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Hi evryone,
My name is adekunle, m making some scratch calculations for a brake design system and I m wondering whether the front or rear axle max deceleratn is twice that per front/rear corner or its the same on the axle and corner.
Sorry if the question sounds lame.
The question doesn't make sense (to me at least), can you re-phrase?
Ok,I want to have a max -ve acc of 1.5. So dividing my dynamic weight by force generated by front tires give me 3.0G that mean that the G force at each front corner is 1.5G,ryt
But I was told that its not so, its suppose to b 1.5G for the front axle nt 3G and 1.5G per corner. Do u understand me nw.
adekunle,Originally Posted by adekunle
Draw a free body diagram.
Car needs to decelerate a 3 g (whoa). You have 4 tires. Each tire contributes to the deceleration. You have a cg height, wheelbase, and static F/R weight distribution. Calculate the wheel loads after weight has been transferred due the deceleration. Assuming the car is symmetrical left to right, you have a coefficient of friction (mu) for the front tires, and a mu for each rear tire. You now have each tire's contribution to your deceleration. Now you do some addition..
@ motorcity1,
Wel I actually did all dat before I got to my max G final result, c.g height, wheel base, weight transfer and al. Already don dat. So u sayn I need is dat 1.5G is the max G dat each corner tire contributes to the axle dat brakes at 3.0G.ryt?
Yes, i believe that's 'ryt'
I'd check your numbers though...3g braking implies something like 60mph to 0mph in under a second. I'd imagine that is quite hard for your average fsae car.
Last edited by MotorCity1; 08-14-2015 at 04:40 AM.
adekunle,
I'm still not sure I understand your exact issue.
But in reference to the wording of your above comment specifically, I think there are some fundamentals that you do not convey an awareness of.
There is a single mass (your vehicle), so by F=ma, a single vehicle acceleration. There are four sources of force that act to generate the acceleration of your mass - these are the tyres. If you were to call the Brake Force from your Front Left tyre "BFFL", by summing the forces from all four of your tyres, the acceleration of your car would be found in the equation:
(BFFL+BFFR+BFRL+BFRR) = m.a
Note again - four force values, one acceleration value. You do not sum accelerations at each corner.
I hope this clears up any confusion you may have. If it does not, perhaps if you tell us your 'inputs', or the values you do have (tyre/vehicle parameters?), the values you are trying to attain (vehicle acceleration/braking performance?) and tell us the way you are trying to attain these outputs, we can check whether you are right or wrong, and give specific advice where needed.
Unless your car is made of slinky's, the whole thing should be decelerating at the same rate. The FORCES from the front and rear tires can be different, but the ACCELERATION at the front is the same as the rear, and same as the CG, and same as any other point on the car.
Last edited by JT A.; 08-14-2015 at 05:58 AM.
For a car with a static weight balance of 50/50, during deceleration of 1.5G, the normal force on the tyres maybe 80/20 depending on your center of gravity height. As it works out with simple calcs and no aero, if you can brake at 1.5G and no quicker, your tyre mu is also 1.5
University of Tasmania (UTAS)
I once had a car that would stop at 1G on the left side but only 0.5G on the right side, it would turn left really well but turning right was a different story.... I had another car that one day the car stopped at 0.8G but the left front wheel stopped at 0G, that was the day my car turned into a trike....
Joking aside, adekunle, what does the "G" you're asking about refer to? How is that "G" calculated?
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