Let's talk about brakes

Discussion in 'Car Modding' started by jtbo, Aug 2, 2012.

  1. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    With regards to inertia keep in mind that only the brake disc is worn down, not the bell, ventilation fins etc. So "BrakeDiscInertia" should only be based on the area that wears away. Like you said, use this equation for the disc area that wears:
    I(z-Axis)=1/2*PI*Thickness*Density*[(OuterRadius^4)-(InnerRadius^4)]
    The rest of the inertia (bell+ventilation fins, axles etc) should be added to the Iz of the wheel in the PM file (actually Ix if you go by rFactor coordinate sys). I also have some data on the wear-rate of the disc as a function of torque and temperature if you're interested.
     
  2. jtbo

    jtbo Registered

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    Thanks, will be interesting to see what kind of inertia software says bell area to have.

    This is aftermarket disc, but there are only minor differences, ~0.1mm in thickness etc.
    http://bremboaftermarket.com/En/Car_Disc_Catalogue/Catalogue_Detail.aspx?ApplicationIDMaster=25816

    From that aftermarket site one can find really lot of discs, even for trucks/buses.

    It is interesting that official drawing has weight of 3.66kg, while my calculation was 4.14kg, differences at least with bolt holes omitted in my calculation and also bell area material thickness was 0.4mm greater in my calculation, all those differences does add up I believe.
     
  3. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    With that geometry I get a mass of 3,754 kg (using a density of 7,150) and Iz 28632,621 kg mm^2 when I draw it with CAD software
     
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  4. jtbo

    jtbo Registered

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    That is probably 2.9187177370030581039755351681957 of rfactor units? (28632,621 / 9.81 / 1000), maybe I should start rounding these long numbers :p

    Any data is interest of mine, so of course I'm interested ;)

    Difference in weight is probably partly because density, I did use 7850 for density as mentioned earlier, maybe it is bit high if lower density gives closer results?

    It sounds quite logical to have bell area to be wheel inertia, or maybe spindle as that has own inertia and it is closest to what bell area is?
     
  5. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    Sorry, its in millimeters. So in SI units its 0,028632621 kg*m^2 for that geometry.

    Brake discs are usually grey cast iron with a good amount of carbon so the density is usually around 7100-7200. 7850 sounds more like (mild) steel.

    Spindle is not a good idea for inertia, because it does not spin. Its better to add to the wheel. Then you get the correct gyroscopic contribution from the rotation of the bell.

    Here is one number for the Wear rate (2.36e-12 m3/N/m):
    http://www.iaeng.org/publication/WCE2010/WCE2010_pp2322-2326.pdf

    In reality it's highly temperature dependent. I normally go by this relation:
    wear_rate=3e-18*exp(0,0189T) , where T is temperature in Kelvin.
    To calculate the wear rate I suggest you pick a temperature that is in the upper range of the operating condition when driving the car.
     
    Last edited by a moderator: Aug 4, 2012
  6. JJStrack

    JJStrack Registered

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    View attachment 3399
    ok, here are my calculations for the two parts of the disk: the dark-grey Bell, which won't wear off, and the light-grey disc, which will get worn off. The disk that gets worn off has an inner diameter of 139.5mm, outer of 239mm and thickness of 12,85mm
    my calculations were done for grey cast iron with a density of 7.150 g/cm³ (EDIT: make that 7150 kg/m³ for RF2 units)
    My complete model has a Mass of 3,631kg (close enough to 3.66kg !?) and Inertia of 0.028957 kg*m².
    The Bell has a Mass of 0.914 kg and Inertia of 0.002942 kg*m²
    The Disk has a Mass of 2.718kg and Inertia of 0.026015 kg*m²
    and the per-meter values for the disk are 211.488 kg/m and 2.024498 kg*m² / m
    i hope this helps!
     
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  7. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    Correct JJStrack. I had a wrong dimension. I got exactly the same as you now.
     
  8. jtbo

    jtbo Registered

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    That is interesting and very useful, thx!

