WHY Blank rotors are better than cross drilled
02-12-2010, 07:22 AM
#21
Elementary physics here, grade-school children learn this concept by measuring the force required to drag a block of wood on a larger-area side and then on a smaller-area side with a spring scale. Normal force and coefficient-of-friction are the same, hence frictional force is the same, regardless of the contact area.
02-12-2010, 03:23 PM
#22
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Brake Pad Area
Even if the pad has a smaller area the braking force can be the same as a pad with a larger area.
As an example.
If you apply a 100lb force on the brake pedal with the mechanical advantage/leverage of the pedal say at x4, the force at the master cylinder will be 400lb and let's say the master piston has an area of 1sqin so you have a pressure of 400lb/sqin.
That pressure is then transferred down the pipes in (an enclosed system).
(For this example and simplicity we will only consider one slave cylinder and piston as pressure in an enclosed system transfers to all surfaces anyway, it doesn't matter apart from if you were calculating flow rates, which is fairly negligible in a car braking system anyway, unless you have a leak.)
So now at the slave cylinder you have a pressure of 400lb/sqin, if the piston has an area of say 4sqin then you have a total force of 1600lb.
That force is then transferred to the back of the pad, and as the back of the pad is not part of the hydraulic system the force is applied simply as a force. If the pad has an area on the disc of say 8sqin then divide the force by the area to get pressure, so the pressure is 200lb/sqin.
8 x 200 = 1600
But even if you reduce the area of the pad on the disc to 4sqin the total force will still be at 1600lb, so the pressure will be 400lb/sqin.
4 x 400 = 1600
So no loss involved.
(Did I mention that I worked at Mintex for 8 years as an engineer on the hydraulic systems, it helps.)
Copyright 2001-2010. Drivers Technology Ltd. All rights reserved.
Even if the pad has a smaller area the braking force can be the same as a pad with a larger area.
As an example.
If you apply a 100lb force on the brake pedal with the mechanical advantage/leverage of the pedal say at x4, the force at the master cylinder will be 400lb and let's say the master piston has an area of 1sqin so you have a pressure of 400lb/sqin.
That pressure is then transferred down the pipes in (an enclosed system).
(For this example and simplicity we will only consider one slave cylinder and piston as pressure in an enclosed system transfers to all surfaces anyway, it doesn't matter apart from if you were calculating flow rates, which is fairly negligible in a car braking system anyway, unless you have a leak.)
So now at the slave cylinder you have a pressure of 400lb/sqin, if the piston has an area of say 4sqin then you have a total force of 1600lb.
That force is then transferred to the back of the pad, and as the back of the pad is not part of the hydraulic system the force is applied simply as a force. If the pad has an area on the disc of say 8sqin then divide the force by the area to get pressure, so the pressure is 200lb/sqin.
8 x 200 = 1600
But even if you reduce the area of the pad on the disc to 4sqin the total force will still be at 1600lb, so the pressure will be 400lb/sqin.
4 x 400 = 1600
So no loss involved.
(Did I mention that I worked at Mintex for 8 years as an engineer on the hydraulic systems, it helps.)
Copyright 2001-2010. Drivers Technology Ltd. All rights reserved.
02-15-2010, 02:58 PM
#24
Registered User
Originally Posted by ZDan,Feb 12 2010, 07:31 AM
Braking force is a function of coefficient of friction and clamping force, NOT contact area.
Among other factors, the swept area of the rotor is the only place for the heat generated by the friction to develop. There is a limit for the speed at which the heat can conduct into the rotor. Also, one of the biggest sources of heat rejection is radiation, and that is directly proportional to the area of the surface.
Thus, if you restrict the swept area, the surface temp of the rotor is going to go up faster. This may actually give you more friction when the rotor is cold, because it brings it up into the operating temperature faster. But it will lead to problems when you are trying to push the rotor/pad up at the high end of its temp range.
It does give more surface area for heat rejection from convection, but it would take a detailed heat transfer analysis to determine exactly how much heat is being removed by what mechanism in any given case.
02-16-2010, 05:38 AM
#25
There are certainly secondary effects beyond the elementary "frictional force = normal force X coefficient of friction".
Regarding surface area for radiating heat away, with holes you will have more surface area as long as hole diameter is less than twice the face thickness. Granted, a large portion of the "view" of the inside surface of the hole will be the other side of the hole which is at the same temp, but at least *some* of the view will be outside the hole.
In any case, whether you have holey or blank-faced rotors, you will of course want to *ensure* that you run pads/rotors/fluid that don't overheat for your usage. As long as that's true for both cases, you aren't losing any braking force with holes.
Anyway, I'm certainly not trying to talk anyone out of blank-faced rotors and into holed rotors, I only run blank rotors myself.
Regarding surface area for radiating heat away, with holes you will have more surface area as long as hole diameter is less than twice the face thickness. Granted, a large portion of the "view" of the inside surface of the hole will be the other side of the hole which is at the same temp, but at least *some* of the view will be outside the hole.
In any case, whether you have holey or blank-faced rotors, you will of course want to *ensure* that you run pads/rotors/fluid that don't overheat for your usage. As long as that's true for both cases, you aren't losing any braking force with holes.
Anyway, I'm certainly not trying to talk anyone out of blank-faced rotors and into holed rotors, I only run blank rotors myself.
