Hi,
the acceleration calculation goes like this:
dw/dt= T / J
total
where J
total is the first moment of inertia of all components, both rotating and linear.
w is angular velocity in radians/s
and T is the applied torque in Nm.
The calculation is really about finding the first moment of inertia.
J
total = J
armature + J
ballscrew + J
linear axis
J
armature is easy, just read the manufacturers spec. For example:
[url=https://www.leadshineusa.com/productdetail.aspx?type=products&category=stepper-products&producttype=stepper-motors&series=cm&model=d57cm31[/url] Object reference not set to an instance of an object.[/url
webpage
This is a typical medium/high torque 23 size stepper and has a J=0.84 x 10
-4 kg.m
2.
You don't say how long your ballscrew is, I assume 300mm:
J
ballscrew =m . R
2/2
where m is the mass of the ballscrew, R is the radius of the ballscrew
m(mass) =l x pi R
2 x rho where rho is the density of steel, 8000kg/m
3
=0.3 x 0.008
2x PI x 8000
=0.482 kg
therefore:
J
ballscrew= 0.482 x 0.008
2/2
=0.15 x 10
-4 kg.m
2
J
linear axis=m . (pitch/2 PI)
2
where m is the axis mass (22lb=10kg), pitch is the ballscrew pitch:
J
linear axis=10 x (0.01/2.PI)
2
=10 x 0.00159
2
=0.253 x 10
-4 kg.m
2
We now combine the individual moments:
J
total= (0.84 + 0.15 + 0.25) x 10
-4 kg.m
2
=1.24 x 10
-4 kg.m
2
Firstly note that the dominant term in the total momentum is actually the armature of the stepper (0.84) followed by the linear axis (0.25), at about 1/3 that of the
stepper alone! The ballscrew is the smallest but not insignificacnt term (0.15). Note also that the first moment of inertia of the ballscrew depends on mass, ie proportional
to the square of the radius and also the radius of gyration, again proportional to the square of ballscrew radius. The net result is that the ballscrew moment is dependent
on the ballscrew radius to the
fourth power. Ballscrew radius ( equivalently diameter) is critical in this calculation.
Lets now calculate the angular acceleration of such a system with 1Nm torque applied:
dw/dt= 1 / 1.24 x 10
-4
=8064 rad/s
2
and given that the axis velocity is related to angular velocity:
v= pitch x w/2 .PI so
dv/dt = pitch x dw/dt/2.PI
= 0.01 x 8064/6.282
=12m/s
2 or 1.2g which is amazing.
1Nm =192 oz.in. Remember that steppers lose torque bigtime the faster they go so if you choose low inductance steppers then you might expect to retain 60% of the torque
at 500 rpm then 192/ 0.6=320 oz.in should be enough torque to accelerate your axis at 1.2 g. Note that if you want thrust in addition to acceleration then you need more torque,
so 500oz.in is looking like your best bet.
Craig