Their are three variables that all affect centrifugal force, speed of rotation, distance from axle to lure and the mass of the lure.

Speed and distance are fairly easy to understand, the faster the spin, the more outward force (centrifugal force) exists. The larger the wheel radius, the faster the lure is traveling, thus the outward force increases.

Try a simple experiment: tie a length of thread to a weight. Hold the string short and see what speed you have to rotate it (vertically, like the drying wheel) to get it to spin. Now lengthen the string and repeat.

When you increase the length of the string, you do not need to spin the weight as fast as you do with a short string.

When the weight is

**just**able to spin, the outward force (centrifugal force) is equal to the mass of the weight, so the weight essentially becomes weightless at the top of the cycle, but at the bottom of the cycle, the downward force is doubled (outward centrifugal force plus gravity effect on the weight). At this speed, at the bottom of the cycle, the epoxy would be pushed downwards to the back of the lure (back facing out). At the top of the cycle, the epoxy would be weightless and remain on the back of the lure. Clearly not a good situation, as with each rotation, the epoxy would gather to the back of the lure.

The ideal wheel speed for even distribution would be next to zero with no centrifugal forces, but of course, the epoxy would fall and harden before it had a chance to distribute. So what

**is**the ideal wheel speed? Well, you need as many rotations in the epoxies fluid state as possible, to aid leveling, with the least force applied centrifugally, in other words, the solution is going to be a compromise.

But we do not need to get involved in words and numbers like centrifugal forces. In fact, we don’t even have to know the weight of the lure or more relevant, the weight of the wet epoxy. We could let the above condition be 100% and decide what percentage of the above ‘spin’ example would be acceptable. One percent seems like a reasonable place to start. This means that the forces on the epoxy would be 1% heavy at the bottom of the rotation and 1% light at the top of the rotation.

I did some calculations based on the 1% scenario. What this means is that each of these conditions should produce the same epoxy distribution results.

Wheel radius/rpm combinations for 1% centrifugal force condition:

1” radius = 18.8 rpm / 2” radius = 13.3 rpm / 3” radius = 10.8 rpm / 4” radius = 9.4 rpm / 5” radius = 8.4 rpm

6” radius = 7.7 rpm / 7” radius = 7.1 rpm / 8” radius = 6.6 rpm / 9” radius = 6.3 rpm / 10” radius = 5.9 rpm

15” radius = 4.8 rpm / 20” radius = 4.2 rpm / 30” radius = 3.4rpm

Conclusions, it seems that the years of trials and experience of the TU members, are borne out by the calculations, wheels upto 6rpm are good up to 10” radius. What we do not know is at what percentage does the epoxy distribution become unacceptable, a future experiment with a variable speed motor.

For those interested, for more details on the calcs and spreadsheets, PM me your e-mail address. It has been a while since I posted a technical thread, I just had to get it out of my system, I feel better now.

Dave