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Braking Calculation for Go-Kart

11/25/2015 2:25 PM

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.

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#1

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:41 PM

Ask your advisor to be transferred to a different team.

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#2

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:42 PM

So 28mph to zero in "0.06second over a distance of 1.38feet."

Now you can contact the safety harness department!

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#4
In reply to #2

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:48 PM

Or call Takata!

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#10
In reply to #2

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 4:30 PM

40ft/sec to nothing in 1/16 of a sec is 21G; the calculation is nonsense.

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#11
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Re: BRAKING CALCULATION FOR GO KART

11/25/2015 4:37 PM

Stop, it hurts when I laugh! Ouch!

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#16
In reply to #11

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 3:47 AM

A 21G deceleration is no laughing matter.

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#3

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:47 PM

Are you saying you have 10283.8 pounds of force on the rear tires? That is one hefty go-cart.

I would assume that 319.67 is the weight of the go-cart in lb. You would only multiply by 32.17 if the mass were in slugs (which weigh 32.17 pounds each). Get rid of g!

How can you have 100% of the weight on the rear tires and still steer with the front tires up in the air? You weight distribution probably is higher on the back than the front because that is where the engine is and close to where the driver's seat is, but not 100%.

You should look at the numbers and see if they make sense before proceeding.

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#6
In reply to #3

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:58 PM

Ok i got it how stupid calculation i posted over here . first thing i learn is . weight distribution shpuldnt be 100 it should 40% front and 60% rear right ??

Second thing is i dont have to use g while calculating ormal force right ??

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#8
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Re: BRAKING CALCULATION FOR GO KART

11/25/2015 3:10 PM

60% - 40% sounds about right. g converts slugs to pounds, but most folks use pounds for mass. Your initial calculation threw the whole thing off. Run the numbers through again and see what you come up with.

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#5

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 2:52 PM

Use a dead axel, you won't need brakes.

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#7

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 3:03 PM

Take the computation the other way around: which will be the decceleration of the body (driver) and how much could he support.

A second aspect is the the adherence to the ground. The maximal friction between tire and ground occurs by a limted slip. Look in the docs for the tires how big it should be. If you brake with a too high moment the tire will slip and loose adherence and the friction will decrease, the consequence is a loss of guidance since tires guide the vehicle via ground adherence.

Do not ask to be put in an other team your calculations are ok but not from the reality point of view. Change the point of view and you will get good results. If you want more support after you did a new approach you can contact me on the private channel to avoid disturbances for non interested readres.

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#15
In reply to #7

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 2:31 AM

Mass of car = 145kg

Max speed = 65 kmph

Coefficient of friction between tyre and road = 0.7

Front tire size : 10* 4.5 -5 , rear tire size : 11*7.1-5

Fluid pressurr : 1565.65psi

Caliper size : 44.45mm

Master cylinder : 15.87 mm

How can i find the deacceleration and stopping time . . can u please help

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#19
In reply to #15

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 11:29 AM

Mass of car = 145 kg

Max speed = 65 km/h

Coefficient of friction between tire and road = 0.7

Front tire size: 10* 4.5 -5 , rear tire size : 11*7.1-5

Fluid pressure: 1565.65 psi

Caliper size: 44.45mm

Master cylinder: 15.87 mm

How can i find the deceleration and stopping time . . can u please help

here are some comments and guide lines:

1- The coefficient of 0.7 is the sliding coefficient but not the rolling one.

2- According to the "engineering tool box" the rolling resistance can be computed with the equation:

c = 0.005 + 1/p (0.01 + 0.0095(v/100)2)

Where c = rolling coefficient or you can consider values as 0.03 (on asphalt) and 0.2-0.4 on solid sand.

p = tire pressure (bar)

v = velocity (km/h)

3- For a "good feeling" a human body should not have to support more than 0.3-0.4g as acceleration. Being a go-cart for young men this limit can be multiplied by about 2-2.5.

4- Your breaking moment should be so that in no case the sliding makes your vehicle uncontrollable. This happens when for instance you break too much on a wet or icy ground.

5- When you break the distribution of loads on the wheels is not any more the static one you considered. The upper structure inertia increases the load on front wheels and decreases the part of weight supported by rear wheels. This MUST be taken into consideration.

6- On a dry ground you can accept that from 50km/h to zero a length of at least 7m is required. Your velocity is 65 km/h so that you should consider a breaking length of 7*(65/50) ^ 2=12[m].

Assuming a constant breaking resistance you have the equation:

S= a*t^2/2 with a= deceleration [m/s²]= 0.3*2.25=0.675 [m/s²]

T=(2*S/a)^0.5=5.96 [s] which much more realistic.

