# Drum Sander #2: Calculations

 Blog entry by Andrew posted 07-31-2012 02:50 AM 1329 reads 0 times favorited 2 comments
 « Part 1: Planning Part 2 of Drum Sander series no next part

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.