    I fail to find unit converter that would convert those inertia units

    0.026015 kg*m² becomes 2.024498 kg*m² / m, it is not 100 times, but something else, there is /m difference in unit, but I fail to get conversion factor, 77.82 times larger, but relationship is perhaps not linear?

    Nice looking model, soo you can export that in 3ds? ;)


    Knut Omdal Tveito , wear_rate=3e-18*exp(0,0189T) what is exp there for? Sorry it is very long since I had my mathematics education so I have forgotten most, I put most formulas to excel and let program worry how to calculate :)


    I have read how wear can be affected a lot by brake pads, some of links I posted earlier had experience from some driver that new pads ate brake discs just in 6 track days, other kind of pads did not eat rotors away while still provided similar grip.

    This thread has lot great info to put Wiki now already, thanks from that too, such information as this will help future modders to make much better vehicles :)
     
  9. JJStrack

    JJStrack Registered

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    hehe mate, you really need to learn what those values mean ;-)
    the 0.026015 kg*m² is the exact inertia for your 12.85 mm thick brake disc. the 2.024498 kg*m² PER METER is a factor, which gets multiplied with the momentary thickness of your brakedisk! so 12.85mm are 0.01285m, and those 0.01285m multiplied with the 2.024498 kg*m² PER METER give you the 0.026015 kg*m²!
    yeah i can easily give you the *.ipt file and you can import that into 3ds max. when importing you can choose how many segments a circle gets, but otherwise i don't know how good the model will look after importing to max...
     
  10. JJStrack

    JJStrack Registered

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    to explain it another way: it's like buying froots in the supermarket: you have a price PER KILO, but you only buy 150g...so your price per kg gets multiplied with 0.150 kg to get the price you have to pay. thats the way this system works, so that your brake discs inertia will decrease as your disc gets thinner (because wearing off)
    EDIT: one more thing: as far as i can see that, most things in RF2 are calculated in SI-Units...those untis define, with wich values you calculate - for an example: you are not supposed to calculate in mm length, but in m length. here, have a look at the wikipedia site: http://en.wikipedia.org/wiki/International_System_of_Units
    so basicly, you have to convert every value you use to meter, kilograms, seconds etc.
    i think there should be an official document which states for every value in RF2 in which unit it gets evaluated....
     
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  11. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    The inertia value of 0.026015 kg*m^2 is for the disk with thickness 12.85 mm as in the original geometry. 2.024498 is per meter of disc thickness.
    Inertia_per_thickness * Thickness = Inertia_of_disc
    2.024498 kg*m^2/m * 0.01285m = 0.026015 kg*m^2
    EDIT: Ninjad:)


    Exp, see here:
    http://en.wikipedia.org/wiki/Exponential_function
    Basically its the number "e" to the power of 0,0189T:
    wear_rate=3e-18*2.718281828^(0,0189*T)

    I have some more information on the heat transfer of brakes in rFactor as well. I'll just have to dig up some old files.
     
  12. JJStrack

    JJStrack Registered

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    oh i'm affraid thats wrong! there is a BIG differencs between e² and 1e2! the thing here is NOT to the power! it is an abbreviation of numbers with many zeros in it. for an example, one million is a 1 with 6 zeros following, so it gets written 1 e6. on the other hand a millionth is a 1 that FOLLOWS 6 zeros, so it gets abbreviated 1 e-6.
    so mathematically 1 e3 is 1*10^3 which is 1*1000
    for an example: 1mm is 1 e-3 m which is 0.001m which is 1*10^(-3)
    Edit: you can read about that here: http://en.wikipedia.org/wiki/Scientific_notation in the section "E-Notation"
     
  13. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    Ah sorry for the confusion JJStrack:) Its just faster to write like that

    When I write 1e3 or 1E3 I mean 1*10^3 as you said.
    When I write exp(3) I mean e^3

    Hopefully there should be no confusion now:
    3.0*10^(-18) * 2.718281828^(0,0189*T)
     
