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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
by 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 agoMy 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 agoBad 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.
- thaumasiotes 8y ago> 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`.) ...is the LHC track a circle?
- rnestler 8y agoNot a perfect circle. It consists of multiple arc segments and the detectors which are straight. See also https://home.cern/science/accelerators/large-hadron-collider https://home.cern/science/accelerators/large-hadron-collider
- JamesBarney 8y agoYour highschool physics education was way better than mine.