[wahern]

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.

[thaumasiotes]

> 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]

Not 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