Cyclotron Info

Path Length Calculator

Between injection and extraction a cyclotron beam spirals through tens to thousands of turns. Given the dee voltage this calculator finds the energy gained per turn, the turn count, the total path length, and how far apart successive orbits sit — the numbers that decide your vacuum requirement and whether an extraction septum can slip between the last two turns.

N = 17.55 turns Δr, turn 1→2 = 10.28 mm Δr last = 2.964 mm dee dee
The accelerating spiral, drawn from the computed turn radii (rn ∝ √n): turns crowd together approaching the extraction radius (thin outer circle), and raising the dee voltage spreads them apart. Tick marks show where the first two and last two turns cross the mid-plane axis. Dee outline schematic — not to scale.

Results

Final kinetic energy—
Energy gain per turn—
Number of turns N—
Total path length L—
First-turn radius r1—
Turn separation, first → second orbit—
Turn separation at extraction (dr/turn)—
Time from center to extraction—

The math

In the idealized model every gap crossing happens at the voltage peak, so each turn adds

ΔT = ngaps · q · Vp

and the beam needs N = Tfinal / ΔT turns, where Tfinal = q2B2r2/2m (see the energy calculator). After n turns the energy is nΔT, and since r = √(2mT)/qB, the orbit radius grows as the square root of the turn number:

rn = rfinal · √(n/N)

The total path length is the sum of the orbit circumferences, which the calculator sums directly:

L = Σ 2πrn ≈ (4π/3) N rfinal

Because r ∝ √n, orbits crowd together as the beam moves out: the separation per turn near extraction is dr/dn = r/2N. This is why higher dee voltage (fewer turns) makes extraction easier — the last-orbit spacing must exceed the septum thickness plus beam width. With a constant orbital period 1/f, the whole spiral takes N/f seconds.

Why path length sets the vacuum requirement

A beam particle is lost when it strikes a residual gas molecule and scatters or captures an electron (charge exchange), and the loss probability grows exponentially with the total distance traveled — not the machine radius. A 10 cm cyclotron running at low dee voltage can easily rack up tens of meters of path. The vacuum calculator takes the path length computed here and turns it into a required operating pressure.

Assumptions and limits

  • Every crossing occurs at the exact voltage peak. Real particles cross off-peak — a sinusoidal drive with a phase excursion φ delivers only cos φ of the peak gain, so real machines take more turns than this ideal count.
  • No phase slip, no field imperfections, no orbit-center drift; non-relativistic throughout.
  • The fractional final turn is dropped from the path-length sum.

Worked check

Protons, B = 0.582 T, r = 0.104 m (so Tfinal = 175.5 keV), Vp = 5 kV, two crossings per turn: ΔT = 10 keV/turn, N ≈ 17.5 turns, L ≈ 7.6 m (direct sum 7.58 m vs. (4π/3)N r = 7.64 m), first-turn radius 24.8 mm, last-turn separation ≈ 3.0 mm, total time ≈ 2 µs at f = 8.87 MHz.

Sources

  • J. J. Livingood, Principles of Cyclic Particle Accelerators, Van Nostrand, 1961 — ch. 5 (orbit radius vs. turn number, turn separation).
  • M. S. Livingston & J. P. Blewett, Particle Accelerators, McGraw-Hill, 1962 — ch. 6 (phase and energy gain per turn).

Educational reference, not an operating procedure: results reflect the stated model and assumptions. Verify anything safety-critical against primary sources, and read the safety fundamentals before applying numbers to real hardware.