Inquisition

I'm learning about brakes and I'm doing some fiddling with calculations and I have a question/concern.

Question: Can we use a 1.12" single piston caliper on our fsae car? The problem is we are not sure if it going to stand up to the pressures needed to stop the car/brake line pressures in general so I want to run a ---- ton of calculations to cover a wide variety of circumstances to really see the limits of the brake setup. So the first thing I did was vary the force placed on the pedal and multiplied it by a variety of pedal ratios. To aid in my question, we will keep it simple and say the force is 10lb and the pedal ratio is 2. That means the force out would be 20lb. Now that force has to go to the MC. Now here is my question. Assuming we are using a .5" MC(again, this will be another variable to find what is actually needed), to find the force to the caliper, one would multiply the force out(20lb) times brake piston diameter and divide it by MC diameter(.5") and you'd get 44.8 lb. Now this calculation is obviously assuming that there is one brake. Now if we are using 4, that means the force to each brake would be divided by 4(this is assuming each brake gets equal stoping force) correct? So again, acting under the assumption you are providing equal force to each brake, that means with the given variables, one caliper will provide 11.2lbs of force? Now line pressure is another variable that needs to be thought of. Obviously given the force one can calculate the pressure as we have the various diameters of the pistons. So the total pressure would be 101.8591636psi. If we were to calculate the individual pressure at each caliper, would this number again, be divided by 4 to get pressure at the caliper of 25.4647909psi?

fork

are you a fresh man that is way off. brake systems use way more pressure than that. The hard part is once you multiply the line pressre times the piston area to get the axial force (along the axis of the axel) how much torque can that exert on the rotor this will probably an intergral which can be found in some book of standards. think about squeezing the rotor an inch away from the center then 7 inches from the center, one will be much more effective than the other. Its frinday I"m done engineering for now.

Inquisition

It was just a numerical example. A pedal ratio of 2 is extremely low, just like a pedal force of 10lb is extremely low. I believe most pedal ratios are 6 or greater which would change things enough. These calculations are all found on Wilwood/Howstuffwork's website. Also I truely don't believe your comment on axial force is very valid. If we know the force on the pedal and the pedal ratio, we know the force placed on the piston of the MC. Now we have a force and the area of the MC, we have a pressure. So if you draw a free body diagram pre and post MC(assuming it hasn't broken off into different lines) it HAS to be the same unless the system isn't in equilibirum and then we'd have a fucked up brake system. Again, this is all explained on how stuff works. Overall I've found their site to have incredibly accurate information, so I'd doubt this article would be any different.

EDIT: I clicked the wrong cell in Excel and I was doing the area of the Caliper piston which is why the psi was so low. So you were right about the numbers being off. 101.8591636psi is the correct value not the 20.3psi.

jagojon3: To me this is pretty general automotive. I figured someone here could help me out here. I'm just doing these calculations to catch me up so I know what the ---- is going on and can comment on things in a relatively informed manner. I'll do a search on the FSAE forums though and see if I can find anything there.

Donald125

why don`t u post up the power and weight of the car so we can give some sugeestions LOL

Inquisition

Approximately 500lbs and 90hp. Peak torque is something like 50ft lbs at around 9k rpm. http://www.wilwood.com/Products/001-...-PS1/index.asp That is the potential caliper to be used and I'd assume the matching brake pad. Rotors are custom designed. I think I fucked up the calculation and the piston diameter should be multiplied by 2 which will change quite a few of the numbers. I'm not sure though. Kind of confused...

fork

wow that caliper uses a rotoe less tan a quarter inch thick. souds weak but I don''t know about that. and I think you miss understood me whatever pressure you get in the line is irrelevant to the exact braking power just because you have 100 psi and 10 square inches of caliper piston area does not mean you have 1000 pounds of backward force on the car It is how the force is applied into a brake pad thik of the rotor as a crank shaft, if you put 5pounds of force omn a 1 foot lever it is a lot different than 5 lbs of force on a 10 foot lever. the rotor is the lever which acts on the axel

88b16civic

There was a complete write-up a couple of months ago in Race-Tech. Simple math but tidius unless you set-up an excel worksheet. So heres the link.
http://www.racetechmag.com/user/soft...are.asp?menu=4
It alows you to enter pedal ratio, number of calipers, MC size, piston size, weight dist of the car, proportioning valve, etc, etc.

Tell people you made the worksheet and drink beer all weekend.

Donald125

why not using counter parts from production cars?

Inquisition

fork, I agree its not the braking force, but it is the force of the caliper re: http://auto.howstuffworks.com/brake.htm. The braking force I'm sure you'd have to include the moment as you implied due to the fact the force is acting on a different axis than the rotor is rotating. Basically the first calculations are to see if the caliper/brake pad can tolerate that kind of press/stress. From what I understand the caliper can only tolerate about 1200psi of brake line pressure before it fails. According to some, the caliper/brake pad is simply too weak to handle the fluid required. I don't know if thats the case. Thats what I'm working on calculating.

88b16, thanks for that link. I'm definitely going to look it over.

Donald, we do use counter parts from production cars but also we manufacture our own stuff. Some teams make their own calipers. We do not. We use Wilwood calipers. On the otherhand, it seems we like to make custom rotors. I just joined the team so I don't know the reasoning behind why we do certain things. From what I can gather every year we get a little better by fine tuning previous years work rather than always redoing everything. The end result is the cars do get driven and need to perform, but its also an exhibition on showing engineering techniques and theories.

Thanks for the help everyone.