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There’s really no reason positive energy output should be impossible though? I guess it’s just a really inefficient design?

webkike 8y ago
Doesn’t that answer your own question?
hutzlibu 8y ago
Well, you are free to provide a better design. Problems are many, mostly the very high temperature and pressure needed to achieve fusion ... no material can stand that so you need to have your fusion happening floating which needs energy etc.
DennisP 8y ago
It'e been proven that this particular design can't achieve net power, but other designs probably can. Scientists are especially confident about the tokamak, for example. (And of course, thermonuclear bombs definitely achieve net power.)
- UnpossibleJim 8y ago
  Thermonuclear bombs work on nuclear fission not fusion, just to clear up any confusion. Nuclear power plants also achieve power from fission, which is the breaking of nuclear bonds.
  - Yetanfou 8y ago
    No, thermonuclear bombs [1] are fusionbombs. Nuclear bombs can be of both types but the term thermonuclear specifically means (partly) fusion-powered bombs. These bombs always use a fission reaction as a starter for the fusion reaction, depending on the type of bomb the power output can be anywhere from mostly fusion to largely fission. [1] https://en.wikipedia.org/wiki/Thermonuclear_weapon https://en.wikipedia.org/wiki/Thermonuclear_weapon
  - robotresearcher 8y ago
    No, the defining characteristic of thermonuclear bomb is a second stage fusion device. The first stage fission device sets off the second stage. https://en.m.wikipedia.org/wiki/Thermonuclear_weapon https://en.m.wikipedia.org/wiki/Thermonuclear_weapon Edit: just acknowledging the funny set of identical replies!
    - neutronicus 8y ago
      No, the person you're replying to is right, the bulk of the energy is from fission. What the fusion does is serve essentially as a "tamper" - it's in a ring around the fissile core and as it explodes it compresses the fissile core. Since the fission rate increases exponentially this tamping dramatically increases the overall energy yield.
      - deleted 8y ago
        [deleted]
      - robotresearcher 8y ago
        I think we all agree that fission AND fusion are involved in a thermonuclear weapon.
      - grkvlt 8y ago
        No, that is NOT what the 'tamper' is or does. See my (or various other) nearby comments where I describe the actual process, or just read the Wiki article linked in another comment...
    - UnpossibleJim 8y ago
      Oh! My bad =) I should have looked it up first. I always thought it was take the unstable uranium isotope and surround it with a horrible amount of explosive material which was set at a controlled time to pressure the uranium into a cascading reaction. Turns out, I'm running on the 1940's Popeye cartoon explanation =[
      - robotresearcher 8y ago
  - deleted 8y ago
    [deleted]
  - Denvercoder9 8y ago
    No, thermonuclear bombs actually work on nuclear fusion (started by a smaller fission bomb).
    - msds 8y ago
      Well, GP didn't mean this, but most thermonuclear bombs derive a large amount of their yield from fission of a U238 "tamper" and "sparkplug", driving by neutrons from D-T fusion.
      - blackflame7000 8y ago
        Technically, there is no pure fusion bomb so I guess in a way you could say everything is fission powered, but that's like saying your gasoline car is battery powered because that's what started the engine.
        msds 8y ago
