 Open Access
 Total Downloads : 452
 Authors : Akshaya Kulkarni, Kunal Bhandari, Pranoti Panchwagh
 Paper ID : IJERTV4IS120348
 Volume & Issue : Volume 04, Issue 12 (December 2015)
 DOI : http://dx.doi.org/10.17577/IJERTV4IS120348
 Published (First Online): 17122015
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Design of Mechanical Drives for a Parabolic Radio Antenna
Akshaya Kulkarni1, Kunal Bhandari1, Pranoti Panchwagp Department of Mechanical Engineering,VIIT, Savitribai Phule Pune University, Ganeshkhind,
Pune7, Maharashtra, India.
AbstractThe paper proposes mechanical drive systems for rotating a radio antenna having dish diameter 5 meters, in the azimuth and elevation axes. The antenna dish is parabolic in shape and is estimated to have a maximum weight of 350 kilograms. Conventional drive systems comprise of a gear pair mechanism for turning the dish through the requisite angle of rotation. These systems are precise and highly efficient, thereby facilitating an extremely high margin of accuracy of the observed data. However, the most significant drawback of a gearpair drive system is its high initial investment and subsequently, high maintenance cost. To overcome this limitation, a rope and pulley drive system is considered, which reduces the cost exponentially without affecting the accuracy of the data to a large extent.
Keywords – Rope drives, azim uth drive sys tem , elevation drive s yst em , wind torque.

INTRODUCTION:
Radio telescope antennas are an important tool in radio astronomy, which relies heavily on observational data. The radio telescope for which the drive system has been designed is used to track celestial bodies and is therefore, required to turn through a maximum rotation of
Â±270Â° in the azimuth axis and through 0Â°to 90Â° in the elevation axis in a continuous, uninterrupted and smooth manner.
The most important consideration that has to be taken into account while designing the drive is the combined torque exerted by the wind on the dish and that due to the mass moment of inertia of the dish itself. The schematic models of the drive assemblies are specified as follows.

AZIMUTH DRIVE
Power is supplied by a hand driven winch to the driving pulley, which then transmits the same to the driven pulley. The driven pulley is connected by a shaft to the top flange on which the hub of the dish rests. Rotational motion is thus imparted to the parabolic dish antenna.

ELEVATION DRIVE
The general procedure to be followed while designing the drive is as follows:

Total torque exerted on the parabolic dish and hence, the azimuth drive, including the wind torque and the torque due to the mass moment of inertia of the dish, is calculated.

This torque is used to find the actual and design power required to turn the dish through the requisite degree of rotation.

The power calculated gives the belt tensions exerted on the driving shaft, which in turn is used to determine shaft diameter.

Furthermore, the dimensions of other supplementary components such as bearings, flanges, bolts, etc. can be found out.
The design process is largely iterative, varying according to the material used and the factor of safety required by individual requirement. The material usedfor this particular design process is M.S. (Sut = 400N/mm2) and the factor of safety is assumed to be 1.5.


DESIGN OF AZIMUTH DRIVE

Calculating wind torque on dish:
F=12PA
Where,1=porosity/solidity ratio=0.31
2= factor taking into account inclination of wind w.r.t. antenna
P=Pressure on dish A=Area of dish P=1/2**vw2
= air density (1.123 kg/m3) vw = wind velocity (80kmph)
P = 277.228 Pa
For perpendicular attack of wind,
Distribution of Wind Load in Perpendicular Direction
2= 1
d=depth of antenna + height of hub= 1.055 m
Let,O be the point about which moment is to be calculated.
A be the center of the dish
be the angle subtended by differential element at O
be the angle subtended by elemental strip at center of circular crosssection A
dA= 2R2sin2d dF= Â½ **vw2*1*dA
tan0 = R/d= 67. 12Â°
dup= dFsin * r (Since r=d/cos)
= dF*tan*d
dup= 85.94*(2R2) *sin2d*tan*d*d
= 1074.25*sin2d*tand Integrating over limits 0 to up
0
up dup = 1074.25
0,/2
sin2d tan d
=0,=0
Solving, up= 826.225 Nm – Wind torque for upper half of dish
Assuming same torque on lower half, total torque due to wind = 1652.45 Nm.

