Im currently looking into kart brakes, but I have difficulty to match the calculations.

I have datalogged in the past a Cadet kart and that was able to brake with 0.7G (7m/s^2). Also I have articles that confirm this value as a realistic one for karts.

However, when I calculate the maximum possible braking force at the rear axle.
A typical kart can only generates around 750N of brake force before the wheels lock up. I did account for the weight transfer to the front and use a tyre to road friction number of 0.8. (F=M*g*u=97*9.81*0.8)

To put a kart to a stop with 0,7G, you would need 1700-1800N at the wheels. (F=m*a=220kg*0.7*9.81=1750N). Which is not unrealistic to produce when you look at the pedal force, pedal ratio, mastercylinder, piston caliper, disc diameter etc. of the kart.
But again the rear wheels supposedly lock up at 750N.

So what is the maximum possible braking force at the rear wheels?

Sorry for overflow of numbers.

Anyway, an nice brake calculator and a nice picture for your trouble!

http://www.jakelatham.com/radical/info/brake_calculators.shtml

Catch is Tyre friction is non linear without a great deal of detail about the tyres friction characteristic under varying load you can get a proper result.

Thats true, but I cannot believe that the tyre friction will go near the number of 2.



[Edited on 11/8/10 by ettore bugatti]

I know this could be a silly question, but you did do the calcs to include all 4 wheels !!

What do you actually mean?

There's no front brake.

Or could I dissmiss half the weight of the kart, because the front wheels never lock up?

[Edited on 11/8/10 by ettore bugatti]

Is the 220kg a laden weight? Remember that some of the deceleration will be due to friction and engine braking. How fast does the kart slow if you just lift off the throttle?

Out of interest,

why do you need to know?

220kg is indeed the laden weight, it is not a lightweight race kart.

The deceleration by engine braking is just a part of the total sum of the forces that are required to stop a kart.
Im after the total value and the relation to locking the rear tires.

I want to know this for the stress analysis of the brake caliper bracket and chassis mounting point.
And secondly because I want to understand the physics of braking.

I would guess (unless you've checked of course) maybe you're underestimating the rear axle load (driver's mass is mainly at the rear and the engine is over the axle, and also the friction of the tyres. If you had a coefficient of 1.2 and an axle load of 140kg you be pretty close to your desired 1700N.

friction coefficient for a tyre is dependant on slip angle and normal forrce.
If you read this article:
http://alexandria.tue.nl/repository/books/627108.pdf

you see that thar avon give the following formula for their cart tyres:

u=2- (1.85e-4)F dry
u=0.87-(0.64e-4)F Wet

where u = coef of friction, and F is the normal force in newtons E.g with approximately 50kg on the rear (each tyre), u~1.9dry, 0.84 in the wet.

Regards
Hugh

ETA, so your friction/slip angle graph above would apply for a competition tyre loaded to about 4KN as this would give a maximum u of approx 1.2 as shown on your graph - e.g more typical for a road car.
hugh

[Edited on 11/8/10 by hughpinder]

To increase the braking effort just do what Karters do and use the Kart in front to slow you down.

quote:

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Originally posted by hughpinder
friction coefficient for a tyre is dependant on slip angle and normal forrce.
If you read this article:
http://alexandria.tue.nl/repository/books/627108.pdf

you see that thar avon give the following formula for their cart tyres:

u=2- (1.85e-4)F dry
u=0.87-(0.64e-4)F Wet

where u = coef of friction, and F is the normal force in newtons E.g with approximately 50kg on the rear (each tyre), u~1.9dry, 0.84 in the wet.

Regards
Hugh

ETA, so your friction/slip angle graph above would apply for a competition tyre loaded to about 4KN as this would give a maximum u of approx 1.2 as shown on your graph - e.g more typical for a road car.
hugh

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