If a bulk capacitor will do, its value is determined as follows:
The energy in a capacitor is E = (1/2) C V^2. We are moving from one voltage to another, say 40 to 50V (so, just reducing the overshoot modestly from what it is now, so it would seem), so we have a change in energy that is the subtraction of two cases of this relation.
E_regen = (1/2) C (Vmax^2 - Vmin^2)
60J = (1/2) C ((50V)^2 - (40V)^2)
C = 120 / 900 = 0.133 F = 133 mF = 133,000 uF
So, that's pretty substantial. I would guess on the order of 10mF is in the circuit; which seriously brings into question the 60V figure (actual overshoot would be ~100V, if everything works as expected). Perhaps the controller won't backfeed above that point and you're not getting full performance, perhaps something else is absorbing that power (destructively?). Or, if that's with the braking resistor in circuit, which I'm guessing just turns on once in a while depending on applied voltage or something?
The figure for a capacitor fully discharging (0 to 40V, what would be used in the complicated capacitor-multiplier booster) is about half that: 75mF, still pretty substantial.
Note that going from 0-40V takes such-and-such amount of energy, but going to sqrt(2) times higher (+41% or 56.6V) takes exactly the same amount again -- so it shouldn't be surprising that we don't gain much (i.e. the value is only about half) by going to a higher voltage swing.
In any case, these figures are in the range where a typical power supply will have some warnings (perhaps it won't start up smoothly, or at all?). So, a bulk cap might work, might not, maybe it'll just be fiddly.
And of course, this assumes 60J is the worst-case maximum. If it can spin faster, that's more energy -- again, quadratically so, as energy goes as velocity squared. You'll exceed the predicted overshoot, and be back with the same problem, except more electrical energy to dump into the controller in the event it fails (satisfying sparks, but do wear eye protection
).
The energy can also be calculated from the mass and velocity of everything, including the motor's rotor, gears, shafts, etc. which, as rotating loads, need to be accounted by inertia rather than mass (E = (1/2) I \$\omega^2\$). Easy enough if they're solid cylindrical parts, more or less, but you need to sum up everything.
Tim