tube80z Posted December 1, 2014 Share Posted December 1, 2014 What I've found to work best is Neil Robert's recommendation. Find a constant corner and apply brakes medium hard. If it pushes you have too much front bias, if the rear steps out too much rear bias. If the cornering balance stays the same they you're set. Cary Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 2, 2014 Share Posted December 2, 2014 (edited) What I've found to work best is Neil Robert's recommendation. Find a constant corner and apply brakes medium hard. If it pushes you have too much front bias, if the rear steps out too much rear bias. If the cornering balance stays the same they you're set. Cary Cary, I'm not so sure about that. It seems to contradict "the books" (Fred Puhn's Brake Handbook, for example). When braking as you describe, hardly any weight transfer takes place. In other words, front and rear tires stay loaded as if the car was parked (almost), except for the fact that there is weight transfer from right to left or vice versa, which is unrelated to front/rear bias. The bias that allows for maximum deceleration is the bias that allows all wheels to lock up simultaneously when weight transfer (F/R) is as high as the tires and track allow. A situation in which a street car (high CG) could have a weight distribution of 80/20 (F/R), for example. Dynamic weight distribution in the situation that you describe seems more like 60/40, for example. The best practical test is still the one in which you brake in a straight line until "something" locks up. Let a helper outside the car observe whether it’s the front or rear that locks up first. Adjust bias until all four lock up simultaneously and then SLIGHTLY dial bias forwards. Adjust bias rearwards when driving on tracks with less grip (less weight transfer) and forwards on tracks with more grip (more weight transfer). Arne Edited December 2, 2014 by BrakePower Quote Link to comment Share on other sites More sharing options...
Leon Posted December 2, 2014 Share Posted December 2, 2014 Cary, I'm not so sure about that. It seems to contradict "the books" (Fred Puhn's Brake Handbook, for example). When braking as you describe, hardly any weight transfer takes place. In other words, front and rear tires stay loaded as if the car was parked (almost), except for the fact that there is weight transfer from right to left or vice versa, which is unrelated to front/rear bias. The bias that allows for maximum deceleration is the bias that allows all wheels to lock up simultaneously when weight transfer (F/R) is as high as the tires and track allow. A situation in which a street car (high CG) could have a weight distribution of 80/20 (F/R), for example. Dynamic weight distribution in the situation that you describe seems more like 60/40, for example. The best practical test is still the one in which you brake in a straight line until "something" locks up. Let a helper outside the car observe whether it’s the front or rear that locks up first. Adjust bias until all four lock up simultaneously and then SLIGHTLY dial bias forwards. Adjust bias rearwards when driving on tracks with less grip (less weight transfer) and forwards on tracks with more grip (more weight transfer). Arne Nicely done Arne. The Brake Handbook is a great read for those interested, and cheap to boot! Quote Link to comment Share on other sites More sharing options...
SleeperZ Posted December 2, 2014 Share Posted December 2, 2014 Cary, I'm not so sure about that. It seems to contradict "the books" (Fred Puhn's Brake Handbook, for example). When braking as you describe, hardly any weight transfer takes place. In other words, front and rear tires stay loaded as if the car was parked (almost), except for the fact that there is weight transfer from right to left or vice versa, which is unrelated to front/rear bias. The bias that allows for maximum deceleration is the bias that allows all wheels to lock up simultaneously when weight transfer (F/R) is as high as the tires and track allow. A situation in which a street car (high CG) could have a weight distribution of 80/20 (F/R), for example. ... Arne I agree with most of what you wrote, but I don't understand your claim of "hardly any weight transfer." When braking, weight transfers forward, whether in a turn or not. Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 2, 2014 Share Posted December 2, 2014 (edited) SleeperZ, good question and thanks for responding. You are right that every brake action causes forward weight transfer. But the maximum possible amount of forward weight transfer (before the tires “brake loose”) is not the same for different situations. The reason that brake tests are done in a straight line, is that a tire can only “transmit” so much friction force. In a corner, part of its total “friction force capacity” will be used up by centrifugal force, which leaves less to deal with brake force. In other words, forwards weight transfer while threshold braking in a corner will always be less than when threshold braking in a straight line. Put differently, in a straight line, you can hit the brakes a lot harder (before the tires brake loose) than while driving thru a turn. Because it is “maximum weight transfer” we are after when dialing in the brake bias that allows for maximum possible deceleration, it makes sense to do the test in a straight line. As a side note, testing your brakes in a corner still is a very informative and fun exercise that gives you lots of important information (experience wise) that will be useful when tracking your car at the limits. Arne Edited December 2, 2014 by BrakePower Quote Link to comment Share on other sites More sharing options...
