i thought me, as big master to small slave = greater quantity of fluid being pushed through? = lower powered brake. Also a bigger slave means that because the same pressure is being exerted throughout the system, a greater pressure is exerted through the slave, as this same pressure is pushed through the whole piston. For example Right, bastards. You've made me get my lecture notes out now Pascals principle If pressure is applied to one part of a closed system it is transmitted without loss in all directions Pressure = Force / Area So if you apply a force of 100N (again I don't know how much 100N is like but bare with me) over a 14mm diameter piston: ************************************************************* 14mm Piston Area = 7mm2 (radius) X π (pi) = 153.938mm2 = 0.1539metres2 100/0.153m2 = 649.612 Pa ************************************************************* Whereas... ************************************************************* 16mm piston Area = 8mm2 (radius) X π (pi) = 201.062mm2 = 0.2010metres2 100/0.201m2 = 497.359 Pa ************************************************************* Therefore a smaller MASTER piston provides a greater pressure. Take note the same isn't the same for slaves because they are being driven by these values given. And combine these results with pascals principle; the slave cylinder(s) will be driven by the same pressure. **Just to clarify** Remember that Pressure = Force / Area and this can be re-arranged to give: Force = Pressure X Area For the example, I'll use slaves with err 20mm and 22mm diameter with the 14mm master slave being compressed at 100N So my Pressure (using previous results) is: 649.612 Pascals ************************************************************* 20mm Piston Area = 102 (radius) X π (pi) = 314.1592mm2 = 0.314m2 Force = 649.612 X 0.314 = 204.081 Newtons ************************************************************* ************************************************************* 22mm Piston Area = 112 (radius) X π (pi) = 380.132mm2 = 0.380m2 Force = 649.612 X 0.380 = 246.938 Newtons ************************************************************* These results in newtons represent the force that pushes the slave pistons outwards and make the pads come into contact with the rim I hope this helps, its helped me revise anyway, if someone can tell me how many newtons can can exert upon a brake lever, ill be more than happy to give a more trials accurate answer by inco-operating levers + moments into it too Yes I know its quarter to 2 but im bored night all EDIT: Yeah that's wrong, I've miscalculated the areas on all the pistons. But just assume the area is right and the ratio is right too, it still works.