I found on german wikipedia (

here) that since 1960 german trams must be capable of 2.73 m/s

^{2} deceleration, so that gives a braking force of F=m×2.73, using newtons and kilogrammes (or kN and tonnes). m is the mass of the vehicle, which must be the maximum mass (i.e., when overcrowded to the limit). That would be the braking force using the emergency brake, which is what we need for the drive by sight speed limit. Of course, when braking for a station stop, the tram would brake less hard, as a sudden 2.73 m/s

^{2} deceleration would certainly lead to injured passengers.

If we have actual braking distances, then instead of assuming the tram follows german law we can calculate its braking force as:

F=0.5×m×v

^{2}/d

with F force, m mass, v speed, d braking distance. There is some air drag and magnetic friction in this, and those are speed dependent, so this is at best an approximation.