Hi,

sorry about the delay, I've had a busy couple of days.

Lets do some simple calculations:

1) The first moment of inertia of the X axis ballscrew:

diameter = 32mm or radius =0.016m

Length=686mm or 0.686m

J_{ballscrew}=mass. (radius/2)^{2}

and mass = pi .Radius^{2}.length. density of steel

m=3.14 (0.016)^{2}.(0.686). 8000 (8000kg/m^{3}=denstiy of steel)

m=4.41kg

so:

J_{ballscrew}=(4.41). (0.016/2)^{2}

=2.8 .10 ^{-4 } kg.m^{2}

2) The first moment of inertia of the servo/stepper armature:

Lets assume that you have a 34 size servo or steppper direct coupled (or 1:1 belt drive) and the armature has a first moment of inertia of

1 . 10^{-4 } kg.m^{2}. For instance the spec sheet of my 750W servo list the armature inertia as 1.13 . 10^{-4} kg.m^{2}

so you can see the number I've suggested is a practical number.

Should we decide that we need a gear or belt reduction we will revisit this assumption.

J_{armature}=1 . 10^{-4} kg.m^{2}

3) The effective first moment of inertia presented by the axis mass.

Pitch=6mm or 0.006m

Assume axis mass of 100kg

J_{linear}=(axis mass). (pitch/2.PI)^{2}

=100 . (0.006/2 . 3.141)^{2}

=0.911 . 10^{-4} kg.m^{2}

The total effective first moment of inertia is:

J_{total}=J_{ballscrew}+J_{armature}+J_{linear}

= (2.8 + 1 + 0.911) . 10^{-4} kg.m^{2}

=4.71 . 10^{-4} kg.m^{2}

Take a look at the proportions of the components that make up the total moment, the ballscrew is 60%, the armature is 21%

and the axis is the remaining 19% Clearly the ballscrew dominates the equation. Even if we over (or under) estimate the axis mass,

or over (or under) estimate the pitch of the ballscrew would still introduce minimal error in the total. Note how even the moment of the

armature exceeds the moment induced by the axis mass. This result is the norm and surprises many people, me included when I first

did the calculation, but physics does not lie.

Note that I haven't made any allowance for gear reduction (as yet) nor have I made any allowance for other rotating components like

pulleys and couplers. Making allowances for these items is simple enough but I did not wish to obscure the main ideas in the equation

by small correction terms.

Not also that I did not make any allowance for ballscrew efficiency or friction. Ballscrew efficiency is an easy correction, friction however

is not.

So in trying to keep the calculation as simple and illuminating as possible **without** introducing a whole bunch of small corrections

and confusing everyone.......please accept that this calculation is close, say within 10-20%.

Lets now do the acceleration calculation with a 1Nm torque:

dw/dt=torque/J_{total} (w= angular velocity in rad/sec, dw/dt= angular acceleration)

=1/ 4.7 . 10^{-4}

=2127 rad/sec^{2}

Converting that to linear:

A_{linear}=dw/dt . pitch/2.PI

=2172 . (0.006/2 . 3.141)

=2.03m/sec^{2}

Which is 1/5 of a g.....which is not too shabby!

Note that a standard 400W servo with 3000rpm rated speed will have rated torque of 1.27Nm, so would exceed 0.2g, which

is pretty impressive.

My 750W servo has 2.4Nm rated torque and 7.2NM peak for accels of 0.5g (rated) and 1.4g (peak).

If you want to consider gear reduction let me know and we can rework this calculation.

The Z axis on a knee mill is often very draggy and so leaving out friction is unrealistic when applied to the Z axis. You really

need to replace the trapazoidal screw with a ballscrew if you wish to drive it at any realistic acceleration. Otherwise you will

need to use a really outsize Z axis motor or major gear reduction. Gear reduction means that even having fast X and Y axes is moot because

to stay coordinated they must slow to the pace of the Z axis.

Craig