BELT CONVEYOR CALCULUS ( ACCORDING CEMA )
BELT CONVEYOR # 1
A) BELT WIDTH AND BELT SPEED
1 Determine initial data
a) Client information and site restrictions
MTPD = 14,000 MTPD = Metric tonnage per day
HPD = 20.00 HPD = Hours of operation per day
Use factor = 75%
Load factor = 75%
34.00
168.60
Process data = Heavy, hard, sharp-edged ore, coarse-crushed stone
Lump size ( in ) = 20.00
Lump proportion = 10%
L = 27.90 m = 91.54 ft Length of conveyor between center of terminal pulleys
H = 8.60 m = 28.22 ft Height of conveyor between center of terminal pulleys
17.95
MTPH = MTPD MTPH = Metric tonnage per hour
HPD x Use factor
MTPH = 933
b)
See: Table 3-1: Flowability - angle of surcharge - angle repose
Table 3-2: Material class description
Table 3-3: Material characteristics and weight per cubic foot
Material Name = Ore
Code = P-2357
34.00
19.00 to 29.00
f ( °sex ) = f ( °sex ) = Repose angle
dm ( lb / ft3 ) = dm ( lb / ft3 ) = Apparent density
g ( °sex ) = g ( °sex ) = Inclination. Site restrictions
f ( °sex ) = f ( °sex ) = Repose angle
a ( °sex ) = f ( °sex ) - [ 5 to 15 ]( °sex ) a ( °sex ) = Surcharge angle
a ( °sex ) =
1
20.00
168.60
18.00 to 20.00
Final values:
34.00
17.95
2 Choose the idler shape.
See: Chapter 5: Belt Conveyor Idlers
35.00
3
(For Metric Tonnage)
12,206
4 Select a suitable conveyor belt speed
See: Table 4-1: Recommended maximum belt speeds
Table 4-2: 20 degree troughed belt - three equal rolls standard edge distance = 0.055b + 0.9 inch
Table 4-3: 35 degree troughed belt - three equal rolls standard edge distance = 0.055b + 0.9 inch
Table 4-4: 45 degree troughed belt - three equal rolls standard edge distance = 0.055b + 0.9 inch
Table 4-5: Flat belt capacity standard edge distance = 0.055b + 0.9 inch
Belt Speed Belt Speed Belt Equivalent Equivalent
actual reduced Width Capacity Capacity
(fpm) (inches) Step 5 Table 4-3
350 262.50 18 4,650 1,274 NO
500 375.00 24 3,255 2,438 NO
400 300.00 30 4,069 3,976 NO
280 210.00 36 5,813 5,886 OK
600 450.00 42 2,713 8,169 OK
600 450.00 48 2,713 10,826 OK
600 450.00 54 2,713 13,855 OK
600 450.00 Over 60 2,713
fpm (actual) * Load factor Belt Speed reduced in fpm
fpm (actual) = 280.00 fpm (actual) = Belt Speed actual in fpm
5
a ( °sex ) = To < a: < load, so for = Q, > V or > b. It is site safe.
d ( lb / ft3 ) = d ( lb / ft3 ) = Average weight
g ( °sex ) = g ( °sex ) = Inclination. Maximum recommended
f ( °sex ) =
g ( °sex ) =
b ( °sex ) = b ( °sex ) = Degree troughed belt
Convert the desired tonnage (MTPH) to ( ft 3 / h )
f t 3 / h = MTPH x 2205 / dm
f t 3 / h =
(fpm)
fpmo = fpmo =
Convert the desired capacity ( ft 3 / h ) to the equivalent capacity at a belt speed of 100 fpm
2
3
4
5
Equivalent Capacity = x 100 / fpm (actual)
Equivalent Capacity = 5,813
6 Using the equivalent capacity, find the appropriate belt width
36.00
7 Check the belt width with the material lump size
See: Item a) to Item c)
a) In USA: b (in) = 18, 24, 30, 36, 42, 48, 54, 60, 72, 84 and 96
b) f ( lump size )
For: i) 20.00 For: i) 20.00
ii) 10 % lumps and 90% fines ii) 1.00 lumps
b min = 3 x ( maximum lump size ) b min = 5 x ( maximum lump size )
b min = 60.00 in b min = in
For: i) 30.00 For: i) 30.00
ii) 10 % lumps and 90% fines ii) 1.00 lumps
b min = 6 x ( maximum lump size ) b min = 10 x ( maximum lump size )
b min = in b min = in
c) See: Figure 4-1 Belt width necessary for a given lump size
58.00
d) For belt conveyor existing
The belt will be new
Conclusion:
58.00
8 Revise the belt speed for final belt width.
Final Equivalent Capacity = check
x 100 / Final Equivalent Capacity
#VALUE!