    Last edited by a moderator: Aug 4, 2012
  14. jtbo

    jtbo Registered

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    I once lost my head and they gave me this wooden one, it is quite good when you hit your head or need to knock the wood, but sometimes I wonder if it has something important bits missing ;)

    If I know not how much inertia is per meter of thickness but know how much it is for 0.01285 meter of thickness, then I do 1/0.01285*0.026015 and I get some error margin, but end up with 2.024513619. That should be same as 2.024498, imo, but maybe some rounding introduces some minor difference. My bet is that 0.026014799 being rounded to 0.026015. My spreadsheet however seems to be in agreement with this and gives me now same values you have got when I use same numbers as input, so I can be sure that formulas are correct.

    That is how I calculated rolling resistance in rF1 as I knew contact patch length and I know tires rolling resistance with that load, I just needed to get it to per meter of deflection, I think that is quite same thing here.



    For wear rate then, 3e-18*exp(0,0189T) vs 3.0*10^(-18) * 2.718281828^(0,0189*T), these should be same things from my understanding, but where did that 2.7..... came from, disc weight in kg?
    Something like this 3e-18*2.718281828exp(0,0189T) might be same maybe? But then why using exp instead of ^? I guess first version exp includes also that 2.718281828?

    When learning new stuff it is of course easier when formulas are in long format, so one don't get overwhelmed with new things, but it really is challenging to remember when scientific notation is as clear as plain english, challenge when anyone writes a tutorial is to actually remember to tell those things that are obvious for writer, but one learning has never heard about such. That is why these discussions are good that all those different angles come up and we can make modding more accessible so that new modders can produce better quality instead of wrestling with same issues everybody has had at the beginning, well maybe.

    What I would like to be in official documentation is how to convert real world data to SI units, or how when you know this much data from real car, how to convert that data to HDV. But it is ok if resource don't allow that, we can learn and write it up in simple terms also :)

    Anyway, using that formula, I got 6.16355E-16, I have used ISI ZR value of 1.50e-013, at least I believe that I have not touched that as of yet. that is hundreds of times smaller value however that I got unless I misunderstood something.
     
  15. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    Yeah it's nice to discuss these things. It's hard to come by information and usually you have to find the answers by testing yourself.

    I think maybe you used temperature in degrees Celsius? You have to use absolute temperature (Kelvin):
    T [K]=T [C] +273.15

    Anyways, I attached an Excel sheet with a quick calculation of the wear rate for different temperatures.

    View attachment 3422
     
  16. JJStrack

    JJStrack Registered

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    to clear things up: that number 2.718281828 is the number e (same as 3.14159... gets abreviated as the number pi)....so when you write e^4 you could also write 2.71...^4 which again is the same as EXP(4). In Excel you have to write EXP(x) to calculate e^x
     
  17. jtbo

    jtbo Registered

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    Thanks, spreadsheet is very useful :)

    My language version twists that that handy exp to insane EXPONENTTI, yes it is in capitals too in program, silly (I really dislike localized versions because of this reason). But it seems to be somewhat different in other ways too.

    As I compared formulas I could not see difference, then I typed both here and noticed I had forgotten one zero after 0. making tenfold difference and causing positive numbers of E, yesterday I got negative ones, but I used disc weight instead of exp, have I mentioned I have difficults with engineer and barely manage plain english? :D

    But at other hand, I'm quite good stumbling to all possible problems so that more gets solved, so making information 'wider' in a way, I hope, has to be some advantage being dense :)


    Thanks, this makes issue I had very clear indeed :)

    I need to still check exact range, but from my memory FE rated brake pads had cool CoF range of 0.36-0.45 and hot range of 0.1 less both ends and if I'm not mistaken that is something around 29% drop and if drop is linear somewhere around 405C it was 50%, so my optimum range ends to 150C as that is temperature where low temp is rated, my overheat temp is 405C as that is where probably is 50% of friction. If I have not made huge mistake, this should fit for all FE pad equipped vehicles.
    BrakeResponseCurve=(-350,80,150,405)

