## Question:

We need a vibratory finishing machine for deburring parts that are drilled and face ground. The parts are 6x4x½ inches. They are ground on both sides and have ¼-inch and ½-inch diameter tapped through holes. The production rate is 2,000 pph. We prefer a through-feed (continuous) design. What size machine will meet our requirement, and how is that figured? C.J.

## Answer:

You can approach the arithmetic from different directions, but you wind up with the same answers if your assumptions are the same. The assumptions here will be the number of parts/cu ft of mass (media and parts) and the time cycle required to get the desired finish. Let us assume that 15% of the load (by volume) will be parts and that it takes 20 min to deburr the parts.

In general, through-feed vibrators have time cycles equal, in minutes, to about one-half their cubic foot capacity. In other words, a 20-cu ft machine will have about a 10 min time cycle; a 50-cu ft machine will have about a 25 min time cycle. That figures to about two cfm. From the standpoint of time cycle, you can already see that you may wind up with a 40-cu ft machine.

A 40-cu ft machine with a 20-min time cycle will empty itself three times an hour, for a total of 120 cu ft per hour. That will be 15% of 120, or 18 cu ft of parts an hour. Your parts are 12 cu inches each, and that works out to 144 parts per bulk cubic foot. The 18 cu ft equals 2,592 parts, meeting your requirement.

Look at this from another direction: Through-feed machines in the 10-50 cu ft size range will process about two cfm of parts and media. A 20-cu ft machine will process the same volume as a 40-cu ft machine, but the time cycle will only be 10 min, disqualifying the machine for you on the basis of time cycle. Suppose you can stretch the time cycle out to 20 min in that 20-cu ft machine. It will, then, yield 60 cu ft per hour of which 15%, or 9 cu ft, will be parts. That calculates to 1,296 parts per hour—again, not for you.

Very large machines improve on the delivered cubic feet per minute, with 100-cu ft machines discharging up to four or five cfm. This is because the discharge chute is considerably larger. Increases in machine size (and cost) are, however, difficult to justify unless you need the longer time cycles or have parts larger than can be processed in smaller machines. For example, the parts you described could be processed in a 100-cu ft machine. The time cycle might be 30 min, yielding 30 cu ft (4,320 parts) per hour. This is only a production increase of 66% for a machine that is 250% larger. Acquisition and operating cost will be in proportion to machine size. You can also see that production volumes are greatly affected by the ratio of parts to media. We had assumed 15% of the load would be parts. If you can actually achieve 20%, the production rate goes up by 33% and so on for different ratios. Be sure, however, to make valid assumptions or you may have some real surprises when the machine is installed.

There are some so-called multi-pass machines (having concentric or parallel bowl designs) that give longer time cycles without proportionate increases in machine capacity. For example, some machines as small as 20 cu ft can give time cycles up to an hour. These designs are usually limited to smaller parts, and the production volumes are proportionately lower. In this example, the hourly gross output is only 20 cu ft compared with the 120 cu ft in the example for a 40-cu ft through-feed machine. The parts ratio will be about the same in either machine. These multi-pass machines offer a good answer for lower parts volumes requiring longer time cycles. Be aware that the action in these styles is considerably less aggressive, often making the longer time cycle necessary.

**A word of caution:** those tapped holes will present a media lodging problem. Be sure to specify a good media sorting device for undersized and broken media. Some through-feed machines are known for breaking media.