6.0 RESULTInput Speed, N in = 1000 rpmInput angular velocity, W in = 104.72 rad/s

6.1 Spur Gear TransmissionTransmission ratio= Output speed, N out= 74.07 rpmOutput angular velocity, W out= 7.76 rad/s

Table 1 Data for Spur Gear TransmissionExcitation Current,[mA]Input Force, [N]Input Torque, [Nm]Input Power, [W]Output force, [N]Output Torque, [Nm]Output Power, [W]Efficiency,

01.75000.08759.16300.00000.00000.00000.0000

503.25000.162517.01704.00000.40003.10400.1824

1003.50000.175018.32607.00000.70005.43200.2964

1254.50000.225023.562011.00001.10008.53600.3623

1505.25000.262527.489016.00001.600012.41600.4517

1755.50000.275028.798021.00002.100016.29600.5659

2006.25000.312532.725026.00002.600020.17600.6165

2257.25000.362537.961033.00003.300025.60800.6746

2508.00000.400041.888037.00003.700028.71200.6854

2758.50000.425044.506045.00004.500034.92000.7846

3009.25000.462548.433050.00005.000038.80000.8011

Average0.4928

6.2 Worm Gear TransmissionTransmission ratio= Output speed, N out= 71.43 rpmOutput angular velocity, W out= 7.48 rad/s

Table 2 Data for Worm Gear TransmissionExcitation Current,[mA]Input Force, [N]Input Torque, [Nm]Input Power, [W]Output force, [N]Output Torque, [Nm]Output Power, [W]Efficiency,

02.10000.105010.99561.00000.10000.74800.0680

503.20000.160016.75527.00000.70005.23600.3125

1004.10000.205021.467615.0001.500011.22000.5226

1255.10000.255026.703622.00002.200016.45600.6162

1506.10000.305031.939628.00002.800020.94400.6557

1757.20000.360037.699232.00003.200023.93600.6349

2008.20000.410042.935240.00004.000029.92000.6969

2259.20000.460048.171248.00004.800035.90400.7453

25010.10000.505052.883652.00005.200038.89600.7355

27511.20000.560058.643262.00006.200046.37600.7908

30012.40000.620064.926468.00006.800050.86400.7834

Average0.5965

7.0 DISCUSSIONI. Show calculation of output speed and output angular velocity for both transmissions and fill in the result section :

Spur gear:Output speed, Nout = 1000 rpm x 2/27= 74.07 rpmOutput angular velocity, =74.07 rpm x (2/60)= 7.76 rad/sWorm gear:Output speed, Nout = 1000 rpm x 1/14= 71.43 rpmOutput angular velocity, =71.43 rpm x (2/60)= 7.48 rad/s

II. Four sample calculations for input torque, input power, output torque and output power of each excitation :Where Linput: 0.050 mLoutput: 0.100 mSpur gear:Input torque, Tin for 0 mA= FL= (1.75 N) (0.05 m) = 0.0875 NmTin for 50 mA = FL= (3.25 N) (0.05 m)= 0.1625 NmTin for 100 mA= FL= (3.50 N) (0.05 m)= 0.1750 NmTin for 125 mA = FL= (4.5N) (0.05 m)= 0.2250 Nm

Output torque, Toutput for 0 mA= FL= (0.00 N) (0.100 m) = 0.0000 NmToutput for 50 mA = FL= (4.00 N) (0.100 m)= 0.4000 NmToutput for 100 mA= FL= (7.00 N) (0.100 m)= 0.7000 NmToutput for 125 mA = FL= (11.00 N) (0.100 m)= 1.1000 Nm

Input power, Pin for 0 mA= T= (0.08570 Nm) (104.72 rad/s) = 9.1630 W

Pin for 50 mA = T= (0.1625 Nm) (104.72 rad/s)= 17.0170 WPin for 100 mA= T= (0.1750 Nm) (104.72 rad/s)= 18.3260 WPin for 125 mA = T= (0.2250 Nm) (104.72 rad/s)= 23.5620 W

Output power, Poutput for 0 mA= T= (0.0000 Nm) (7.76 rad/s) = 0.0000WPoutput for 50 mA = T= (0.4000 Nm) (7.76rad/s)= 3.1040 WPoutput for 100 mA= T= (0.7000 Nm) (7.76 rad/s)= 5.4320 WPoutput for 125 mA = T= (1.1000 Nm) (7.76 rad/s)= 8.5360 W

