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Vehicle dynamics: Calculation of Anti-Roll-Bar roll rates [Help needed]

Discussion in 'Assetto Corsa' started by Georg Siebert, Dec 5, 2014.

  1. Georg Siebert

    Georg Siebert

    Hi everybody,

    been diving into vehicle dynamics for setup testing in AC recently. I've figured out Aerodynamics, Spingrates & Damper settings but I need some help with the approach and execution of determining ARB roll rates. If you want to read up on the material first, I'd recommend the pdfs from the link at the end. AC gives and uses a lot of values & data derived from real life physics. Prior work on aero, springs & dampers has proven the accuracy of this crossover approach to the game.

    This post has a lot of math involved, so take it as it is. We calculate in SI units, so no lb.ft or something. Because we need most of the solutions for follow up equations, rounding and valid numbers are ignored. FYI, I totally don't get the aversion some people have against math, but let's get to business shall we.

    mid-engined GT car which generates ~1000 kg of downforce and ~600 kg of drag at ~280 km/h
    Total mass = 1100 kg
    Total weight = 10791 N
    CG = 0,48 @ front
    Wheelbase = 2,770m
    Sprung mass - front (single wheel) = 226,56 kg
    Sprung mass - rear (single wheel) = 245,44 kg
    Total sprung mass = 944 kg
    Springrates - front = 122500 N/m, rear = 125600 N/m
    Ride frequencies - front = 3,7 Hz, rear = 3,6 Hz (undamped natural frequency in ride)
    Desired total roll gradient = 0,7 deg/g (degrees of body roll per g of lateral acceleration)
    Trackwidths - front = 1,620 m, rear = 1,550 m, average = 1,585 m
    Tirerate front = 313524 N/m, rear = 321516 N/m, average = 317520 N/m
    Magic Numbers = 55% @ front, 45% @ rear (Total lateral load transfer destribution, called by Milliken, or percentage of roll gradient of front suspension in normal English)

    Motion Ratio
    Motion Ratios (MR) are an enigma in AC. There are 1 for Spings, but I have no idea what to choose for ARBs - just set 1 for them as well for now. Because they are 1 for Springs, that means in AC: Springrate = Wheelrate.
    The formula for MRs in the pdfs (e.g. pdf 3) cited below is wrong. I've derived the right one as follows:

    WR = SR * (MR)^2

    WR = Wheelrate
    SR = Springrate
    MR = Motion ratio

    The actual ARB roll rates come from:

    Formulae 1: ARB stiffness

    FARB MR & RARB MR = 1

    Total roll rate (K_phi A) needed for Formulae 1 calculations is:

    Formula 2: Total ARB roll rate

    Note, that K_W in the equation is wheelrate at a single wheel. I've taken the average of 124050 N/m here. This big one also takes the springrate of the tyres into account.
    We have every other value at hand apart from the desired total roll rate:

    Formula 3: Desired total roll rate

    H, the vertical distance from the roll center axis to the CG I don't have and had to guess.
    H = 0,1 m
    Being a factor in the divident of the fraction, this value has a huge influence on the solution of this equation.

    The roll rates without taking the spring rate of the tyres into account are much simpler:

    Formula 4: Front roll rate (w/o tyres)

    Formula 5: Rear roll rate (w/o tyres)

    done by hand traditionally with pen & paper

    Formula 4 (F4)
    K_phi F = [pi * (1,620 m)² * (122500 N/m)²] / [180 * (122500 N/m +122500 N/m)]
    K_phi F = 2805,52 Nm/deg

    K_phi R = [pi * (1,550 m)² * (125600 N/m)²] / [180 * (125600 N/m + 125600 N/m)]
    K_phi R = 2633,30 Nm/deg

    K_phi DES = (10791 N * 0,1 m) / (0,7 deg/g)
    K_phi DES = 1541,6 Nm//deg/g

    K_phi A =
    pi / 180 * { [1541,6 Nm//deg/g * 317520 N/m * (1,585m)² / 2] / [317520 N/m * (1,585m)² / 2 * pi / 180 - 1541,6 Nm//deg/g } - [pi * 124050 N/m * (1,585 m)² / 2] / 180
    K_phi A = pi / 180 * (6,149 * 10^14 (Nm)²/deg) / (5419,49 (Nm)²/deg) - 2719,59 Nm
    K_phi A = 1,9801 *10^9 Nm

    what?? 2*^10^9 ?

    in F1:
    K_phi FA = 1,9801 *10^9 Nm * 55 * 1/100
    K_phi FA = 1,089 *10^9 Nm

    again, something completely crazy which can't be real

    In summary
    The solutions for F3, F4 & F5 are understandable, but F2 & F1 are out of the ordinary to say the least. Values that should be looked at are H from F3 and K_W from F2. Any other value is either a constant, a given or a chosen value, all of whom should be solid.
    I wanted to take the solutions from F1 and set them as ARB values in the setup menu, but the ones above are useless. The unit for ARB stiffness in the game is Nm.

