Note that this is the acceleration achieved with just the rated torque of the servo....not its overload rating which would be 0.45g!!!
My machine is somewhat lighter than yours and I get 0.27g at rated servo torque, but that is too scary fast for me. I dial it back to 0.15g and it's still scary bloody fast. Thus depending on what you want
to do even a 900kg axis can be accelerated by a 750W servo to 0.15g with a 32mm diameter screw and 5mm pitch. Is that enough?