Torque due to rotation of dish
= I*
Mass moment of inertia of dish I = 2
dm/dA = surface density () = M/A Hence, dm = dA= *dx*(Hax2)
Calculating limits for x, for vertical parabola, y=ax2
x=y/a
for y=H i.e. maximum depth of dishand a=8 from general equation of parabola
x=Â±0.655/8 = Â±2.5
Therefore,
2.5 0
I = +2.5 2( 2) = 2 +2.5 2( 2)
Solving for I, I =0.04585 kgm2 Maximum rotation speed = 1 rpm = 0.104719 rad/s
Angular acceleration, = 2/2 = 270Â°
=0.01111 rad/s2
Hence calculating torque, = I*=5.094*104 Nm
Total torque exerted on drive = Wind Torque + Torque due to dish
= 1652.45 Nm.
Hence, power required to drive the parabolic dish (P) ,
P = 2 = 173.044 W
60

Design of Crossed Belt Drive
The torque exerted on the driven pulley is thesummation of the torque due to wind and that due to the mass moment of inertia of the dish.Given that,
Torque on driven pulley (T1) = 1652.45 Nm.
Power required to drive the pulley (P1) = 173.044 W Speed of driven pulley (N1) = 1 rpm
Velocity of belt v = r*= 0.0131 m/s
Let, Driven pulley diameter (D1) = 250 mm. Driving pulley diameter (D2) = 50 mm.
Center to center distance (C) = 1000 mm.Since velocity of the driving and driven pulleys is the same,
Design Power = Service Factor(cs) * Required Power Hence, Pdes = 1.5*173.044= 259.566 W
Belt
Tensions,
(P1 P2)*v = Power (W) P1 P2 = 19814.2
Coefficient of friction () = 0.2 For a crossed belt drive,
= sin1 ( 1+2) = 8.627Â°
2
Angle of lap,
1 = 2 = 180Â° + 2 = 3.4427 rad
P1 = 2.92
2
Calculating belt tensions P1 and P2 P1 = 22877.95 N ;P2 = 3063.75 N
Calculated Rope Length =
L0= (1+2 ) + 2 + ( 12)=2.4713 m
2 4
Shortening the belt length by 1%, L0 = 2446.58 mm

Shaft Diameter Calculation for Axial load. Let weight of dish = 350 kg = 3433.5 N
Dead weight of flanges= 12 kg = 117.72 N Total load on shaft= 3551.22 N
Area of dish = 19.256 m2 Material: M.S. Plain Carbon Steel
Syt = 380N/mm2; Sut= 680 N/mm2 FOS= 2
Max shear stress = 0.5Sut/FOS= 95N/mm2 = F/A
A = 37.38mm2 =d2 */4

Shaft Diameter Calculation for Bending load
Total bending load on shaft = F1 + F2= 25941.7 N Minimum distance between bearings = 200 mm and shaft length = 1000 mm
Calculating reactions at bearings:
R1 = R2 = (F1 + F2)/2= 12970.85 N Mt = (P1 P2)*r= 2476775 Nmm
Mb = R1*100 = 1297085 Nmm
The material selected for the shaft is M.S.
Sut =400 N/mm2 FOS = 1.5
= 0.5*Sut/FOS =133.33N/mm2
= 16()2+()2
3
Where, Fatigue factor () = 1; Shock factor () = 1.5
Hence,
d = 49.366 mm 50 mm

Design of Rigid Flange Coupling
Outer Diameter of hub (dh) = 2*d = 100 mm Length of hub (lh) = 1.5*d = 75 mm
PCD of bolts (Db) = 3*d = 150 mm Thickness of flange (t) = 0.5*d = 25 mm
Protecting rim Thickness (t1) = 0.25*d = 12.5 mm Spigot recess Diameter (dr)= 1.5*d =75 m
Outer diameter of flange (D) = (4*d + 2*t1) = 225 mm.

