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Calculating the force in suspension as a car hits the brakes

  1. Feb 3, 2016 #1
    Firstly thank you for reading my post.
    This is an interesting question I was wondering if anyone here could teach me how to work out.

    Say if you had a large go-kart (FSAE/Formula Student if anyone knows of that) and you were driving at 120kph and hit the brakes and stopped in 3 seconds, how much force is going through the suspension spring/dampener.

    Similarly what if it is going around a corner with a radius of 7.7kph at 80kph.

    Here is a diagram of what I am talking about. Its double A-Arm wishbone suspension, along with any data you will need, such as weights and dimensions.


    cough cough http://brage.bibsys.no/xmlui/bitstream/handle/11250/183058/Master oppgave-endelig.pdf?sequence=1
  2. jcsd
  3. Feb 4, 2016 #2

    Ranger Mike

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    Ben..nice question...you are not addressing one HUGE part of the question...
    in racing, its all about tires, Tires, TIRES..
    tires are what interfaces between the immovable asphalt ( could be dirt but i digress) track and the moving object ( in this case..the race car).
    also the tire compound must be addressed. these things are covered in the slip angle charts for tire adhesion by the tire manufacturer. Other factors involved are track conditions..i.e. is track slick due to sum light or is track cold due to cloudy conditions..all impact the brakes...and what compound brake pads are you using or is it drum brakes?

    so you can not properly calculate the force going thru the suspension until you know the force countering the momentum...
    i suggest you read Race CAR Suspension Class in Mechanical Engineering section of this fine forum
  4. Feb 4, 2016 #3
    Force = mass x Acceleration (in this scenario, deceleration)
    A= (120km/h)/3s = 11.1 m/s^2
    F= (350kg )11.1m/s^2
    = 3889N
    Now that's the total force from the car stopping from 120kph to 0 in 3 s. How it gets distributed throughout the suspension, I wouldn't be able to tell you.
  5. Feb 5, 2016 #4

    Ranger Mike

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    from a crew chief prospective

    Longitudinal Load transfer (LT) – Braking at 1.2 G retardation
    Vehicle weight x Center of Gravity Height / wheelbase
    An Indy car weighing 1760 lbs. total with 100” wheelbase and CG at 13” would have
    LT = 1.2g [ 1760x 13 / 100] = 275 lbs.

    the 1.2 G needs to be known..
    this is considering LOAD transfer as the front suspension would "feel" if it was possible to place this car on scales at speed..it is used to calculate spring rate required to reach maximum handling. Other factors are diagonal load transfer in a turn or while cornering but that is a whole new topic.
  6. Feb 8, 2016 #5
    Tire data on request from manufacturer.
    Tire PSI of 22.
    A value of 3,500N to 4,000N from braking is roughly what I was expecting. Perhaps a bit lower.

    Ranger Mike here is what data I have for the tires and I'm working on getting extra information from a team mate. Yes how would I calculate the diagonal load transfer in a 7m radius turn during flat cornering. I read your MEGA-thread. Excellent writing. I couldn't find any section about diagonal loading though, especially on actually calculating , aside from the software you mentioned. Is the software free to download?

    Last edited: Feb 8, 2016
  7. Feb 8, 2016 #6

    Ranger Mike

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    Race car suspension class, page 37 post # 676 tells you how to figure the total amount of " weight transfer" as well as the amount going forward. Subtracting the amount going forward from the total will approximate the diagonal amount of " weight " transfer. This calculation is used to approximate the required spring rate to handle this.. " weight". It is a good way to begin to understand the dynamics of the suspension in a turn.

    On page 41 the true definition of LOAD transfer is for the purist who wants to understand the physics behind the process.
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