Explain to me how replacing a lighter pulley with a smaller diameter will magically increase bhp to the engine?
The short answer? See what I can do, probably fail...
Strictly speaking, neither a less massive damper nor a smaller diameter pulley can or do increase the actual power manufactured by the engine (i.e. they don't). It's not remotely similar to say fitting different cams or a supercharger (which
will increase actual power). This begs the question, so what is the rationale behind reducing rotating mass or wanting to reduce pulley diameter?
It's not magic, but because doing so in pratice and / or in theory reduces the amount of a given power output that is 'parasitically lost', and therefore is
unavaliable as power to accelerate the car forward, and also improves throttle response for much the same reason.
Reducing rotating mass:
Whether by fitting a lighter pulley / damper or a lighter flywhel etc. this doesn't increase power, sort of depending on how that power is measured.
If we were to measure power on a 'brake' dynomometer (with which engine power is determined with the engine held at a given steady rpm against a 'braking' force that is 'trying' to stop the crank from rotating, i.e.
brake horsepower, or
brake KW if you like), then at any given steady rpm the as measured power would not increase with a lower rotating mass. This is because at a steady crank rpm we are not accelerating the the speed of the rotating masses (which takes some power to do, 'robbed' from the engine).
However, if we were to measure power with an 'inertial' dynomometer, then with less rotating engine mass there
will be an apparent increase in 'measured' power. Note that with an inertial dynomometer the power is not directly measured at a given steady rpm (as with a 'brake' dyno), but rather is 'inferred' by the time it takes for the engine power (at continuously increasing rpm with a WOT) to increase the rotational speed of a heavy mass of known proportion (part of the dyno typically being a heavy cylinder of known weight and diameter, hooked up to sensors feeding data to a computer). Engine power hasn't really increased as a result of the reduction in rotating mass, but the inertial dyno will 'say' that power is increased. This is because the inertial dyno is 'seeing' the reduction in the time that it takes to increase the rpm of the rotating known mass, which is less time because of the decrease in parasitically 'lost' power (due to the reduction in rotating engine mass). The dyno interprets this reduction in time (needed to accelerate the known mass) as an increase in power.
Inertial dynos can't determine steady rpm power because at steady rpm the known mass is not changing its' rate of rotation. They give a 'measure' of power at any given rpm based on the time it takes to rotationally accelerate the known mass. The known mass and the rotating engine mass both affect the time it takes to increase the rpm of the known mass (the attached computer can't tell where the difference in reduced mass is coming from, all it 'sees' is the time it takes for the known mass to change rotational speed). My understanding is that an inertial dyno measures this change in rotational acceraration over very short time increments, which it computationally 'joins together' to create a 'graph' of inferred power over the engines' rpm range (or can detect the rate of change in acceleration of the known mass). Inertial dynos don't directly measure torque, but infer it mathematically from the measured power and the engine rpm at any given moment.
A 'brake' dyno is different, it directly 'measures' measures torque, and only infers power via a mathematical calculation. It quantifies torque by applying a mechanical resistance to engine rotation (i.e. a 'braking' force), and in some manner measures the force required to hold the engine at a given rpm. This can either be a resistance directly applied to the crank (engine dyno), or less directly at the wheels (chassis dyno, somewhat less accurate). With a brake dynomometer, time doesn't come into it, it is an 'instantaneous' measure of torque that is not time dependant (though time is an element used in the calcuation used to infer power, because crank rpm is used in the calculation, and the 'm' in 'rpm' represents 'per minute', i.e. time). A reduction in rotating engine mass isn't directly 'seen' by a brake dynomometer, because it doesn't use time to measure torque.
What we learn from understanding this (I hope), is that a reduction in rotating mass improves acceleration (to whatever degree) because of reduced inertial loss. But, it won't affect top speed when the forces acting to limit top speed hold the engine at a power limited maximum steady rpm, because the reduction in rotating mass doesn't make any more steady rpm power (rotational inertia is not a factor at steady rpm, only the aerodynamic and rolling resistances).
So, it takes power and
time to increase the rotational speed of the engines' rotating masses. When rotational mass is reduced, this doesn't affect the power that the engine is actually making, but does decrease the time it takes for a given amount of power to increase the rotational speed of the rotating masses, and so acceleration improves.
This explains why a reduction in rotating mass has a more evident affect on acceleration in the lower gears, i.e. in a lower gear the crank rpm increase more rapidly, and any improvement is more subjectively evident than in a higher gear when the rpm increase more slowly, and, rotational inertia is less relative to the increasing aerodynamic and rolling resistances at higher speeds in taller gears.
Reducing rotating mass is most commonly achieved by lightening the flywheel, or fitting a flywheel that is lighter to start with. Reducing the mass of the crank pulley has a similar if much smaller affect on rotating mass.
Reducing the diameter of the pulley:
Lessens the rpm of the pulleys attached to the auxilliary components, at any given crank rpm. Supposedly this reduces the power required to drive the auxilliary components, which it might do in certain circumstances. I explained my doubts about how true this might be, or how much of an effect it might actually have, in an earlier post.
Regards,
John.