#VALUE!
f t 3 / h
bo ( in ) =
b1 ( in ) =
a (°sex) = a (°sex) =
a (°sex) = a (°sex) =
b1 ( in ) =
b2 ( in ) =
b ( in ) =
fpm1 = f t 3 / h
fpm1 =
fpm2 = fpm1 / Load factor
fpm2 =
5
7
8
6
V = 280.00 fpm Elected speed
B) POWER
hp = Te x V
33,000
Te = L .Kt (Kx + Ky.Wb + 0,015.Wb) + Wm ( L . Ky +/- H ) + Tp + Tam+ Tac
Frictional resistance of the carrying and return idlers (lbs)
Tx = L . Kx . Kt
Kx = 6.80 1.00E-04 ( Wb + Wm ) + Ai / Si
58.00 168.60
See: Table 6-1: Estimated average belt weight ( lb/ft )
Wb = 0.00 lb / ft
Wm = ( Q x 2205 ) / ( 60 x V )
Wm = ( MTPH x 2205 ) / ( 60 x V )
Wm = 122.50 lb / ft
See: Table 5-1: Idler classification
Description = Medium Duty
Classification = D5
See: Table 5-2: Suggested normal spacing of belt idlers ( Si )
Si = 4.00 ft
Ai = 1.80 lb
Wb + Wm = 122.50 lbs / ft
Kx = 0.53
Tmin = -15.00 °C = 5.00 °F
Kt = 1.20
Tx = 58.58 lb
Resistance of the belt to flexure as it moves over the idlers (lbs)
b ( in ) = dm ( lb / ft3 ) =
1
2
Tyb = L x Wb x Kt x (Ky + 0,015)
See: Table 6-4: A and B for equation Ky = …
Ky =
Average belt tension = 1,000 lb
A = 2.20
B = 2.25
Ky = 0.05
Tyb = 0.00 lb
Resistance of the material to flexure as it rides the belt over the idlers (lbs)
Tym = L x Ky x Wm
Tym = 554.49 lb
Force needed to life or lower the load (material) (lbs)
Tm = +/- H x Wm
Load upwards: 1.00
Load downwards: -1.00
sign: 1.00
Tm = 3,456.36 lb
Tp = 950.00 lb
See: Table 6-5: Belt tension to rotate pulleys
# Pulleys: Tight side = 2.00
# Pulleys: Slack side = 3.00
# Pulleys: All others = 1.00
Pulleys: Tight side = 200.00 lbs/pulley
Pulleys: Slack side = 150.00 lbs/pulley
Pulleys: All others = 100.00 lbs/pulley
Force to accelerate the material continuously as it is fed onto the belt (lbs)
( Wb + Wm ) x A x 10(-4) + B x 10(-2)
Resistance of the belt to flexure around pulleys and the resistance of pulleys to rotate on their bearings (lbs)
2
3
4
5
6
Tam = (V - Voj )
3600 x 32,2 60.00
0.00 -100.00
794.44 MTPH 138.89 MTPH
Tam = 87.25 lb
Resistance generated by conveyor accessories (lbs)
Tac = Ttr + Tpl + Tbc + Tsb
Ttr = Trippers and stackers
Tpl = Frictional resistance of plows
Tbc = Belt-cleaning devices
Tsb = Skirtboard friction
Ttr = 0.00 In this case
Tpl = 0.00 In this case
Tbc = e x b x # devices
Estimated ( e ) = 8.00 lb / in
# devices = 2.00
Tbc = 928.00 lb
Tsb =
Lb = 49.21 ft Lb = Skirtboard length, two skirtboard
hs = 4.00 in hs = depth of the material touching the skirtboard
Cs = 2 dm x
288.00
Cs = 0.33
Tsb = 555.92 lb
Tac = 1,483.92 lb
Te = 6,590.60 lb
hp = 55.92 HP
S Qj x 2205 x
Vo1 = Vo2 =
Q1 = Q2 =
Lb ( Cs x hs2 + 6 )
( 1 - sinf )
( 1 + sinf )
7
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 17.95 ° = 0.31 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 360 fpm
g = 32.20
= 1.29 = l
e = 108.0 ° = 1.88 rad
r0 = 1.04 ft X0 = -0.32 ft Y0 = 0.99 ft
r1 = 1.25 ft X1 = -0.39 ft Y1 = 1.19 ft
r2 = 1.55 ft X2 = -0.48 ft Y2 = 1.47 ft
