Consider a case where the wheels of the vehicle are not locked. So the kinetic energy of the vehicle is dissipated through heat from the disc brakes and a little amount of energy as frictional work(probably rolling or kinetic, not sure) done between tires and the ground.
Consider the vehicle to be moving at very high speeds, say 100m/s. So how can I calculate the stopping distance for this high velocity? because if brakes are applied at this high velocity, temperature drastically increases at the brake pads which would make them ineffective.
Which of the following ways of braking gives lesser stopping distance?
1. Wheel getting locked and the kinetic energy of vehicle is dissipated by working against kinetic friction between locked tires and road.
2. Wheel are still rotating but the relative motion between the discs of brakes and brake pads leading to heat dissipation. I think little amount of vehicle's kinetic energy is also utilized to do work against friction between rotation tires and the road
I will give a few more specific details so that it might help you understand the constraints I have. We are working on the design of pod for Elon Musk's hyperloop pod competition. So the expected top speed of the pod would be 100m/s.
We are still in the basic design stages and are estimating the stopping distance of the pod. No braking system is developed yet, but assuming that I will employ a braking system similar to that of current four wheelers in the market, I would like to make a rough estimate of stopping distance.
Stopping distances from real world testing of four wheelers in the market are as follows:
Velocity: 26.8 m/s - Stopping distance: 35 m
Velocity: 44 m/s - Stopping distance: 100 m
A very rough calculation would be to consider that stopping distance is directly proportional to the square of velocity (the maximum velocity at which brakes are applied and then vehicle comes to rest). This holds good with a small error if the velocities I am considering are in the similar range but I am calculating the stopping distance for such high velocity i.e., 100 m/s. At such velocities, the physics behind the brakes would completely changes as the relative velocity between the brake pads and disc would be very high leading to higher temperatures and very low frictional coefficients. So the following are my queries:
1) is there a way to compare the two conditions, velocities of (26.8m/s & 44m/s) and 100 m/s to get a good estimate of stopping distance?
2) What would be an efficient braking system for such high velocities? using multiple brake pads and more number of wheels so that energy dissipation is distributed and done at multiple locations?
Competition track itself is 1 mile, so we would want a minimum possible stopping distance.
I think, the vehicle kinetic energy is dissipated mostly through the tires-to-ground static friction rather than through the disc brake heat. You can check it by trying to stop a car on the icy road, where the contribution of road to the energy dissipation is zero. Moreover, the tire-to-ground friction should always be static, which is provided by the antilock brake system and, what is important for you, the static friction is constant and does not depend on speed, of course if the wheel is not sliding or flying.
For estimation of stopping distance using assertion above we can use the data you presented and simple physics. All vehicle kinetic energy (Ek=mv^2/2) is transformed into a work done by friction forces (F), which is equal to W=Fx, where x is stopping distance. The equations of energy balances we can get from for data set you presented for two different speeds (v1 and v2 ): mv1^2/2=F1*x1 and mv2^2/2=F2*x2. Dividing the second equation by the first one and using numerical values from your data sets, we will get (44/26)^2=(F2/F1)(100/35) and interesting ratio: F1/F2=1 with 0.2% accuracy! The obtained result proves the idea, that the static friction is responsible for the stopping distance regardless of the speed of the car (or the angular speed of the wheel).
Now, the final answer of your problem stems from equation 2 (or 1). Calculating the constant C=2F1/m=2F2/m=v2^2/x2=v1^2/x1=19.36 we obtain the stopping distance X for the speed of v=100m/s: X=v^2/C=100^2/19.36=517m.
Bocking wheels lead to a reduced braking acceleration in comparison with not blocking wheels, and thus to a longer stopping distance. The reason can be found in the lower friction coefficient of a sliding wheel in comparison with a rolling wheel at the limit of static friction. Therefore, ABS braking has advantages in comparison with non-ABS braking and sliding tires.
To calculate the stopping distance, several factors of longtudinal driving dynamics have to be considered. Details can be found in the attached publication.
Kind regards,
Mario
Conference Paper Basics of longitudinal vehicle dynamics
From conservation of energy, you're converting kinetic energy primarily into thermal energy in the brake discs. You will also dissipate some in the air (but less in a reduced pressure), tyre rolling resistance, or as recuperated energy. The friction between tyres and roadway only limits the amount of force that you can apply - note that if you use magnetic levitation this reduces the load on the tyres, so you can't brake as hard.
So assuming that you're using the wheels for braking, you're still going to have convert that kinetic energy into thermal. The mass doesn't change the braking distance, because you have moire energy to convert, but more braking effort available - but more mass will mean hotter brakes. That's why big vehicles have big brakes - not so much for more effort, but to dissipate more heat. And the braking distance from 100m/s will be about 500m, unless you use some non-friction system (retro rockets, magnetic coupling, air fans, whatever). This isn't a problem for a 350 mile journey, but is a limit for a 1 mile test loop.
If you used heavy vehicle tyres, they are designed for long life. Better tyres mean better braking and a shorter stopping distance.
Interesting problem - good luck!
Please, look at the Braking distance article in wikipedia (link below) and references therein. As I wrote above, the only tires-to-ground static friction is responsible for a dissipation of car kinetic energy regardless of the mass of car. In difference with wikipedia, my result (517m) is obtained on the basis of testing data set.
https://en.wikipedia.org/wiki/Braking_distance
Hello Siddharth!
Generally braking distance C=(V^2)/ (26*a)
V-initial velocity,
a- deceleration during braking (depend on many things)
If you know Abaqus then try this approach: https://things.maths.cam.ac.uk/computing/software/abaqus_docs/docs/v6.12/books/exa/default.htm
Mr Samvel-> you are wrong, What have static friction coefficient to do in case of braking with ABS?
Thank you everyone for the answers.
1. Firstly I am still confused with the mixed response to this query. Will most the Kinetic Energy of a vehicle turns into heat at the disc brakes or converted into work done against friction at Tire-road contact? A clear explanation or source of reliable information would be great.
2. Mr. Bartosz, it was hard for me to imagine but according to many websites static friction will be present in case of ABS where the tire will still rotate while the brakes are applied and it will be kinetic friction if the wheels are locked. It would be of a great help if someone explains how it is static friction but not rolling friction in case of ABS. Professor Samvel, your explanation would be of a great help.
3. Mr. Bartosv, can you briefly explain what factors to be consider for determining the deceleration of vehicle and how to calculate it?
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