Template:Wheelset vs single wheels in curves

Abstract drawings of a rail vehicle with gauge $g=1$, wheelbase $b=5$ and curve radius $r\approx 5.59$ (e.g. meters)

The distance of the vehicle center from the curve center is $h=5$, and $r$ follows from that:     $r={\sqrt {\left({\frac {b}{2}}\right)^{2}+h^{2}}}={\sqrt {2.5^{2}+5^{2}}}\approx 5.59$

Typically, $h$ would be calculated from $r$:     $h={\sqrt {r^{2}-\left({\frac {b}{2}}\right)^{2}}}$

With the independent wheels the gauge becomes narrower in the curve:

$h_{o,i}=h\pm {\frac {g}{2}}=5\pm 0.5~~;~~~~h_{o}=5.5~~;~~~~h_{i}=4.5$

$r_{o,i}={\sqrt {\left({\frac {b}{2}}\right)^{2}+h_{o,i}^{2}}}~~;~~~~r_{o}={\sqrt {2.5^{2}+5.5^{2}}}\approx 6.041~~;~~~~r_{i}={\sqrt {2.5^{2}+4.5^{2}}}\approx 5.148$

So the narrowed gauge for this curve radius is $g_{n}=r_{o}-r_{i}=0.894$ (red) rather than 1 (green).
$g-g_{n}=0.106$ (offset 5.3 cm for each rail)

(same for a more realistic curve radius $r=15$)

$h={\sqrt {r^{2}-\left({\frac {b}{2}}\right)^{2}}}={\sqrt {15^{2}-2.5^{2}}}\approx 14.79$ $h_{o,i}=h\pm {\frac {g}{2}}=14.79\pm 0.5~~;~~~~h_{o}=15.29~~;~~~~h_{i}=14.29$ $r_{o,i}={\sqrt {\left({\frac {b}{2}}\right)^{2}+h_{o,i}^{2}}}~~;~~~~r_{o}={\sqrt {2.5^{2}+15.29^{2}}}\approx 15.493~~;~~~~r_{i}={\sqrt {2.5^{2}+14.29^{2}}}\approx 14.507$ So the narrowed gauge for this curve radius is $g_{n}=r_{o}-r_{i}=0.986$ rather than 1. $g-g_{n}=0.014$ (offset 7 mm for each rail)

wheelsets

single wheels

with straight track