rpm = 55.00 rpm end of pulley
Vo = 6.00 fps Vxo = 5.71 fps Vyo = 1.85 fps
V1 = 7.20 fps Vx1 = 6.85 fps Vy1 = 2.22 fps
V2 = 8.93 fps Vx2 = 8.49 fps Vy2 = 2.75 fps
Yo / Xo = -3.09
F = 0F > 0 1.88F < 0
= 0.95 =#VALUE! °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 1
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 7.94 ° = 0.14 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 360 fpm
g = 32.20
= 1.29 = l
e = 97.9 ° = 1.71 rad
r0 = 1.04 ft X0 = -0.14 ft Y0 = 1.03 ft
r1 = 1.25 ft X1 = -0.17 ft Y1 = 1.24 ft
r2 = 1.55 ft X2 = -0.21 ft Y2 = 1.54 ft
rpm = 55.00 rpm end of pulley
Vo = 6.00 fps Vxo = 5.94 fps Vyo = 0.83 fps
V1 = 7.20 fps Vx1 = 7.13 fps Vy1 = 0.99 fps
V2 = 8.93 fps Vx2 = 8.84 fps Vy2 = 1.23 fps
Yo / Xo = -7.17
F = 0F > 0 1.71F < 0
= 0.99 =#VALUE! °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 2
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 17.83 ° = 0.31 rad
a1 = 2.50 inches h = 6.10 inches
R = 9.00 inches Pulley e = 0.50 inchesespesor faja
V = 550 fpm
g = 32.20
= 4.16 = l
e = 107.8 ° = 1.88 rad
r0 = 0.79 ft X0 = -0.24 ft Y0 = 0.75 ft
r1 = 1.00 ft X1 = -0.31 ft Y1 = 0.95 ft
r2 = 1.30 ft X2 = -0.40 ft Y2 = 1.24 ft
rpm = ### rpm end of pulley
Vo = 9.17 fps Vxo = 8.73 fps Vyo = 2.81 fps
V1 = 11.58 fps Vx1 = 11.02 fps Vy1 = 3.55 fps
V2 = 15.05 fps Vx2 = 14.33 fps Vy2 = 4.61 fps
Yo / Xo = -3.11
F = 0F > 0 1.88F < 0
= 0.95 =#VALUE! °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-0.7
9-0
.71
-0.6
3-0
.55
-0.4
7-0
.40
-0.3
2-0
.24
-0.1
6-0
.08
0.00
0.08
0.16
0.24
0.32
0.40
0.48
0.55
0.63
0.71
0.79
0.87
0.95
1.03
1.11
1.19
1.27
1.35
1.43
1.50
1.58
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-0.7
9-0
.63-0
.47-0
.32-0
.16
0.00
0.16
0.32
0.48
0.63
0.79
0.95
1.11
1.27
1.43
1.58
1.74
1.90
2.06
2.22
2.37
2.53
2.69
2.85
3.01
3.17
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 3
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y = 5.405.00
Y = 0.400.12
Y = ( Yo / Xo ) * X
F = 14.39 ° = 0.25 rad
a1 = 2.13 inches h = 5.19 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 550 fpm0.85
g = 32.20
= 2.93 = l
e = 104.4 ° = 1.82 rad
r0 = 1.04 ft X0 = -0.26 ft Y0 = 1.01 ft
r1 = 1.22 ft X1 = -0.30 ft Y1 = 1.18 ft
r2 = 1.47 ft X2 = -0.37 ft Y2 = 1.43 ft
rpm = 84.03 rpm end of pulley
Vo = 9.17 fps Vxo = 8.88 fps Vyo = 2.28 fps
V1 = 10.72 fps Vx1 = 10.39 fps Vy1 = 2.67 fps
V2 = 12.97 fps Vx2 = 12.56 fps Vy2 = 3.22 fps
Yo / Xo = -3.90
F = 0F > 0 1.82F < 0
= 0.97 =#VALUE! °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
2-0
.69
-0.3
5-0
.02
0.31
0.65
0.98
1.31
1.65
1.98
2.31
2.65
2.98
3.31
3.65
3.98
4.31
4.65
4.98
5.31
5.65
5.98
6.31
6.65
6.98
7.31
7.65
7.98
8.31
8.65
8.98
9.31
9.65
9.98
10.3
1
10.6
5
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 4
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 15.78 ° = 0.28 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 120 fpm
g = 32.20
= 0.14 = l
e = 8.2 ° = 0.14 rad