andyhorror Posted December 2, 2014 Author Share Posted December 2, 2014 during the rebuild of my stock Z front calipers a proper measurement of the caliper pistons was taken. 240z caliper piston 42mm diamter 21 rad surface area 1385mm sq 2770mm s2 front axl 5540 sq front axle rx7 caliper piston 30mm diameter 15mm rad surface area 706mm sq x4 piston 2824mm sq 5652mm sq front axl not a huge leap in piston area, but an rx7 caliper weighs about the same as one half of a stock z caliper this coupled with the option to go with larger vented rotors or even a two piece rotor down the road makes these a viable option for me down the road and maybe others. the toyota 4x4 "upgrade" hardly seems conducive to a performance car in which keeping unsprung weight down is a valid issue. and in an area "brakes" where most of that weight resides seems to me this is where the most savings car be made. if that can be done without spending 1000$ on a wilwood kit would be nice. Quote Link to comment Share on other sites More sharing options...
SleeperZ Posted December 2, 2014 Share Posted December 2, 2014 I agree with the rationale for using the RX-7 brakes; the ability to use a vented rotor with minimal unsprung weight seems advantageous. The popular Toyota option seems inappropriate in several ways, one in that it is a much heavier caliper and two that it provides so much more clamping it is difficult to find an appropriately balanced rear disc option. That is the reason I am currently using the S130 front calipers with Z31 rotors, and plan to install the S130 master and Maxima rear calipers (with S130 rotors). I figure that will be easier to balance front to rear without radical differences in pad materials. Quote Link to comment Share on other sites More sharing options...
andyhorror Posted December 2, 2014 Author Share Posted December 2, 2014 Even a rather small clamping force can still produce high torque numbers when the effective radius (distance from rotor-center to halfway the swept area of the rotor) is big enough and/or friction coefficient between pad and rotor is high enough. could you elaborate on this? this is something that has been bouncing through my head. my thoughts: a large rotor equates to a larger lever to slow the hub. also fighting against this theory is that the larger the rotor has a higher speed that the caliper needs to slow. Am i way off base? if not, where is the happy medium or are they equal and balance themselves out? Quote Link to comment Share on other sites More sharing options...
andyhorror Posted December 2, 2014 Author Share Posted December 2, 2014 another plus with the rx7 calipers is that all the pads are available. Quote Link to comment Share on other sites More sharing options...
Leon Posted December 2, 2014 Share Posted December 2, 2014 could you elaborate on this? this is something that has been bouncing through my head. my thoughts: a large rotor equates to a larger lever to slow the hub. also fighting against this theory is that the larger the rotor has a higher speed that the caliper needs to slow. Am i way off base? if not, where is the happy medium or are they equal and balance themselves out? The linear speed of the rotor has nothing to do with braking torque. A longer lever arm (larger diameter rotor) will stop a car faster, given it has enough traction to sustain that deceleration. Yes, the linear speed will be higher and you would think that it would generate more heat but to generate the same brake torque from a smaller diameter rotor you must squeeze it harder, also generating more heat. Heat generated (friction) has more to do with the amount of kinetic energy the brakes are being asked to absorb. In other words, if we're talking speed, the important one is the speed of the car itself. Weight is the other part of the kinetic energy equation but this takes us into a whole different topic... Quote Link to comment Share on other sites More sharing options...
andyhorror Posted December 2, 2014 Author Share Posted December 2, 2014 makes sense to me. so identical calipers one on a 11in rotor and one on 13 rotor. the caliper fixed to a 13in rotor will have more brake torque in theory? i need to sit down and do the calculations. Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 2, 2014 Share Posted December 2, 2014 The linear speed of the rotor has nothing to do with braking torque. A longer lever arm (larger diameter rotor) will stop a car faster, given it has enough traction to sustain that deceleration. Yes, the linear speed will be higher and you would think that it would generate more heat but to generate the same brake torque from a smaller diameter rotor you must squeeze it harder, also generating more heat. Heat generated (friction) has more to do with the amount of kinetic energy the brakes are being asked to absorb. In other words, if we're talking speed, the important one is the speed of the car itself. Weight is the other part of the kinetic energy equation but this takes us into a whole different topic... Leon, thank you for the excellent explanation. Arne Quote Link to comment Share on other sites More sharing options...
Leon Posted December 3, 2014 Share Posted December 3, 2014 makes sense to me. so identical calipers one on a 11in rotor and one on 13 rotor. the caliper fixed to a 13in rotor will have more brake torque in theory? i need to sit down and do the calculations. Yes. Leon, thank you for the excellent explanation. Arne Cheers! You made some great contributions in your first few posts, welcome to HybridZ. Quote Link to comment Share on other sites More sharing options...