    For my track day pads (FF) I figure it should be something like this, but I have no idea from temperature where Ferodo DS2500 starts to loose friction, probably fairly high? But as I know only that range at 315C or 600F, I set fading to start after that point but still have really lot of range until friction is half of optimum. Must refine when I find out more as is case with majority of parameters.
    BrakeResponseCurve=(-350,80,315,805)


    From Brazilian bus study I did read how temperatures go above 300C when stopping 120kph to 0, which took around 6 seconds. Problem is that passenger cars have quite lot smaller bits that generate heat, but also less mass so they heat easier, stopping distance and time is also less as are forces that push the pad and vehicle mass/inertia that resists reduction of speed.
    Without further data I have been thinking that as materials are similar, they are probably meant for similar temperatures and as brakes are designed to have decent stopping power for each vehicle, temperature ranges probably are similar, maybe heating too is relatively similar ?
    I guess better try and dig bit more of this :)

    From brake heating, there is Nissan 350Z testing saying:
    Source:http://www.zeckhausen.com/Testing_Brakes.htm

    But I don't know if that applies to my case much, those brakes in Nissan are probably designed for far higher temps and are able to cool much better while having more mass than my brakes, but anyone making sport cars that gives good insight and as there are different brakes tested it gives chance to make realistic upgrades too.


    Things currently missing from brake model from my understanding are these
    Brake pad wear
    Brake fluid properties, water contamination, DOT5...DOT3 or racing brake fluid

    Pads wear more than discs, in general, I wonder if pad wear should be compensated in disc wear? But then again, I doubt that anyone can race enough long to wear out pads or discs from street car, it would take several thousand of km even with very hot brakes.

    Brake fluid can boil, result is brake pedal goes to floor and very poor braking, does happen surprisingly easy with older street cars.
    That is no concern with proper race cars, but important aspect with street cars, boy racers have no idea how poor that 10 year old dot3 fluid is when they choose to have little fun on road, around 3rd and 4th hard braking fluid starts to be useless.
     
  18. jtbo

    jtbo Registered

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    I put here one more calculation of brake torque, from real car manufacturer data, which I also put here and I put my spreadsheet also as attachment, which might help new guys to know what they need to know and how real values become usable rF2 units.

    This is all the data I have for another car I'm making, data is 1967 Camaro, but I use it for 1967 Firebird as they are very close to each other, both are F-body, kind of sister models, but different sheet metal and engines. Besides I can find so much more from Camaro, haven't found brake stuff for Firebird yet or it has not met my eye :)
    View attachment 3425
    View attachment 3426

    I do here optional disc brake model as I still need to learn more of how to make drum brakes. Drum brakes might need very narrow optimal temperature and are in kind of fade all the time, except first use or two, but I haven't tested if that results correct emulation of them, it is however pretty much how they are in reality from what I have read.


    Wheel cylinder bore is 1.875" which is 0.047625m brake caliber piston size, piston is slightly smaller than that of course, 0.1-0.2mm I believe, but to keep things simple, I don't bother with that detail now.

    There are four pistons so 0.047625 * 4 = 0.1905m total, I use 190.5mm in my spreadsheet as it converts mm to m automatically.

    Master cylinder bore is 1", which is 0.0254m, I use 25.4mm in my spreadsheet.

    Radius for center of forces = 0.111633, calculated as follows:
    Brake disc diameter is 11" which is 0.2794m
    Brake pad height/width is 2.21" which is 0.056134m
    Center of pad is 0.056134 / 2 = 0.028067
    Forces center from edge of disc is 0.2794 / 2 - 0.028067 = 0.111633 Remember that disc diameter need to be converted to disc radius, which is done by 0.2794 / 2
    I use 111.633mm for my spreadsheet

    Coefficient of friction, I go with 0.36, but I don't have much more data from this, molded asbestos is material, maybe it is possible to find something from asbestos brake pads, but as those were banned before Internet became widely popular, there might not be much to be found from CoF of that material. Need to refine if I find something.