7- Breaking is equivalent to the reduction of the kinetic energy the vehicle has when the process is started. In fact during this period the rolling resistance is still active but decreases rapidly since the vehicle speed figures at the square so in the first step it has to be neglected. The problem can be set this way: how big should be a resistance torque placed at the rear axis so that in 6[s] the whole kinetic energy will be transformed in heat. The working done by such a moment is W= T*Φ=E=M*v²/2

Φ= angle of rotation for the breaking time Φ=ω*t = velocity *time or Φ=ε*t²/2 with ω=ε*t ε being the angular acceleration of the rear wheels. ε= a/r with a= above linear acceleration = 0.675 [m/s²] and r= rear wheel radius.

Knowing Φ you can compute T= E/ Φ take care to have Φ in radian and NOT in degrees!

8- Now you know the torque you must apply to break your cart. Its value is well computed by the product T=2*F*µ*R with F- applied force by the hydraulic cylinder/ µ the friction between disc and brake and R the mean acting radius of the caliper on the disc. F is as you computed F= p*A cyl.

9- Now comes the problem how do you define "p"?

10- The driver will press with his foot the pedal and its force will be applied considering also the ratio of distances in a pressure in the master cylinder.

To reach the correct breaking torque range you have 2 possibilities to adjust:

- Distances at the pedal level.

- Acting radius of caliper on disc.

After you did all what is listed above you MUST check that in an emergency situation under panic when the driver will press the pedal with his full weight the torque will not be over the sliding limit!

This is what I meant by changing the point of view in my comment.

If other questions appear continue to ask. Same way you do the job as you think and present it after and you get the correction.

After you did this I shall ask you to consider the decreasing effect of deceleration on the rear wheels and check their adherence for the safe limit check.

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#20
In reply to #19

Re: BRAKING CALCULATION FOR GO KART

12/01/2015 2:01 AM

thanks for your helping hand in my calculation . i did calculation again and got these values which i think IT SHOWS realistic value this time . please correct me where im wrong

deacceleration = -6.86m/s^2 = 0.7g

stopping time = 2.02 sec

stopping distance = 14m

i calculated the fluid pressure also = 107.94 bar (1565.56psi)

taking master cylinder of 15.87mm and caliper of 44.45mm

now i wanna know the clamping force and braking torque .

so according to me for clamping force cal. we first multiply fluid pressure to are of caliper .

and for braking torque we need to multiply coeff. of rolling friction to clamping force and then multiply the resultant by distance of caliper from centre of rotor dics.

am i going on the right trail nickname ?

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#21
In reply to #20

Re: BRAKING CALCULATION FOR GO KART

12/01/2015 11:55 AM

Let us make some assumptions first we accept that the deceleration is constant from braking start to end. This means that the velocity is decreasing linear V(t)=v(0)-a*t. Il follows -if we note the raking time with "τ" → a=v(0)/τ. The braking length will be L= 0.5*v(0)*τ. → τ=2*L/v(0).

You assume L=14[mm] with a v(0)= 65[km/h]= 18.06[m/s] →τ=2*14/18.06=1.55[s] and a= 18.06/1.55=11.65 [m/s²] = 1.187 g and not as you write 0.7g. Independently of previous values this seems to be a too high deceleration. It means that we have to accept a longer braking stroke.

Starting with a= 0.7 g=6.87 [m/s²] →τ= 18.06/6.87=2.63 [s] and L= 0.5*18.06*2.63= 23.75 [m].

Braking torque should be so that the kinetic energy E (0) = M*v(0)²/2 will be transformed in heat by the friction with a torque "T" over an angle Φ [radian]= L/(π*d) where d= wheel diameter (about 0.3 [m]). You get the equation → M*v(0)²/2=T*L/( π*d) → T= M*v(0)²*π*d/(2*L) =938 [Nm].

You have to check if this torque is less the sliding limit T*= G*µ*d/2 =145*9.81*µ*0.3/2=213.37*µ.

This means that if the ground is wet for instance and µ≤ 938/213.37=4.4 the sliding limit will be reached and a loss of control can be expected. If µ>4.4 then driving is safe.

Now for the brake itself:

Caliper size: D=44.45[mm] → A= 1.55 E3 [mm²]

Master cylinder: dc=15.87 [mm] → Ac=1.978 E2 [mm²]

The caliper should develop a braking moment of 938 [Nm]. The force with which it applied on the disc is F=A*p *ηb and the torque will be M=2*F*µb*Rb =2*µb*Rb*A*P*ηb.

The master cylinder has to have applied a force Fp= Ac*p/ ηc. → p= Fp* ηc/Ac

It follows that M=2*µb*Rb*A* ηb* Fp* ηc/Ac =2*µb*Rb * ηb* ηc * Fp*(A /Ac).

But Fp= Ffoot*C pedal where Cpedal is the ratio of the distances between pedal axis and foot or master cylinder.

The braking should not be > 213.37*7=1494 [Nm] sliding friction coefficient assumed being µ*=7.

An average human has a weight of 750…850 [N]. A normal foot force is around 1/6 of it and an emergency one is about 2…3x higher.

