Department of Design and Manufacture Technology

PE4112- Production Technology Coursework #1

Student ID: 14144662

Vehicle Assignment:

Preliminary Design

Submitted: 20/02/2015

Design Procedure:

1. Estimating the weight of the Vehicle:

Body:

220x100 Flat piece of aluminium (as per design drawing) with two 5x220 and

two 5x100 stripes and 116x43 hole for soft drink can (as per design):

(22*10*0.1) + (2*0.5*22*0.1) + (2*0.5*10*0.1) (11.6*6.3*0.1) = 19.1cm3

Assume aluminium density of 2.7g/cm3

2.7 * 19.1 = 51.57g

Known Parts:

-330ml can of soft drink (using diet drink is preferable as it can be up to 50g

lighter than the normal version): ~350g

-9V PP3 battery (45g) with ~4g connector: 49g

-6V D.C. motor with an integral 9:1 gearbox with output speed of the shaft at

330RPM: 125g

Pulleys

-10mm: 0.4g

-20mm: 4.6g

-40mm: 7.6g

-60mm: 11.8g

Wheels

40mm: 14.7g

56mm: 15.2g

75mm: 12.7g (x4 = 50.8g)

90mm: 40.5g

Other:

Bearing block: 1.3g (x4=5.2g)

Battery clip: 4.9g

4mm Rod x1000mm: 98.7g(x2=197.4g)

5mm Rod x1000mm: 154.1g

+Star lock fasteners ~1g

Misc.: ~10g

Total: 952.6g (~950g) = 9345.006N (9319.5N)

Body 52g

Known Parts

330ml Can 350g

9V Battery 49g

6V DC Motor 125g

Subtotal 524g

Pulleys 10g

Wheels 50.8g

Other 208.5g

Misc. 10g

Total(grams) 856.6g

Total(Newtons) 8403N

Fig 1.1: Tabular representation of masses of the components.

2. Estimate the rolling resistance.

Assume 0N for forces due to air resistance

Assume 0.4N per each kilogram of mass of the vehicle

Estimated mass of the vehicle: 850g

0.85 * 0.4N = 0.34N

3. Calculate total resisting force to motion

Fig 2.1: Gravitational force on the vehicle.

Assume rolling resistance of 0.38N

Assume vehicle weight of 850g

Gravitational forces:

0.850(9.81) = 8.339N

= 5

8.339*Sin5 = 0.727N

Total of resisting forces: FR = 0.34N+0.727N=1.067N

4. Assume the value for the power efficiency

Considering the possibility of drive belt slippage, the expected value for power

efficiency

= 0.85

5. Optimise the ratio of motor shaft pulley to axle pulley

Nominal reduction ratio of the integrated gearbox of the motor: 1:9

Shaft Pulley

Size Axle Pulley

Size Gearbox

Reduction Overall

reduction

10 10 (1:9) (1:9)

20 (1:9) (1:18)

40 (1:9) (1:36)

60 (1:9) (1:54)

20 10 (1:9) (1:4.5)

20 (1:9) (1:9)

40 (1:9) (1:18)

60 (1:9) (1:27)

40 10 (1:9) (1:2.25)

20 (1:9) (1:4.5)

40 (1:9) (1:9)

60 (1:9) (1:13.5)

60 10 (1:9) (1:1.5)

20 (1:9) (1:3)

40 (1:9) (1:4.5)

60 (1:9) (1:9)

Fig 1.2: Tabular representation of combined reduction of gearbox and pulleys.

Assume pulley reduction of 1:2.

Shaft pulley diameter: 10mm

Axle pulley diameter: 20mm

Torque at 10mm motor shaft pulley: 25mNm (as given in the specifications table in the brief)

Torque at 20mm axle pulley: 50mNm = 0.05Nm

0.05Nm gives 0.05N of force over 1m

DW = 0.075m Dr = 0.0375m

0.05/0.0375 = 1.33N

1.33N * = 1.33(0.85) = 1.13N (FD)

1.13N 1.06N = 0.07N Most efficient use of torque of all pulley combinations

Results for similar pulley ratios (available from pulley sizes provided):

Pulley reduction 1:4 = 2.26N (1.2N of net Drive Force)

Pulley reduction 1:3 = 1.78N (0.72N of net Drive Force)

Pulley reduction 1:1.5 = 1N (-0.6N of net Drive Force, insufficient drive force)

It is evident, assuming vehicle weight of ~850N and initial torque of 25mNm,

that pulley ratio 1:2 provides the most efficient use of torque and hence

maximises the maximum speed the vehicle can achieve.

6. Calculate the overall speed of the vehicle

Speed of the motor = 330revs/min

5.5revs/sec

u = pi * DW * NM * DM/DA *

u = pi * (5.5) (0.075) (0.5) (0.85)

u = 0.175pi m/s

u = 0.55 m/s -or- u = 1.98km/h

7. Estimate the initial acceleration based on the net force at v=0

At v = 0 TM = ~130mNm (see above graph)

Fig 2.2: Total Resisting Force and Drive Force

FR = Total resisting force = 1.06N

FD = 130/25 (1.13) = 5.876N

FD FR = 5.876 -1.06 = 4.87N

Fig 2.3: Free body diagram of the vehicle at v=0

s = 0.4

FFRIC = sN

N = mg * sin85 = 8.307N

FFRIC = 0.4(8.307N) = 3.32N

F = 4.87 3.32 = 1.55N (Net Force after overcoming maximum static friction)

F = ma

a = F/m = 1.55/0.85 = 1.83m/s2 at v=0

Therefore, the vehicle will overcome the maximum static friction and achieve

initial velocity of 1.83m/s2.