r0 = 1.04 ft X0 = 1.03 ft Y0 = 0.15 ft
r1 = 1.25 ft X1 = 1.24 ft Y1 = 0.18 ft
r2 = 1.55 ft X2 = 1.53 ft Y2 = 0.22 ft
rpm = 18.33 rpm end of pulley
Vo = 2.00 fps Vxo = 0.29 fps Vyo = -1.98 fps
V1 = 2.40 fps Vx1 = 0.34 fps Vy1 = -2.38 fps
V2 = 2.98 fps Vx2 = 0.43 fps Vy2 = -2.95 fps
Yo / Xo = 0.14
F = 0F > 0 0.14F < 0
= 0.96 = 1.4381.8 °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 5
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 15.23 ° = 0.27 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 260 fpm
g = 32.20
= 0.67 = l
e = 42.2 ° = 0.74 rad
r0 = 1.04 ft X0 = 0.77 ft Y0 = 0.70 ft
r1 = 1.25 ft X1 = 0.93 ft Y1 = 0.84 ft
r2 = 1.55 ft X2 = 1.15 ft Y2 = 1.04 ft
rpm = 39.73 rpm end of pulley
Vo = 4.33 fps Vxo = 2.91 fps Vyo = -3.21 fps
V1 = 5.20 fps Vx1 = 3.49 fps Vy1 = -3.85 fps
V2 = 6.45 fps Vx2 = 4.33 fps Vy2 = -4.78 fps
Yo / Xo = 0.91
F = 0F > 0 0.74F < 0
= 0.96 = 0.8347.8 °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 5A
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 0.00 ° = 0.00 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 180 fpm
g = 32.20
= 0.32 = l
e = 18.8 ° = 0.33 rad
r0 = 1.04 ft X0 = 0.99 ft Y0 = 0.34 ft
r1 = 1.25 ft X1 = 1.18 ft Y1 = 0.40 ft
r2 = 1.55 ft X2 = 1.47 ft Y2 = 0.50 ft
rpm = 27.50 rpm end of pulley
Vo = 3.00 fps Vxo = 0.97 fps Vyo = -2.84 fps
V1 = 3.60 fps Vx1 = 1.16 fps Vy1 = -3.41 fps
V2 = 4.46 fps Vx2 = 1.44 fps Vy2 = -4.23 fps
Yo / Xo = 0.34
F = 0 0.33F > 0F < 0
= 1.00 = 1.2471.2 °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 5B
X
Y
DISCHARGE TRAJECTORIES
Y =
Y =
Y =
Y =
Y = ( Yo / Xo ) * X
F = 13.36 ° = 0.23 rad
a1 = 2.50 inches h = 6.10 inches
R = 12.00 inches Pulley e = 0.50 inchesespesor faja
V = 420 fpm
g = 32.20
= 1.75 = l
e = 103.4 ° = 1.80 rad
r0 = 1.04 ft X0 = -0.24 ft Y0 = 1.01 ft
r1 = 1.25 ft X1 = -0.29 ft Y1 = 1.22 ft
r2 = 1.55 ft X2 = -0.36 ft Y2 = 1.51 ft
rpm = 64.17 rpm end of pulley
Vo = 7.00 fps Vxo = 6.81 fps Vyo = 1.62 fps
V1 = 8.40 fps Vx1 = 8.17 fps Vy1 = 1.94 fps
V2 = 10.42 fps Vx2 = 10.13 fps Vy2 = 2.41 fps
Yo / Xo = -4.21
F = 0F > 0 1.80F < 0
= 0.97 =#VALUE! °
SQRT ( r02 - X2 )
Yo + ( Vyo / Vxo ) * ( X - Xo ) - ( g / 2 Vxo2 ) * ( X - Xo )2
Y1 + ( Vy1 / Vx1 ) * ( X - X1 ) - ( g / 2 Vx12 ) * ( X - X1 )2
Y2 + ( Vy2 / Vx2 ) * ( X - X2 ) - ( g / 2 Vx22 ) * ( X - X2 )2
ft / s2
V2 / g r
e ( rad )
l > 1 l = 1 l < 1l>=cosF l<cosF
cos F G ( rad )
-1.0
4-0
.94
-0.8
3-0
.73
-0.6
2-0
.52
-0.4
2-0
.31
-0.2
1-0
.10
0.00
0.10
0.21
0.31
0.42
0.52
0.63
0.73
0.83
0.94
1.04
1.15
1.25
1.35
1.46
1.56
1.67
1.77
1.87
1.98
2.08
-0.01
-0.01
-0.01
-0.01
-0.01
-0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.01
0.01
0.01
0.01
-1.0
4-0
.88-0
.71-0
.54-0
.37-0
.21-0
.04
0.13
0.29
0.46
0.63
0.79
0.96
1.13
1.29
1.46
1.62
1.79
1.96
2.12
2.29
2.46
2.63
2.79
2.96
3.13
3.29
3.46
3.63
3.79
3.96
4.13
4.29
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
DISCHARGE TRAJECTORIES FAJA 6
X
Y