We have selected Disc type of brakes for the above project. As these are more powerful than the remaining two. As braking power is high this will be of great help for disabled person. They also perform better in water compared to rim brakes. Disk brakes are able to operate at a higher mechanical advantage than rim brakes because disc rotors in good condition are more true than rims in good condition, and as such do not need to retract as far from the rim when released.
The first basic problem (Annan,James) of a disk brake is that the position of the disk calliper means that the frictional force of the brake pads on the disk acts largely to push the wheel downwards, broadly in the direction of the open fork ends. The force that the pads exert on the disk is very large - much much greater than the force from a rim brake (due to the smaller diameter of a disk compared to wheel rim) and it is also at an inappropriate angle.
The relevant forces and dimensions on the wheel are indicated. We have the braking force B and the ground reaction force R acting at the contact patch. The disk calliper exerts a force D as indicated, tangential to the disc at the point of calliper contact. The radii of the disk and wheel are r1 and r2, and finally the angle of the dropout exit is 'A' in front of vertical. For this fork and calliper, the force D is virtually vertical. If it wasn't, there would be another angle 'b' for the angle the disk force makes behind the vertical.
Let's assume we have a bike + rider weighing 90kg in total, The rear wheel applied brakes give decelerating effect of 0.1g (3m/s^2).With the rear brake alone (this is a reasonable estimate for braking). The rearward force is 90 x 0.1 x g = 90N, and the vertical reaction force is 90 x g =900N (all the weight is on the rear wheel). Taking moments around the axle, the force D exerted by the disk is given by D = 270 x r2 / r1. Let us take radius of wheel to be 330 mm(r2).Let us assume r1= 72.5mm is the effective radius of the disk (ie to the centre of the 'force at the pads, rather than the outer edge = 10mm less than the full radius of 165mm / 2). So D = 410N.This force is acting on both rear wheels. So total force =410 +410 =820 N, acting vertically downwards. This acts entirely on the left hand side of the wheel, but the ground reaction force and braking forces are split equally between