        Nope! Essentially all modern large thermonuclear devices are fission-fusion-fission devices, where the last fission stage (again, driven by fusion neutrons) produces the majority of the yield. You can omit that stage, by using a non-fissile material, for easier testing (...do you really want to test a 100MT device?...) with less fallout, but in actual use, they would all have fissile tampers.
        grkvlt 8y ago
        Except that in a thermonuclear bomb, fission also occurs at the end as well as the start... The initial fission explosion starts off a fusion reaction, and then neutrons from the fusion process carry on to start another fission reaction, usually of the LEU 'tamper' encasing the bomb. In fact, a large proportion of the energy output is from this third stage fission reaction.
wahern 8y ago
It is impossible: https://en.wikipedia.org/wiki/Fusor#Radiation https://en.wikipedia.org/wiki/Fusor#Radiation The issue is bremsstrahlung radiation. I only have high-school physics education, but AFAIU basically because the charged particles are constantly turning inward, the constant negative acceleration radiates energy away. The faster the particles move in the device, the harder the deceleration, the more energy they give up relative to the input energy. These losses overcome the system before the break-even point. The same process limits the energies that can be achieved in non-linear particle accelerators, like the Large Hadron Collider; it's why the accelerator ring diameter has to grow to achieve higher particle energies. The LHC has a circumference of 27 kilometers. Fusors are... slightly smaller. You don't really need to do the math or understand the physics to appreciate that Fusors are a dead-end. It's worth reading the whole article. The Fusor design has a long and fascinating history. One of the original inventors, Farnsworth, invented television.
- pontifier 8y ago
  My understanding is that grid collisions, and a region of cool gas that forms in the center are the main limiting factors in this design. Bremsstrahlung is there, but is it the main source of loss?
- cpb 8y ago
  > bremsstrahlung radiation Oh that familiar rascal... Always getting the better of positive energy output.
- coolaliasbro 8y ago
  "...it's why the accelerator ring diameter has to grow to achieve higher particle energies. The LHC has a circumference of 27 kilometers." Please humor me. Does this mean that if the LHC had x diameter, it would perform the same if it was 1km in circumference?
  - wahern 8y ago
    Bad choice of phrasing on my part as diameter is ambiguous. Also I could only find a reference for the circumference and was too lazy to derive the diameter from the circumference to avoid the ambiguity by making it clear which diameter I was referring to. (EDIT: Diameter is ~8.6km: `echo "pi=4*a(1);27/pi" | bc -l`.) It's not the diameter of the tube but the diameter of the circular path the particles follow through the ring. The greater the circumference the less the relative angular acceleration (i.e. shallower curvature). To accelerate particles even faster without giving up most of the extra input energy to bremsstrahlung radiation you must increase the circumference to maintain the same relative angular acceleration. It's Friday and math[1] and physics is not something I can do off the top of my head so at this point it's better if someone else step in. I won't be able to describe it correctly without relearning half of this stuff myself. [1] I mean... the algebra and geometry is rather trivial but it'll take some effort for me to correct and make more precise the terminology and formulas.