Let us assume Rb=0.08 [m] (half of wheel radius), both efficiencies = 0.85, µb=0.3 then

938 = 2*0.3*0.8*0.85²*(1550/197.8)*Fp → Fp= 3452 [N]. If we accept the force on the pedal Ffoot= Weight/ = 125…142 [N] then the ratio of the distances foot to axis and master cylinder to axis should be i=3452/(125…142)=27.6…24.3. The pressure for the "normal" braking p=14.83[N/mm²] which is a pressure in the usual range. Since the pedal force can be multiplied by 2.5…3 in case of emergency reaction the risk of sliding is present since the T*/T= 7/4.4 = 1.59!

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#22
In reply to #21

Re: BRAKING CALCULATION FOR GO KART

12/02/2015 3:46 PM

nickname ... in first three lines u wrote T =2L/V(0)

shouldnt it be T=L/V(0) coz according to 2nd equation L(distance)=v(0)*T +1/2 *A*T^2

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#23
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Re: BRAKING CALCULATION FOR GO KART

12/02/2015 7:34 PM

No, because the velocity goes from v(0) which is the initial velocity to zero. according to the assumtion that the deceleration is constant velocity versus time is a straight line. Displacement is the area under the velocity curve ( I do not know if you are familiar with the integral notion). In this case the area is a triangle with area v(0)*time from start braking to end and divided by 2.Is it clear now?

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#24
In reply to #23

Re: BRAKING CALCULATION FOR GO KART

12/31/2015 3:22 PM

Thank you so much nickname.for your guidance. I made my calculations on what you taught me . and my advisor was impressed . thanks once again :)

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#9

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 4:23 PM
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#12

Re: BRAKING CALCULATION FOR GO KART

11/25/2015 5:16 PM

himankambashta,

You display a common, but alarming, characteristic with many new engineering students.

That is that you seem to think that mathematical formulae can answer any and all questions by simply plugging in the numbers, without adding any logic or real world thought to the reactions and forces involved in the mechanical interactions of your math.

Such as:

rear braking system

weight distribution. Front 0%, Rear 100%

Therefore the vehicle would stop in 0.056 second over a distance of 1.159 ft.

The normal forces acting on front and rear calipers

Finally, think about what your numbers mean IN REAL LIFE!

nick name is a very knowledgable engineer. Take him up on his offer of help. He will be an excellent source of advice and knowledge.

Good luck.

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#13
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Re: BRAKING CALCULATION FOR GO KART

11/25/2015 7:50 PM

So no common core I'm thinking....

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#14

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 12:30 AM

I concur with the comment that you need to do more than juat plug numbers into a formula. You need to understand what you are doing and what the numbers are. One method is to visualise where all the forces are going and how they interact - but maybe that is easier once you have done it a bit.

I had a staff member once show me a formula for the load on the front and rear axles of a vehicle braking to a stop, that was "Pstatic plus or minus X" and on that basis he went on to claim that this showed why the rear brakes were cracking (from overheating), 'cause, as he said, "the load is going to the rear axle under brakes" ! - worse, he was supposed to be qualified. !!!

HINTS:

1. 1g is the rate you fall at and is about as quick as you can stop and is equal to 32 ft/sec. Any accel result too different from this is wrong.

2 Difference in magnitude of a force unit and a mass unit is "g" and the name for a force unit in the "english" system is lbal not lb or lbf. Equivalent in metric is Newton compared with Kg.

3 When the kart stop, what will the force of the driver do to the Kart, and will the weight distribution still be 50% 50% or whatever it was static.

4. Are the tyres hot or cold - huge difference in friction in a kart tyre

Time to start learning.

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#17
In reply to #14

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 3:50 AM

<...32 ft/sec...> is a speed. 32 ft/sec/sec is an acceleration.

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#18
In reply to #17

Re: BRAKING CALCULATION FOR GO KART

11/26/2015 3:57 AM

pwslack how you got tgis result .please care to explain .

And nickname thNks. for your concern but one thing i didnt understood is what do u mean by saying that i should do calculation from diff point of view .

Thanks to all of you , i learned that number should have practical meaning

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#25

Re: BRAKING CALCULATION FOR GO KART

01/20/2016 12:53 AM

Since a brake functions by converting kinetic energy (motion of GO KART and DRIVER) into heat, I would think that the first thing you would have to determine is how much kinetic energy you are going to have to dissipate at maximum speed of the vehicle.

Once you have determined this, you are going to have to determine both the rate of deceleration required, and the maximum rate tolerable by the components of the vehicle, again including the driver.

You assumed (we know what that means) a coefficient of friction for the tires, but that is going to vary with track surface and temperature (note that professional race teams carry multiple sets of tires with different treads and compounds for just such situations).

I see differential, as well as higher order polynomial equations being involved, and I don't think such complexity should be handled on this forum,

Once you determine the equations required for a given set of variables, then it is relatively easy to change one variable at a time.

While this "backwards approach" may seem primitive, you are going to have to design for a limiting variable.

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