G-E Posted December 3, 2014 Share Posted December 3, 2014 Which is exactly why you see hugely expensive german cars with rather ordinary sliding calipers and mediocre high-dust pads, the leverage lowers the heat requirements of the pads, and offers less chance of cracking the rotors too. Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 4, 2014 Share Posted December 4, 2014 (edited) Horsepower is for accelerating, brakes are for decelerating. There is no horsepower rating for a brake system, it makes no sense. Sorry for the late response, but there is a horsepower rating for brake systems: Horsepower = 0.002667WdmaxS (or in SI units: Pb = mcamaxVc) 0.002667 = constant factor to convert the metric SI units into Standard units W = Weight of car in pounds dmax = maximum deceleration in g's S = Speed of the car in miles per hour Pb = Power of brakes in Watts (= J/s = Nm/s = kgm²/s³) mc = mass of car in kg amax = maximum deceleration in m/s² Vc = Velocity of car in m/s More info and examples can be found here. Edited December 4, 2014 by BrakePower Quote Link to comment Share on other sites More sharing options...
NewZed Posted December 4, 2014 Share Posted December 4, 2014 (edited) I actually knew that, and realized the words were wrong after I wrote it. But left it anyway. Horsepower is measured with a big brake. Brake Horsepower. Edit - or torque is measured, horsepower calculated.... I got lazy. My point was really that the horsepower used to move the car is only relevant in that it's used to bring the mass of the vehicle up to a speed where brakes are necessary to stop it. But the horsepower of the engine in the car cannot be used to determine the horsepower required from the brakes. There is no need for a car's engine's horsepower to determine a brake system's horsepower. As you show above. "I have more horsepower, so I must need better brakes" is essentially a nonsense statement, unless you add the condition "because I will be going faster". The last part is usually assumed, but for a street car is probably not that relevant. Edited December 4, 2014 by NewZed Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 5, 2014 Share Posted December 5, 2014 (edited) ... "I have more horsepower, so I must need better brakes" is essentially a nonsense statement, unless you add the condition "because I will be going faster". ... I fully agree. Thanks to your brief moment of “I got lazy”, this thread became even more informative, and also thanks for inspiring me to take a refreshment class “brake-formulas” ( I don’t have these copy + paste ready, you know ). Keep up the good work! Arne Edited December 5, 2014 by BrakePower Quote Link to comment Share on other sites More sharing options...
BrakePower Posted December 6, 2014 Share Posted December 6, 2014 during the rebuild of my stock Z front calipers a proper measurement of the caliper pistons was taken. 240z caliper piston 42mm diamter 21 rad surface area 1385mm sq 2770mm s2 front axl 5540 sq front axle rx7 caliper piston 30mm diameter 15mm rad surface area 706mm sq x4 piston 2824mm sq 5652mm sq front axl I noticed that regarding caliper piston size, the numbers are not always accurate... Usually, when I need information about piston diameters, I go to RockAuto.com that has pictures and measurements posted on most parts they sell. For example: the 4 pistons of the RX7 caliper have a 36mm diameter and the 240Z has two 54mm pistons. See attached pictures of the RX7 FC, RX7 FD and 240Z calipers. With the Total Piston Area Calculator we quickly find that the RX7 has 8138.88mm² and the 240Z 9156.24mm² on the front axle. This information tells us that we would lose 11% total piston area, and therefore clamping force would go down 11% as well. If the rear brakes were not participating enough and pedal effort was light to begin with, this could be a very good swap. On the other hand, if pedal effort was high already, and the rear axle was over-braked or perfectly balanced with the front, the swap would require a serious look at (and possibly redesign of) the rest of the system, in order to make it work right. For example a slightly smaller M/C (to compensate for the reduction in Total Piston Area up front) and adjustable proportioning valve (to reduce hydraulic pressure increase (with increasing pedal pressure) to the rear brakes). Or even a larger rotor and moving the front calipers outwards on that rotor to increase “leverage” to increase torque. Another way to increase brake torque on the front axle would be to go with pads with a higher friction coefficient... As you can see, the possibilities are endless... Hope this helps. Arne Quote Link to comment Share on other sites More sharing options...
skirkland1980 Posted December 6, 2014 Share Posted December 6, 2014 You would have 8 pistons with the rx7 calipers. Quote Link to comment Share on other sites More sharing options...
skirkland1980 Posted December 6, 2014 Share Posted December 6, 2014 Never mind. The math didn't seem right. Quote Link to comment Share on other sites More sharing options...
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