Out of curiosity, I ran some calculations on power requirements:
If I have a 1.5 HP motor, I can use a force of 2.88 lbs on a 25” piece. This would mean a low removal rate for larger pieces. I could not determine how much of a removal rate. Some sources say they would remove at a rate of 1/128” each pass. This seems reasonable for a 3 lb force. On smaller pieces more force can be applied. A 1” piece would stall the motor at 72.12 lbs. This amount of force applied to the material would seem to dislodge the piece from the tray, or destroy the sanding belt.Assumptions :
- A thin slice of sanding belt would come in contact with the material.
- No losses on the pulleys and belt
When I sand by hand using an orbital sander, I do not apply very much force, just a bit more than the sander weighs. My sander weighs 6 lbs over a surface of 19.25 in2 (5 in diameter). This would result in a force of 0.31 lbs/in2. Assuming I apply an additional 10 lbs of force to the sander, totaling 16 lbs of force, this results in a pressure of 0.83 lbs/in2.
Making an assumption that the drum will deform to allow a transverse contact of 0.25 in and utilizing full force potential before stalling the motor, this would allow for a force of 11.52 lb/in2. Therefore, I believe a motor of 1.5 HP would be sufficient to drive the machine and effectively remove more material than I can using an orbital sander. This would also result in a straighter surface.