WE ARE designing a go kart . i am in braking team therefore i did braking calculations accordingly . as i am a rookie therefore i need someone experience to check my calculation . thanks in advance
Calculation
· We are using only rear
braking system over the axle .
First Case: Assuming
Slip occurs.
N = (m*g * C)
Where:
N = normal force on each
tire caused by the weight of the go kart and driver, lbf
m = total mass. (driver + go
kart), lbm
g = constant gravitational
acceleration, ft/(s^2)
C = weight distribution.
Front 0%, Rear 100%
From this formula the normal
forces on front and rear tires are respectively:
N (rear) = (319.67 * 32.17 * 0.100) = 10283.8 lbf
Now that we have the normal
forces acting on rear tires the frictional forces rear can be calculated as
follow:
f(rear) = µ*N(rear)
Where:
f = frictional force, lbf
N = front normal force, lbf
µ = coefficient of friction
* The coefficient of
friction was assumed-to be 0.7 for our tires according to Stratus Karts.
Normally this value varies between (0.1 to 0.9) depending on environment.
Therefore the front and rear
frictional forces are as follow:
f(rear) = 0.7 *10283.8 = 7198.66 lbf
Then:
F= ma
Where:
F = total frictional force,
f(rear)lbf
m = total mass, lbm
a = acceleration, ft/(s^2)
Solve for acceleration it
follows:
a = f(rear)/m
a = - (7198.66) /(319.67
/32.2)= - 725.67 ft/(s^2)
* the negative sign indicates deceleration.
Now we can calculate the
time takes to stop the vehicle
t = u/a
Where:
t = time,
u = initial velocity, ft/s
a = acceleration, ft/(s^2)
t = (41.01/ 72.51) = 0.056 sec
And
d =
(V^2) / (2*a) = (41.01^2) / 2*72.51 = 1.159 ft
Therefore the vehicle would stop in 0.056 second over a distance of 1.159
ft.
Second case: Assuming No slip
occurs:
The fluid pressure that was caused by
master cylinder can be calculated as follow:
P = (FP*R*η) / A
Where:
P = fluid pressure, psi
FP = pedal force, lbf
R = pedal lever ratio
η = Pedal efficiency
A = cross section area of master
cylinder
Fluid
Pressure = (100 * 6 * 0.8) /
(0.3066) = 1565.56 psi
The normal forces acting on front and
rear calipers can be found by following formula:
N = P*A
Where:
N = Normal force, lbf
A = caliper area, in^2
Front:
N(rear) = (1565.56 * 2.4) =
3757.344
*We are using only rear brakes
Once we found the normal forces the
frictional forces could be calculated:
f( Rear)
= µ
N(rear) = (0.4 *3757.344 ) = 1502.94 lbf
*The coefficient of friction was assumed
to be 0.40 for our brake pads.
Now we can calculate the torque cause by
these forces:
Braking Torque = f(
Rear) * D (half of rear track width)
= (1502.937 *1.804) =
2711.20 lbf.ft
Note that "D" is the distance from
each caliper to the center of each moving axle.
Assuming the torque is constant over the
entire length of the axle we can find the forces that are acting on each tire.
F(rear) = (Torque / Radius)
= [2711.2/ (0.4)] = 6022. 54 lbf
Where: "R" is the radius of
rear tires.
The acceleration could be calculated as:
acceleration = F (rear) / mass
=
- [6022.54 /(319.67/32.2)] = -606.5ft/ s^2
t=u/a= 41.01/606.5 = 0.06 s (Stopping time )
And
d
= (V^2) / (a*2)
= (41.01A2) / (606.5 *2) = 1.38ft(
Stopping Distance )
Therefore the vehicle will stop in 0.06second over a distance of 1.38feet.
2
Braking Calculation for Go-Kart
Re: BRAKING CALCULATION FOR GO KART
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