Trying to refine my analysis of this, just because I find it intrinsically interesting...
Resolving force vectors in the X and Y axes involves intimidating mathematics beyond my comfort zone, so to help I’ve used an online vector calculator found here:
Vector Calculator
Again, to provide a real world example (but other pads will be similar in principle) I’ve used the listed and diagrammatic dimensions of a new GTA brake pad (132mm long, 16mm thick). I don’t have a specimen GTA pad to measure so I’m not sure of the actual backing plate thickness, but 6.5mm seems common for relatively larger pads, so I’ve assumed the backing plate to be 6.5mm thick.
The corner / edge of the backing plate face closest to the disc (i.e. “inner edge”) is the ‘pivot’ point defining one end of the force vector, not the centre plane of the plate as I had previously assumed (incorrectly, not that it makes a substantive difference...).
With this particular pad the angle of the self energizing force vector is 5.2° tangential to the plane of the pad / disc faces (ascertained using CorelDraw). This is a product of the lateral offset of the backing plate from the disc, and the distance from the pad leading edge to the farthest end of the backing plate, where it abuts the caliper.
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It should be noted that while the self energising effect depends on the pistons to press the pads against the disc, the strength of the effect is
not directly dependant on the magnitude of piston force, but rather
is directly dependant on the magnitude of the friction / drag at the pad / disc interface. The piston force alone does not dictate the force created by pad friction / drag, which is as much a product of pad material, temperature and etc. It is quite possible to have a weak ‘drag’ at the pad / disc interface, despite a very high piston force, for instance if the pads are too cold or overheated...
The lateral component of the vector force is generated geometrically by the vector angle (5.2° in this example), so the magnitude of the self energized force is always
proportional in ratio to the magnitude of the force generated by friction / drag at the pad / disc interface. This ratio doesn’t change unless the vector angle changes, which it does as the pads wear (the vector angle, and thus the self energizing effect, decreasing as the pad becomes thinner).
The force created by pad friction / drag can range from very slight with gentle brake applications, to substantially strong with heavy brake applications, but the ratio of the strength of friction / drag induced force relative to the strength of the self energising force does not alter with changes in pad friction / drag force.
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So, according to the above vector calculator:
If an arbitrary value of 100 ’units’ of magnitude is assigned to the force generated by pad friction / drag, then with a vector angle of 5.2° (as per new GTA pad) the portion of force generated by pad drag which acts purely in parallel with the pad / disc face (Y axis) is less than 100 ‘units’, becoming 99.58 ‘units’. The other 0.42 ‘units’ is redirected to act laterally, which
tries to push the caliper farther away from disc face (due to the angle of the force vector).
And also according to the vector calculator, another 9.06 ‘units’ of force arises, acting laterally (X axis) to press the pad leading edge against the disc, this force being separate to piston force. This is
additional force generated by the 5.2° vector angle, being the lateral pad force created by the self energizing effect...
This force is not directly proportional to the clamping force created by the pistons, rather it is directly proportional to the force created by pad friction / drag. For X force generated by pad friction / drag, the self generating force will always be Y, but the piston force (component) required to create a given level of pad friction / drag is variable depending on pad material, temperature, pad and disc face condition etc.
We can’t easily define what %age of the total lateral force is created by the pistons, and what %age is generated by the self energizing effect, because we do
not know (well, I don’t...) the magnitude of the force created by pad friction / drag, which is hugely variable anyway. In other words, the amount of piston force required to generate a given magnitude of pad friction / drag is indeterminate, so there is no 'ratio' that can be used in calculations.
However it is posible to calculate the relative magnitude of the self energizing effect as a %age of the force created by pad friction / drag. When this ‘drag’ force is assigned an arbitrary value of 100 ‘units’, then in our example calculation gives a self energising force of 9.06 ‘units’ which is near enough to being 9% of the force strength generated by pad / disc friction /drag.
This could very easily be a quite substantial level of self generated force, given the potentially
very large forces generated by pad friction...
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Lets compare the same pad when worn down to 3mm pad thickness:
The offset of the ‘inner edge’ of the backing plate to the disc face decreases from 9.5mm to 3mm. The chamfer disappears completely, and so the effective pad length increases from 105mm to about 132mm (length of backing plate, or just a bit less). This causes the angle of the force vector to decrease to 1.65° (compared to full thickness pad force vector being 5.2°).
Crunching these numbers in the calculator, with ‘drag’ force magnitude at the pad / disc interface as before being arbitrarily 100 ‘units’, but vector angle being 1.65°:
The portion of the ‘drag’ force which now acts
in alignment with the pad / disc plane (Y axis) increases to 99.96 ‘units’ (new pad 99.58), and the lateral component (X axis) acting to push the caliper sideways becomes 0.04 ‘units’ (new pad 0.4).
More importantly, the
additional lateral force generated by the self energizing effect acting at the pad leading edge (causing the pad to ‘self apply’ in some degree) decreases from 9.06 'units' to only 2.88 ‘units’, i.e. the lateral force generated by the self energizing effect decreases in strength from about 9% equivalance to the pad friction / drag force strength, to less than 3% equilvalence. So, when these pads wear down to 3mm thickness (about 1/3rd original thickness) the strength of the self energizing effect also reduces to about 1/3rd of what it was when the pads were new.
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The force generated at the pad can be HUGE, so if the self energising effect is around 9% of that HUGENESS (give or take), then it can be a far from a trivial force, even if it is far less than the self energizing effect of a typical drum brake. From this, I would suggest that the very widely held belief that it is only drum brakes which generate a significant self energizing effect, and that disc brakes generate no self energising effect whatsoever, is very far from being true...
Again I’ll suggest that this could well be at least one reason why new pads can create a significantly stronger braking effect than well worn pads, i.e. because new pads have a significantly stronger self energising effect than well worn pads, due to geometric effects (i.e. strong enough to be noticeable). And, why pads wear with a longitudinal taper pattern, and not parallel.
I still think that it is probably better to have minimal self energising effect, to minimise taper wear and maximize pad area, and just maybe perhaps reduce possible ‘grabbiness’, even if this might only at most be a very minor issue for disc brakes (if it is at all...?)..
Regards,
John.