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Slowing Down a Bug
Bunnell Double Speed
TAC Hole-In-The-Wall Bug
Lionel J-36 Bug
How to Slow Down a Bugby Walt Fair, Jr., W5ALT
I've seen a lot of debate about how to slow down a bug. There is often an argument whether it's better to add weight or to lengthen the arm. So, in an effort to beat this subject into the ground and answer the question once and for all, here are my observations on the subject - for whatever they are worth.
Actually a bug is an example of a fairly simple mechanical oscillator. As I recall from Engineering Mechanics classes way too long ago, the equation for the frequency of vibration is similar to that for a pendulum and inversely proportional to the square root of the mass times the length. (It's an interesting exercise in basic physics to see if you can figure out why a pendulum frequency doesn't involve the mass, while the spring oscillator does involve mass.) It's also interesting to observe that the bug design involves a fairly ingeneous means to minimize friction and, since the dit contacts are usually on a subsidiary spring, they provide a little restoring force to help maintain isochronous oscillations in the short term.
As a first approximation, looking at a bug, the mass is the combination of the mass of the weight and the mass of the arm, while the length is the center of gravity for the weight and arm. So lets say the bar weighs about an ounce and the weight is 2 oz, the bar is 4 inches long (from the place where it flexes) and the weight is at the end. In that case the total mass would be 3 oz and the center of gravity would be [(1 * 2) + (2 * 4)]/(1 + 2) = 3.33 in, so the frequency would be proportional to 1/sqrt(3 * 3.33) = 0.316.
Lets say we double the weight. The total mass is now 5 oz and the center of gravity moves out a little to [(1 * 2) + (4 * 4)]/(1 + 4) = 3.6 in, so the frequency would be proportional to 1/sqrt(5 * 3.6) = 0.236. Thus, doubling the weight would decrease the frequency by about 25%, so if you were sending dits at 30 WPM before, you'd be at roughly 22.4 WPM by doubling the weight.
Now lets extend the arm to double its original length, from 4 to 8 in, and assume that the extension also doubles the weight of the arm, from 1 to 2 oz. In practice it might well more than double the arm weight, since the extension probably is larger and needs either some overlap or some mechanical means to connect it to the arm, but we'll ignore that for now. We'll keep the weight the same, but move it out to the end of the extension. Thus, the total mass is now (2 + 2) = 4 , instead of 3 oz, and the center of gravity is [(2 * 4) + (2 * 8)]/(2 + 2) = 6 in. Using these numbers, the frequency of oscillation is proportional to 1/sqrt(4 * 6) = 0.204, or about 65% of the original. Therefore, if the bug originally sent dits at 30 WPM, it would now send dits at about 19.5 WPM, which is slower than the 22.4 WPM when the weight was doubled.
So, as we can see, extending the arm had the effect of both adding length and weight, but increasing the weight very slightly affected the equivalent length. Since the dit rate (oscillation frequency) depends on the product of the length and the weight, extending the arm can be more effective, since it affects both length and weight to a greater extent. Of course, extending the arm and adding weight can cause even larger effects.
Please note that I didn't actually take my bug apart and measure the weight of the arm, but if someone is prone to do that, the figures could be improved. I do think that the numbers I used are in the right ballpark, so the conclusions should be fairly good. The net result is that you can either add weight or extend the arm, but extending the arm will probably give more results.
Now, have I beat this subject into the ground enough yet?