Design of Bolts
As the shaft diameter is 50 mm,
Number of bolts (N) = 4 (standard value) c permissible = 200Mpa
Bolt diameter (d1)2 = 8Mt
(DN)
Thus, d1 = 8.37 mm = 10 mm. c = 2*Mt/(N*t1*t*D)
= 22.0157 Mpa < 200 Mpa

Design of Bearings
The vertical shaft is subjected to both radial and axial loads. Hence, taper roller bearings are preferred as compared to ball bearings. The main advantage of using taper roller bearings is that they are able to withstand combined axial and radial loads since the line of action of the resultant reaction on the bearing can be resolved into separate axial and radial components.
Bore of the bearing = 100 mm.
The antenna is to be operated continuously i.e. 24 hrs per day, hence the life of bearing is assumed to be 40,000 hrs. L10h = 40,000 hrs
Consequently, L10 = 60*n*L10h
10e6
Where, L10 = Rated bearing life (in million revolutions) n= speed of rotation (rpm)
Hence, d=6.8988 mm = 10 mm
L10= 60*1*40,000
10e6
= 2.4 million revolutions
For taper roller bearings, P = Fr if e
P = (0.4*Fr) + (Y*Fa) if > e
Where, P = Equivalent dynamic load (N) Fr = Radial load (N)
Fa = Axial or thrust load (N)
These equations are based on the assumptions that, for taper roller bearings,
i. The both bearings are exactly identical and bearings are adjusted against each other to give zero clearance in operation without preloading.
Now, Fr = (P1 + P2)/2= 12970.85 N
0.5*Fr
Mt = 5199833.38 Nmm. P = 2NT= 544W
60e3

Wind torque for dish at 60o Mt= 3700532.75 Nmm.
P =2NT= 387.51W
60e3

Wind torque for dish at 90o
Mt=1652450 Nmm. P =2NT= 173.044W
60e3
From the above values, it is clearly seen that the maximum power required is when the dish is inclined at an angle of30o w.r.t. horizon.
Fa=
= 4988.788 N
Y
Assumptions for design
Fa = 0.3846
Fr
The bearing selected is Taper roller bearing no. 32020X from the bearing catalogue,
Bearing specifications:
Inner diameter of bearing (d) = 100 mm Outer diameter of bearing (D) = 150 mm Axial width of bearing (B)=32 mm Dynamic load capacity (C) = 161,000
e = 0.46 Y = 1.3
WinchSpecifications:
Weight = 50 Kg; Reduction Ratio = 14:1


DESIGN OF ELEVATION DRIVE
The elevation drive consists of a threepulley rope drive system that rests on a yoke and is fixed at one end to the parabolic dish. The main consideration while designing the elevation drive system is the wind torque

the uprooting force that acts on the dish. Ideally, the dish is assumed to be turning through 0Â° to 90Â°. However, for all practical purposes, a clearance of Â±5Â° is maintained.
The first pulley is considered of diameter of 300mm since it is available easily in market for its size. Besides, there is a space constraint layout to be followed which limits the size of the first pulley mounted on the base yoke.
The wire rope used for this drive system is of following specifications taken from IS 2266:2002 manual
Type: 6×36 (14771)
Wire diameter (dwire):10mm Rope grade: 1960
Approximate mass: 41.8 kg/100mm Minimum breaking force: 70 KN Type of lay: Right Regular Lay.
2. Crossed belt drive design:

Stage no.1:
Where D diameter of first pulley, N rpm of firstpulley
P (W) = (P1P2)*V