    Brake pedal lever 7.9:1 (790lb line pressure at 100lb. pedal load, means vacuum assist + pedal lever = 7.9:1)
    100lb is 45.35924kg according to converter, but for brake torque I believe one would need to use pedal load that one can put to pedal reasonably, 80kg being standard weight and generally I think that human can put his own weight to brake pedal, that 80kg is what I use in spreadsheet.

    That gives 3737.424615 for me (Cell H13), I'm bit surprised of such high value, but car is of course pretty heavy, 1600kg range with driver.

    Now there is interesting bit of data, percent brake effectiveness - front 58.9 I wonder if this is brake balance or general effectiveness of brakes? I have taken that as a brake balance, but it is quite a rear, I need to calculate rear brakes too and figure out method how to get it right for drum brakes, but that is how front brakes are done unless I made horrible error.


    I wanted to calculate inertia too, but I can't spot thickness of brake disc from data, also as disc is vented I would need to know thickness of disc and size of vents so I could use thickness only from areas that will wear, then add inertia of vents to wheel, that might provide to be bit of on challenge.

    With imagination produced 0.02m thickness I end up with 3.730009949 inertia.

    I add spreadsheet in a moment, I believe that my session has expired and I have no idea what will happen with my image attachments either, so I try posting and see what happens. Stay logged in seems to be in disorder.

    edit: Added spreadsheet, just remember that it needs quite bit of work to be nice looking and easy to use, but I do that when I integrated that to my all calculators calculator set.
     
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  19. Knut Omdal Tveito

    Knut Omdal Tveito Registered

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    I guess the inertia is ok now, so here is some information on the brake heating.

    We can start with the conservation of energy to explain the heat transfer:
    V*rho*cp*dT/dt=Q_friction+Q_convection
    ,V=Volume of disc
    ,rho=density of disc material (7150 kg/m3)
    ,cp=specific heat of disc material (460 J/kg/K)
    ,dT/dt=Temperature time derivative
    ,Q_friction=heat due to friction
    ,Q_convection=heat due to air cooling

    The friction term is equal to:
    Q_friction=T*omega
    ,T=Torque [Nm]
    ,omega=disc rotation speed [radian/s]

    The convection term is equal to:
    Q_convection=h*A*(T_air-T_disc)
    ,h=heat transfer coefficient (depends on velocity of the car, higher speed-> more cooling)
    ,A=disc cooling area [m2]
    ,T_air=air temperature
    ,T_disc=disc temperature

    The heat transfer coefficient can be written like this to depend on velocity:
    h=h1+h2*velocity


    So we get this equation for the heat transfer:
    V*rho*cp*dT/dt=T*omega + (h1+h2*velocity)*A*(T_air-T_disc)

    Divide by V*rho*cp:
    dT/dt=(T*omega)/(V*rho*cp) + (h1+h2*velocity)*A*(T_air-T_disc)/(V*rho*cp)



    Now this is how rFactor does it:
    dT/dt=BrakeHeating*(T*omega) + (BrakeCooling1+BrakeCooling1*velocity)*(T_air-T_disc)


    If you compare the two equations we see that:
    BrakeHeating=1.0/(V*rho*cp)
    BrakeCooling1=h1*A/(V*rho*cp)
    BrakeCooling2=h2*A/(V*rho*cp)

    This is how you calculate the heating/cooling values in rFactor

    So you already know the volume, cooling area and material (density rho, specific heat cp) so most of the information is known. Now you need only h1 and h2. h1 we can say is around 4 W/m2K (not that important anyway), but h2 depends on how efficiently the brake disc is cooled (cooling ducts etc.). If you want to do any tweaking, h2 value is the one you should tweak, not the other stuff.

    Here is an Excel sheet where I give an example of how to calculate the rFactor coefficients and look at how my calculations compare with rFactor telemetry data:
    View attachment 3427

    Here are my calculated temperatures and rFactor's around Kyalami:
    View attachment 3429
     
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  20. Domi

    Domi Registered

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    Good stuff, Knut. Just one question, how did you extract rFactor telemetry to Excel?
     

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