Calculations for Spindle Drive and Tool Release

Reduction

1. Reduction ratio: 50 / 25 = 2
2. Spindle torque: rated 3.18 × 2 = 6.36 N·m, maximum 11.13 × 2 = 22.26 N·m (excluding belt losses)
3. Spindle speed: rated 3000 / 2 = 1500 rpm, maximum 6000 / 2 = 3000 rpm

The maximum speed at 200V is unverified. For spindle orientation, the same motor angle corresponds to two spindle angles (180° apart), so it will work fine as long as the drive keys are symmetrical.

For the pulley strength, see Spindle Pulley Strength Calculation.

Tool Release

1. Cylinder area: A = π × 100² / 4 = 7,854 mm²
2. Drawbar retention force: 280 kgf × 9.81 = 2,746 N (as listed on the sales page)
3. Balancing pressure: 2,746 / 7,854 = 0.35 MPa
4. 3,927 N (400 kgf) at 0.5 MPa, 4,712 N (480 kgf) at 0.6 MPa

Because a spring's force increases the more it is compressed, it might require more than 2,746 N at the end of the release stroke. We will verify this on the actual machine, including friction. The 4.5 mm stroke is the rating on the spindle head, so a 10 mm cylinder stroke will be plenty.

Thrust and Taper

Drilling thrust pushes the holder further into the taper, so it does not compete with the drawbar retention force. What actually matters is the axial load capacity of the bearings, which remains unverified.