Given:
Diameter of driven pulley (D)= 300 mm Diameter of driving pulley (d)= 150 mm Speed of driven pulley (N)= 1 rpm
Center to center distance between pulleys(C) = 475 mm
V=DN=0.0157 m/s
60e3
P (W) = (P1P2)*V
P1 = 34649.68 + P2

Rope tension ratios: Where, P1 and P2 Rope tensions P1/P2 = ef/sin (/2) = e (2.92*f) Assume, =40o f = 0.2


= 180 + 2sin1(D+d )+ 360 = 10.411 rads.
2C
1. Torque calculations:
Given:Wind torque = 1652.45e3 Nmm.
Mt = (Bending moment of dish*9.81) + Wind torque

Wind torque at 30o
P1 = 437.08 P2
P2 = 79.47 N; P1 = 34729 N
Hence the tensions in the rope wound between 1st and 2nd pulley would be 34729 N on tight side and 79.47 N on slack side.

Design of Shaft 01:
Assumption: The length of shaft = 300mm
Mb = P1 + P2 Ã— 300=3111193.82 Nmm

Design of Shaft 02: Assumption:The length of shaft = 300mm.
Mb = P1 + P2 Ã— 300= 3479980.79 Nmm
2 2 2 2
Mt = [P1 – P2] [R] = 5197452 Nmm
= 16 Ã— M 2 + M 2
Mt = [P1 – P2][R] = 2598694.5 Nmm
= 16 Ã— M 2 + M 2
d3 b t d3 b t
d = 51.935 mm (55 mm) (8)

Bearing selection from the catalogue: Bearing No: 6211
D: 100mm B: 21mm
C: 43600 Co:19600
The antenna is to be operated continuously i.e. 24 hrs per day, hence the life of bearing is assumed to be 50,000 hrs. L10h = 50,000 hrs
60*n*L10h
So, d = 46.49mm (50 mm)

Bearing selection from the catalogue: Bearing No: 6210; L10h = 50,000 hrs.
D: 90mm B: 20mm
C: 35100 Co: 19600
L10 = 60*n*L10h = 6 million revolutions
10e6

Design of key:
Consequently, L10 =
10e6
For a flat key:
L10= 60*1*50,000= 3 million revolutions
10e6

Design of key: For a flat key:

d
= d = 16.25mm 18 mm
4
= d = 10.833mm 12 mm
6
= 1.5d for 2.1d = 136.5mm 138 mm
width () = height () =
4 = 16.25mm
d = 10.833mm
6
= 2Mt For shear forces
db
= 72.07N/mm2< 110
length () = 1.5d for 2.1d = 136.5mm
<permissible
= 2Mt
db
For shear forces
= 4Mt For torsional forces
cdh
= 72.07N/mm2< 110
< permissible
= 4Mt For torsional forces
cdh
c = 216.298 N/mm2 < 220
c < cpermissible
Hence the key is safe. Dimensions of the key are:
b = 18mm h = 12mm l = 138mm

Stage no 2:
Diameter of driven pulley D = 300mm Diameter of driving pulley (d) = 150mm Speed of driven pulley (N) = 2 rpm
Center to center distance between pulleys(C) = 1212mm
V = Ã—150Ã—2 = 0.0157 m/s
60e3
= 180 + 2sin1( D+d ) =3.3067 radians
2C
P = (P1P2)*V
P1= 34649.68 + P2
P1/P2 = e (2.92*f)
P1= 6.6298 P2
Solving the equations,
P1 = 40524.5 N ; P2 = 5875.24 N
c = 216.298 N/mm2< 220
c < cpermissible
Hence the key is safe. Dimensions of the key are:
b = 18mm h = 12mm l = 138mm


CONCLUSION
The ropepulley drive thus designed, is found to be accurate, easy to operate and economically and ergonomically viable. The design is also found to be safe based on analytical calculations performed.

REFERENCES

Analytical Model of Wind Disturbance Torque on Servo TrackingAntennaTusharGolani and Suresh Sabhapathy

Design of Machine Elements – V.B.Bhandari