Are these things safe?
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!!! NO WAY !!!
MAKE & USE AN ENCLOSURE
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You are spinning around a rod at thousands of RPM. If the rod is off center enough and the RPM is high enough, the rod will pull out of the polyethelene arbor and crash into the stationary electrodes and explode. Over time, the friction grip will most likely degrade, and a rotor that was fine last year may fail this year. Also, the arcs give off UV that will damage the eyes.

It is recommended that ANY rotary spark gap of ANY design be enclosed in a box that can withstand a rotor explosion. Make and use an enclosure of at least 3/4" thick plywood. Some would say 1" thick plywood and a double box. If you can make one of these rotary spark gaps, you can make a plywood box. It won't be fun if someone gets hurt. It might be you.

Even 1 mm offset is too much slop in the rotor and this should be minimized. You should be able to get the rotor very close to perfect center. Simply check each side against a sationary electrode, and adjust it until it is nearly perfect. Tap the electrode with a chunk of wood or pry it with a screwdriver, until it is in balance.

Balance issues get much worse as the RPM goes up. That is because of the V^2 term in the force equation (see below). Double the RPM does a lot more than double the forces pulling on the rod. Even though my ARSG has a 15,000 RPM motor, I have never had it much over 3600 RPM. That is 480 breaks per second and plenty for most any occasion. Because of this, I recommend keeping the RPM down and adding more stationary electrodes. If you insist on greater RPMs, I suggest you build a bomb shelter around the thing.

I will attempt to quantify many issues formally, but I cannot even guess at the friction-grip degradation over time.

__Conclusions of the following evaluation__

1) Make an enclosure that surrounds the perimeter of the rotor (at least 3/4" plywood). When the shrapnel hits the enclosure, nobody gets hurt.

2) Take extra care to center the rod. If it is in perfect balance, there will be no force pulling it out.

3) Use more stationary electrodes and less RPM. Speeds over 3600 RPM will uncover any unbalance you have in your rotor.

4) Check the friction grip of the arbor by pushing the rotor against a bathroom scale until the rod moves. I typically get at least 20 pounds of grip.

5) Add more grip. After the rod is centered, scuff it up with sandpaper on both sides next to the arbor, and dab some epoxy on it.

6) Add shaft collars. After the rod is centered, slide on shaft collars with set screws on both sides of the arbor. Keep them tightened.

This is the summary of the data. RPM is across the top and rod missalignment is in mm down the left side.

One of my rotors showed a frictional grip of 20 pounds, so any number greater than 20 lb is a failure.

Not sure exactly where to draw the line, so I'll leave it fuzzy for now.

Use this information to guide what you do.

1800 | 3600 | 7200 | 14400 | |

0.1 mm | 0.04 lb | 0.1 lb | 0.5 lb | 2.1 lb |

0.5 mm | 0.19 lb | 0.7 lb | 2.7 lb | 10.7 lb |

1.0 mm | 0.38 lb | 1.4 lb | 5.4 lb | 21.5 lb |

5.0 mm | 1.94 lb | 6.9 lb | 26.9 lb | 107.6 lb |

Calculations based on a 5/32" x 7" Tungsten rod

D = 5/32 * 2.54 cm/in = 0.40 cm

L = 7 * 2.54 cm/in = 17.80 cm = 0.178 m

Mass of the 5/32" x 7" Tungsten rod

Tungsten density: 19.35 g/cc

Volume of a cylinder: Pi * R^2 * L = 3.14 * (0.2 cm)^2 * 17.80 cm = 2.24 cc

M = 19.35 g/cc * 2.24 cc = 43.25 g = 0.043 Kg

Velocity at the end of the rod

V = Distance / Time = Circumference * (Revolutions / Second) m / S

V = 3.14 * L * (RPM / 60) m / S

Velocity at end of rod at 3600 RPM (120, 240, or 480 BPS)

V = 3.14 * 0.178 * (3600 RPM / 60) = 33.5 m / S

D = 0.180 m (diameter of rotor)

R = D / 2 = 0.09 m (radius of rotor)

M = 0.043 Kg (mass of rotor)

V = 34 m / S (rotor tip velocity at 3600 RPM)

Suppose the rod was off-center by 0.5 cm (which is huge), there would
be an extra 1 cm of mass on one side, that is not on the other.