webkike 8y ago

Doesn’t that answer your own question?

hutzlibu 8y ago

Well, you are free to provide a better design. Problems are many, mostly the very high temperature and pressure needed to achieve fusion ... no material can stand that so you need to have your fusion happening floating which needs energy etc.

DennisP 8y ago

It'e been proven that this particular design can't achieve net power, but other designs probably can. Scientists are especially confident about the tokamak, for example. (And of course, thermonuclear bombs definitely achieve net power.)

UnpossibleJim 8y ago
Thermonuclear bombs work on nuclear fission not fusion, just to clear up any confusion. Nuclear power plants also achieve power from fission, which is the breaking of nuclear bonds.
- Yetanfou 8y ago
  No, thermonuclear bombs [1] are fusionbombs. Nuclear bombs can be of both types but the term thermonuclear specifically means (partly) fusion-powered bombs. These bombs always use a fission reaction as a starter for the fusion reaction, depending on the type of bomb the power output can be anywhere from mostly fusion to largely fission. [1] https://en.wikipedia.org/wiki/Thermonuclear_weapon https://en.wikipedia.org/wiki/Thermonuclear_weapon
- robotresearcher 8y ago
  No, the defining characteristic of thermonuclear bomb is a second stage fusion device. The first stage fission device sets off the second stage. https://en.m.wikipedia.org/wiki/Thermonuclear_weapon https://en.m.wikipedia.org/wiki/Thermonuclear_weapon Edit: just acknowledging the funny set of identical replies!
  - neutronicus 8y ago
    No, the person you're replying to is right, the bulk of the energy is from fission. What the fusion does is serve essentially as a "tamper" - it's in a ring around the fissile core and as it explodes it compresses the fissile core. Since the fission rate increases exponentially this tamping dramatically increases the overall energy yield.
    - deleted 8y ago
      [deleted]
    - robotresearcher 8y ago
      I think we all agree that fission AND fusion are involved in a thermonuclear weapon.
    - grkvlt 8y ago
      No, that is NOT what the 'tamper' is or does. See my (or various other) nearby comments where I describe the actual process, or just read the Wiki article linked in another comment...
  - UnpossibleJim 8y ago
    Oh! My bad =) I should have looked it up first. I always thought it was take the unstable uranium isotope and surround it with a horrible amount of explosive material which was set at a controlled time to pressure the uranium into a cascading reaction. Turns out, I'm running on the 1940's Popeye cartoon explanation =[
    - robotresearcher 8y ago
- deleted 8y ago
  [deleted]
- Denvercoder9 8y ago
  No, thermonuclear bombs actually work on nuclear fusion (started by a smaller fission bomb).
  - msds 8y ago
    Well, GP didn't mean this, but most thermonuclear bombs derive a large amount of their yield from fission of a U238 "tamper" and "sparkplug", driving by neutrons from D-T fusion.
    - blackflame7000 8y ago
      Technically, there is no pure fusion bomb so I guess in a way you could say everything is fission powered, but that's like saying your gasoline car is battery powered because that's what started the engine.
      - msds 8y ago
        Nope! Essentially all modern large thermonuclear devices are fission-fusion-fission devices, where the last fission stage (again, driven by fusion neutrons) produces the majority of the yield. You can omit that stage, by using a non-fissile material, for easier testing (...do you really want to test a 100MT device?...) with less fallout, but in actual use, they would all have fissile tampers.
      - grkvlt 8y ago
        Except that in a thermonuclear bomb, fission also occurs at the end as well as the start... The initial fission explosion starts off a fusion reaction, and then neutrons from the fusion process carry on to start another fission reaction, usually of the LEU 'tamper' encasing the bomb. In fact, a large proportion of the energy output is from this third stage fission reaction.

wahern 8y ago

It is impossible: https://en.wikipedia.org/wiki/Fusor#Radiation https://en.wikipedia.org/wiki/Fusor#Radiation The issue is bremsstrahlung radiation. I only have high-school physics education, but AFAIU basically because the charged particles are constantly turning inward, the constant negative acceleration radiates energy away. The faster the particles move in the device, the harder the deceleration, the more energy they give up relative to the input energy. These losses overcome the system before the break-even point. The same process limits the energies that can be achieved in non-linear particle accelerators, like the Large Hadron Collider; it's why the accelerator ring diameter has to grow to achieve higher particle energies. The LHC has a circumference of 27 kilometers. Fusors are... slightly smaller. You don't really need to do the math or understand the physics to appreciate that Fusors are a dead-end. It's worth reading the whole article. The Fusor design has a long and fascinating history. One of the original inventors, Farnsworth, invented television.

pontifier 8y ago
My understanding is that grid collisions, and a region of cool gas that forms in the center are the main limiting factors in this design. Bremsstrahlung is there, but is it the main source of loss?
cpb 8y ago
> bremsstrahlung radiation Oh that familiar rascal... Always getting the better of positive energy output.
coolaliasbro 8y ago
"...it's why the accelerator ring diameter has to grow to achieve higher particle energies. The LHC has a circumference of 27 kilometers." Please humor me. Does this mean that if the LHC had x diameter, it would perform the same if it was 1km in circumference?
- wahern 8y ago
  Bad choice of phrasing on my part as diameter is ambiguous. Also I could only find a reference for the circumference and was too lazy to derive the diameter from the circumference to avoid the ambiguity by making it clear which diameter I was referring to. (EDIT: Diameter is ~8.6km: `echo "pi=4*a(1);27/pi" | bc -l`.) It's not the diameter of the tube but the diameter of the circular path the particles follow through the ring. The greater the circumference the less the relative angular acceleration (i.e. shallower curvature). To accelerate particles even faster without giving up most of the extra input energy to bremsstrahlung radiation you must increase the circumference to maintain the same relative angular acceleration. It's Friday and math[1] and physics is not something I can do off the top of my head so at this point it's better if someone else step in. I won't be able to describe it correctly without relearning half of this stuff myself. [1] I mean... the algebra and geometry is rather trivial but it'll take some effort for me to correct and make more precise the terminology and formulas.