Worm gear:Input torque, Tin for 0 mA= FL= (2.10 N) (0.05 m) = 0.1050NmTin for 50 mA = FL= (3.20 N) (0.05 m)= 0.1600 NmTin for 100 mA= FL= (4.10 N) (0.05 m)= 0.2050NmTin for 125 mA = FL= (5.10 N) (0.05 m)= 0.2550Nm

Output torque, Toutput for 0 mA= FL= (1.00 N) (0.100 m) = 0.1000 NmToutput for 50 mA = FL= (7.00 N) (0.100 m)= 0.7000 Nm

Toutput for 100 mA= FL= (15.00 N) (0.100 m)= 1.5000 NmToutput for 125 mA = FL= (22.00 N) (0.100 m)= 2.2000 Nm

Input power, Pin for 0 mA= T= (0.1050 Nm) (104.72 rad/s) = 10.9956 WPin for 50 mA = T= (0.1600 Nm) (104.72 rad/s)= 16.7552 WPin for 100 mA= T= (0.2050 Nm) (104.72 rad/s)= 21.4576 WPin for 125 mA = T= (0.2550 Nm) (104.72 rad/s)= 26.7036 W

Output power, Poutput for 0 mA= T= (0.100 Nm) (7.48 rad/s) = 0.7480 W

Poutput for 50 mA = T= (0.7000 Nm) (7.48rad/s)= 5.2360 WPoutput for 100 mA= T= (1.5000 Nm) (7.48 rad/s)= 11.2200 WPoutput for 125 mA = T= (2.2000 Nm) (7.48 rad/s)= 16.4560 W

III. Calculate the efficiency of each excitation current and fill in the Table 1 and Table 2. Show an example of calculation :

Worm GearEfficiency,

Spur GearEfficiency,

IV. Plot two graphs of efficiency, vs input force, Fin for both transmissions. Explain your findings.

From the graph plotted, we can say that the efficiency will increase as the input force increase. At a certain point, the increasing in efficiency is getting smaller as the input force increase and eventually the efficiency will reduce if the input force keeps increasing. From the graphs, we found that spur gear has better efficiency compare to worm gear. This is done by referring to the line of the graph which is directly proportional for spur gear, this show that the increasing in input force and efficiency is at the same pace. Meanwhile, the line for worm gear show a good increasing at the beginning of the experiment and the efficiency is decreasing at the end of the experiment. The worm gear reach its maximum efficiency during input force at 10N.

V. How can the efficiency of the transmission be related to the excitation current?When an excitation current is used in a device, the current is used to operate the device at some point. A certain amount of energy is needed to overcome the internal resistance of the steel core. When the excitation current give the motor its power, the transmission is then convert the power to move the load.

VI. Wat are other sources of errors in this experiment?After discussed with group member, we found that there are few of errors involved that affect our experimental result. One of the errors is parallax error. Its mean that student does not read the scale of spring balance and dynamometer perpendicularly. In order to achieve an accurate result student should read the scale perpendicularly and student can also take multiple reading to minimise the error. Besides that, when taking the reading for output force student have to pull the load back to its middle position and this is hard to be done because the load always moving and cause the reading difficult. The higher the current the higher the force of the load is and more harder for student to maintain the load to its middle position. Therefore the results for output load might differ a bit from the theoretical result. And finally is the error within the equipment itself. The equipment has been used for a long time and some part in the equipment might have wear. This small error will also affect the result. Besides, the longer machine worked, the hotter it will become. Some of the output energy will be release in the form of heat and this will reduce the efficiency of the machine.

8.0 CONCLUSION At the end of the experiment, we have able to determine the efficiency of a spur gear system and worm gear system. After we have done the gear efficiency experiment, we now knew that the effectiveness of a power transmission can be measured by finding their efficiency. Based to the experimental result that we obtained, we can see that for spur gear the average efficiency is 0.4928 while for worm gear the average efficiency is 0.5965. Which clearly state that worm gear has better average efficiency compare to spur gear. But the efficiency is reduces as the input force increase. As conclusion, even though a worm gear has an average efficiency better that spur gear, it can only operate at low input force compare to spur gear which can operate at higher input force.