= Cyclotron energy and orbit radius
{title2=$K=|q|^2B^2r^2/(2m)$}
Combining $r=mv/(|q|B)$ with the classical <kinetic energy> gives
$$
E\equiv K=\frac12mv^2=\frac{|q|^2B^2r^2}{2m}.
$$
Here $E$ denotes <kinetic energy>, not relativistic total <energy>. At fixed <magnetic field>, twice the final <radius> allows four times the classical <kinetic energy>. At fixed final <kinetic energy>, a larger accelerating gap voltage reduces the number of crossings needed, but does not change the required orbit <radius>.
For an illustrative <proton> with $B=1.00\,\mathrm T$ and peak gap voltage $V_0=10.0\,\mathrm{kV}$, take negligible injection <kinetic energy>, on-crest kicks, a zero-width gap, and no losses. Each crossing adds $10.0\,\mathrm{keV}$, so $100$ crossings give $1.00\,\mathrm{MeV}$, using <electronvolts>. With $e=1.602176634\times10^{-19}\,\mathrm C$ and $m_p=1.67262192595\times10^{-27}\,\mathrm{kg}$,
$$
f_{\mathrm{RF}}=15.2452\,\mathrm{MHz},\qquad
\tau_{1/2}=32.7972\,\mathrm{ns},\qquad
r(1.00\,\mathrm{MeV})=0.144497\,\mathrm m.
$$
The $100$th kick arrives after $99$ intervening half-turns, about $3.247\,\mu\mathrm s$ after the first kick in this idealization. The final two curvature <radii> differ by $r(1\,\mathrm{MeV})[1-\sqrt{99/100}]\simeq0.724\,\mathrm{mm}$. That difference is a curvature-radius increment, not a general formula for separation of real extracted turns: successive arc centers shift, and extraction also depends on focusing and the gap geometry. These chosen settings are a calculation, not historical measurements.
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