M (1 cm) = 0.043 Kg * (0.01 m / 0.178 m) = 0.0024 Kg

The centrifugal force pulling on the rod would be

F = MA = MV^2/R = 0.0024 Kg * (34 m/S)^2 / 0.09 m = 31 N (Newtons)

1 lb = 4.448 N

31 / 4.448 = 6.9 pounds of force pulling on the rod

Sounds like a lot. I think that IS a lot. I suspect the rotor would be too out of balance to spin up.

But how bad is it? When I go check one of my rotors, I find that I have to
push on it with over 20 pounds of force (using my bathroom scale), to get
the rod to move. The safety margin in that situation is about 20 / 6.9 = 2.9
factors, or 290%.

So what about a more likely scenario? If the rod is off-center by 1 mm, there
would be an extra 2mm of mass on one side, that is not on the other.

M (2 mm) = 0.043 Kg * (0.002 m / 0.178 m) = 0.00048 Kg

The centrifugal force pulling on the rod would be

F = MA = MV^2/R = 0.00048 Kg * (34 m/S)^2 / 0.09 m = 6.1 N (Newtons)

6.1 / 4.448 = 1.4 pounds of force pulling on the rod

The safety margin in that situation is about 20 / 1.4 = 14 factors or 1400%.

Here are the numbers in a table. As you improve the rod centering, pull goes down, as expected.

Force on non-centered rod at 3600 RPM

5.0 mm | 1.0 mm | 0.5 mm | 0.1 mm |

6.9 lb | 1.4 lb | 0.7 lb | 0.1 lb |

R = 0.09 m (radius of rotor)

M = 0.043 Kg (mass of rotor)

V = 18 m / S (rotor tip velocity at 1800 RPM)

If the rod was off-center by 0.5 cm, the centrifugal force pulling on the rod would be

F = MV^2/R = 0.0024 Kg * (18 m/S)^2 / 0.09 m = 8.6 N (Newtons) = 1.9 pounds

If the rod is off-center by 1 mm, the centrifugal force pulling on the rod would be

F = MV^2/R = 0.00048 Kg * (18 m/S)^2 / 0.09 m = 1.7 N (Newtons) = 0.39 pounds

This shows that at lower RPM, there is very little pull force on the rod.

Force on non-centered rod at 1800 RPM

5.0 mm | 1.0 mm | 0.5 mm | 0.1 mm |

1.94 lb | 0.38 lb | 0.19 lb | 0.04 lb |

R = 0.09 m (radius of rotor)

M = 0.043 Kg (mass of rotor)

V = 67 m / S (rotor tip velocity at 7200 RPM)

If the rod was off-center by 0.5 cm, the centrifugal force pulling on the rod would be

F= MV^2/R = 0.0024 Kg * (67 m/S)^2 / 0.09 m = 119.7 N (Newtons) = 26.9 pounds

That would not stay in the arbor.

If the rod is off-center by 1 mm, the centrifugal force pulling on the rod would be

F = MV^2/R = 0.00048 Kg * (67 m/S)^2 / 0.09 m = 23.9 N (Newtons) = 5.38 pounds

This shows that at higher RPM, pull force on the rod can get too high, if the rod much off center.

Force on non-centered rod at 7200 RPM

5.0 mm | 1.0 mm | 0.5 mm | 0.1 mm |

26.9 lb | 5.4 lb | 2.7 lb | 0.5 lb |

R = 0.09 m (radius of rotor)

M = 0.043 Kg (mass of rotor)

V = 134 m / S (rotor tip velocity at 14400 RPM)

If the rod was off-center by 0.5 cm, the centrifugal force pulling on the rod would be

F= MV^2/R = 0.0024 Kg * (134 m/S)^2 / 0.09 m = 478.8 N (Newtons) = 107.6 pounds

That would not stay in the arbor.

This shows that at even higher RPM, pull force on the rod gets huge, if the rod much off center.

Force on non-centered rod at 14400 RPM

5.0 mm | 1.0 mm | 0.5 mm | 0.1 mm |

107.6 lb | 21.5 lb | 10.7 lb | 2.1 lb |

RPM = 1800, 3600, 7200, 14400

The centrifugal force pulling on the rod is

F = MV^2/R

The force goes up as the square of the velocity (or RPM).

Summarizes how the RPM can affect a somewhat small error in rod centering.

Force on non-centered rod at various RPM

1800 | 3600 | 7200 | 14400 |

0.38 lb | 1.4 lb | 5.4 lb | 21.5 lb |

Go back to the top for the general conclusions.