drill press force multiplier?

July 18, 2006 8:41 AM

Power Tool Filter: Is there some kind of 'force multiplier' in effect when using a drill press? i.e. when you put two pounds of pressure on the handle, is there more pressure applied to the work surface? Perhaps most importantly, how hard would this be to calculate?

Please note: I'm not an engineer, just a carpenter-y type! I'm not looking for a perfect calculation, just a way to approximate. (I've *heard* that there is a 6 to 8 times multiplier, but I'd like some more confirmation)

posted by schwap23 to technology (6 comments total)

It'll be the distance the you move the handle divided by the distance the drill moves.
posted by cillit bang at 8:53 AM on July 18, 2006

Yes. The handles are attached to a pinion which drives a rack attached to the quill. The force multier is the difference between the radius of the pinion gear and the length of the handle.

To measure: Pull the handles through one complete revolution while measuring the diplacement of the chuck(D). Multiply the length of the handle(R) by 2pi. The Ratio of 2piR:D will give you the force multiplier minus the strength of the return spring.
posted by Mitheral at 8:54 AM on July 18, 2006

Guesstimate how far the handle moves, and how far the drill bit moves, for a given movement. (You can use the entire range of motion, if that's most convenient, or a lesser movement of the handle, as long as you're measuring the movement of the drill bit over the same range as what you're measuring for the handle itself.) The ratio between these two distances is also the ratio of force between the two.

So, if you move the handle roughly 12 inches, and the bit moves 2 inches, that's a sixfold difference, and the force you apply to the handle is multiplied six times at the tip of the drill bit. If you want to get more accurate, just measure the distances more accurately.

on preview: what cillit bang said.
posted by DevilsAdvocate at 8:54 AM on July 18, 2006

There's a gear on the shaft that the handle is on. The gear moves a rack connected to the spindle (and hence, the chuck). That's pretty much one-to-one, though. The real increase in force is due to the leverage from the length of the handle. Longer handle = more force.
posted by Kirth Gerson at 8:57 AM on July 18, 2006

Just a guess here, but my first impression is yes. You could either take it apart, and have a look at the gears, or you could calculate the distance travelled by the arc of the arm (using the length and some formula involving pi) and then measure the distance traveled by the bit during a measured amount of arc travel and you would end up with a fraction which would be applicable for whatever force you applied to the arm.
Example Arm arc = 5 inches, Bit travels 2.5 inches downward so.... fraction is 2/1 or 2x so.... 10lbs force on the arm gives 20 lbs force at the bit.
Just a guess though. I may be talking out my ass.

On preview... Props to Cillit Bang for saying the exact same thing in one sentence. I am truly humbled.
posted by TheFeatheredMullet at 9:01 AM on July 18, 2006

cillit bang does indeed have a succinct explanation, but I liked the example given by DevilsAdvocate. At any rate, that was the one that made me slap my forehead from the obviousness of it all!

FYI: the handles are about 6 inches long, and there is 2 1/2 inches of travel on the quill (as indicated by the scale at teh base of the handle). I'm rounding quite a bit here, but I'm going to call that as a 10 to 12 times multiplier. Yikes!

The good stuff always comes down to levers, doesn't it?
posted by schwap23 at 10:51 AM on July 18, 2006