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Reason: None provided.

It depends on what you do in terms of what charge is in the middle plate at the start.

If you connect it to ground, it will be negative, if you connect it to an antenna, it will be positive.

if you use arial electrodes for both (a spark gap), whatever residual charge may or may not be there will start it. Bennet doublers are "self exiting".

If there's absolutley nothing, rub your feet on the carpet and touch a plate

The bennet double is an amazing visual. It's the absolute best way to explain the theory of opperation of an electrostatic influence machine, but it is not the best design of one.

Ideally, you want a rotating design that maximises the size of the charge surfaces and allows for high speed rotation as the ammount current produced is proportional to the surface area of charge carriers passing the collection point.

I'm currently working on a barell based design that tries to maximise the effects from Lord Kelvin's Replenisher. I've decided to switch gears and do this instead because contrarotation is mechanically complex. This will be much simpler and way more reliable.

I have a 180 mm diameter barrel with 30 170 mm long 10 mm wide plates on the barrel.

I'm going to be electrically insulating my design more and elimate coronal losses in the machine so that the voltage can go higher than 15 KV.

Because of the barrel design my surface area is huge.

According to the professors equation, I should be able to provide over a milliamp of current at a rotational speed under 1000 rpm, the only question will be how much I can mitigate the coronal losses incumbent in the original design so I get the voltage to climb to where it should actually be.

1 year ago
1 score
Reason: None provided.

It depends on what you do in terms of what charge is in the middle plate at the start.

If you connect it to ground, it will be negative, if you connect it to an antenna, it will be positive.

if you use arial electrodes for both (a spark gap), whatever residual charge may or may not be there will start it. Bennet doublers are "self exiting".

If there's absolutley nothing, rub your feet on the carpet and touch a plate

The bennet double is an amazing visual. It's the absolute best way to explain the theory of opperation of an electrostatic influence machine, but it is not the best design of one.

Ideally, you want a rotating design that maximises the size of the charge surfaces and allows for high speed rotation as the ammount current produced is proportional to the surface area of charge carriers passing the collection point.

I'm currently working on a barell based design that tries to maximise the effects from Lord Kelvin's Replenisher

I have a 180 mm diameter barrel with 30 170 mm long 10 mm wide plates on the barrel.

I'm going to be electrically insulating my design more and elimate coronal losses in the machine so that the voltage can go higher than 15 KV.

Because of the barrel design my surface area is huge.

According to the professors equation, I should be able to provide over a milliamp of current at a rotational speed under 1000 rpm, the only question will be how much I can mitigate the coronal losses incumbent in the original design so I get the voltage to climb to where it should actually be.

1 year ago
1 score
Reason: None provided.

It depends on what you do in terms of what charge is in the middle plate at the start.

If you connect it to ground, it will be negative, if you connect it to an antenna, it will be positive.

if you use arial electrodes for both (a spark gap), whatever residual charge may or may not be there will start it. Bennet doublers are "self exiting".

If there's absolutley nothing, rub your feet on the carpet and touch a plate

The bennet double is an amazing visual. It's the absolute best way to explain the theory of opperation of an electrostatic influence machine, but it is not the best design of one.

Ideally, you want a rotating design that maximises the size of the charge surfaces and allows for high speed rotation as the ammount current produced is proportional to the surface area of charge carriers passing the collection point.

I'm currently working on a barell based design that tries to maximise the effects from Lord Kelvin's Replenisher

I have a 180 mm diameter barrel with 30 170 mm long 10 mm wide plates on the barrel.

I'm going to be electrically insulating my design more and elimate coronal losses in the machine so that the voltage can go higher than 15 KV.

Because of the barrel design my surface area is huge.

According to the professors equation, I should be able to provide over a milliamp of current at a rotational speed under 1000 rpm, the only question will be how much I can mitigate the losses incumbent in the original design.

1 year ago
1 score
Reason: Original

It depends on what you do in terms of what charge is in the middle plate at the start.

If you connect it to ground, it will be negative, if you connect it to an antenna, it will be positive.

if you use arial electrodes for both (a spark gap), whatever residual charge may or may not be there will start it. Bennet doublers are "self exiting".

If there's absolutley nothing, rub your feet on the carpet and touch a plate

The bennet double is an amazing visual. It's the absolute best way to explain the theory of opperation of an electrostatic influence machine, but it is not the best design of one.

Ideally, you want a rotating design that maximises the size of the charge surfaces and allows for high speed rotation as the ammount current produced is proportional to the surface area of charge carriers passing the collection point.

I'm currently working on a barell based design that tries to maximise the effects from Lord Kelvin's Replenisher

I have a 180 mm diameter barrel with 30 170 mm long 10 mm wide plates on the barrel.

I'm going to be electrically insulating my design more and elimate coronal losses in the machine so that the voltage can go higher than 15 KV.

Because of the barrel design my surface area is huge.

According to the professors equation, I should be able to provide over a milliamp of current at a rotational speed of 800 rpm, the only question will be how much I can mitigate the losses incumbent in the original design.

1 year ago
1 score