    Here are the questions:
    Is this at all the right approach to determine ARB stiffness values?
    Are the solutions from F1 the right ones to apply for the ARBs in the game?
    Are these at all the right formulae to use?
    Are there any mistakes in these calculations?

    If you have knowledge of vehicle dynamics / kinematics and/or have read the excelent book from Millikan on these topics, please share that as well.

    Most equations taken from pdf #2 here:
    (Spings & Dampers section)

    looking forward to your replies,
    Last edited: Dec 5, 2014
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  2. John Cunningham

    John Cunningham

    Sorry I don't speak Klingon, but if you get that teleporter to work I will have a go.
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  3. Georg Siebert

    Georg Siebert

    Derision, ridicule and mockery as responses I didn't had in mind, @John Cunningham , when I was writing this thread.
    Last edited: Dec 5, 2014
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  4. Connor Caple

    Connor Caple
    Slowest Racer in Town...

    I suppose, in part, in depends on what the figure in AC represents, doesn't it?

    As observed from your calculations, the various formulae provide wildly differing results.

    Perhaps the only way to find the correct answer for the AC setting would be to ask Kunos what it represents and how they calculate the displayed value, then reverse engineer it from there?
  5. Stereo


    The units for K_phiDES should be Nm, not Nm deg/g, in order for units to match up in eq2, also when I run eq. 2 I get K_phi A = -739N.m which I guess means for your target scenario you need the opposite, something decreasing spring rates on roll vs. regular compression.

    I'd expect H more like 0.3m which gives K_phi A = 10,000 N.m if I ran things right, don't know the car in question though.

    Converting from N.m / deg back to N/m wheel rate (the ARB in AC does use a 1 motion ratio and is in either N/m or N/mm depending on the setup) N.m becomes N via dividing by half the track (converting torque to force), deg becomes m by multiplying by 180/pi * track, so overall 10,000Nm * pi/(180*1.585^2) = 110N/m ARB. Maybe the car in question just naturally has near the desired deg/g on springs alone; I know that some high downforce cars have been in that situation.
    Last edited: Dec 5, 2014
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  6. John Cunningham

    John Cunningham

    Georg, no offence was meant at all, in fact I am mocking myself; my sad lack of understanding. I appreciate your post as it may raise my utterly poor levels of knowledge. Please assume from now on any of my early morning, before coffee posts, are definitely not meant to upset you in anyway!
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  7. assettonoob


    Your post seems like nonsense to me, but I never did well at Maths! The 'Black Art Dynamics' site has lots of good stuff. Link:
    The same guy is also developing an app focused version of the above site and should make setting up any car easier, as long as you have the right data:
  8. Georg Siebert

    Georg Siebert

    @Nick Murdoch

    Thank you both, that was helpful. The car is a Mazda Furai.
    Will run them again with different values for H.

    Looked at the blackartracing/blackartdynamics sites, turns out this is a failed kickstarter project and hasn't been updated in a year. The equation for spring rate seems wrong and the author uses non SI units mixed with SI units mixed with multiplyer prefixes (kNm) so casually it's hard to follow.
    His text about load transfer & tyres seems alright, tho.
    Last edited: Dec 5, 2014
  9. PhilS13


    K_phi A =
    pi / 180 * { [1541,6 Nm//deg/g * 317520 N/m * (1,585m)² / 2] / [317520 N/m * (1,585m)² / 2 * pi / 180 - 1541,6 Nm//deg/g } - [pi * 124050 N/m * (1,585 m)² / 2] / 180
    K_phi A = pi / 180 * (6,149 * 10^14 (Nm)²/deg) / (5419,49 (Nm)²/deg) - 2719,59 Nm
    K_phi A = 1,9801 *10^9 Nm

    This doesnt work. I get 6.149 x 10^8. Not 10^14.

    That would bring your F1 value around 1000 Nm