1. Find a combination quarter of a circle. 2. Find the value of all 8, 8₁, 8, and y in open and close configuration of link lengths where motion of a point on output link is one

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Open Configuration
Angle
Value
2
28°
उ
84
28°
Y
28°
| AD=r1 AB=r2 BC=r3 CD=r4
r1 is grounded. r2 is given input
Grashof Mechanism
क
Crossed Configuration
Angle
Value
82
28°
उ
272.38°
84
239.62⁰
28°
Y
+
VARIABLES
80
r1 (cm)
r2 (cm)
r3 (cm)
r4 (cm)
e (º)
60
80
60
28
CONTROLS
►
●
Transcribed Image Text:Open Configuration Angle Value 2 28° उ 84 28° Y 28° | AD=r1 AB=r2 BC=r3 CD=r4 r1 is grounded. r2 is given input Grashof Mechanism क Crossed Configuration Angle Value 82 28° उ 272.38° 84 239.62⁰ 28° Y + VARIABLES 80 r1 (cm) r2 (cm) r3 (cm) r4 (cm) e (º) 60 80 60 28 CONTROLS ► ●
1. Find a combination
of link lengths where motion of a point on output link is one
quarter of a circle.
2. Find the value of all 0, 0, 0, and y in open and close configuration
Read the value of link lengths and the input angle 8., then use the
formulae given below to calculate the value of unknowns 03, 0, and y
K₁ = = K₂= d
K2
K3
=
a²-b²+c²+d²
2ac
A = cos 0₂ - K₁ - K₂ cos 0₂ + K3
B = -2 sin 0₂
C = K₁ (K₂ + 1) cos 02 + K3
-B± √B²-4AC
2A
0412 = 2tan-1
d
K₁ = —
K5
=
c²d²a²-6²
2ab
D = cos 0₂ - K₁ - K4 cos 0₂ + K5
E = -2 sin 0₂
FK₁+ (K₁ - 1) cos 02 +K5
0312
2 tan-1
(-E±
-E± √E²4DF
2D
Y = 04-03
Transcribed Image Text:1. Find a combination of link lengths where motion of a point on output link is one quarter of a circle. 2. Find the value of all 0, 0, 0, and y in open and close configuration Read the value of link lengths and the input angle 8., then use the formulae given below to calculate the value of unknowns 03, 0, and y K₁ = = K₂= d K2 K3 = a²-b²+c²+d² 2ac A = cos 0₂ - K₁ - K₂ cos 0₂ + K3 B = -2 sin 0₂ C = K₁ (K₂ + 1) cos 02 + K3 -B± √B²-4AC 2A 0412 = 2tan-1 d K₁ = — K5 = c²d²a²-6² 2ab D = cos 0₂ - K₁ - K4 cos 0₂ + K5 E = -2 sin 0₂ FK₁+ (K₁ - 1) cos 02 +K5 0312 2 tan-1 (-E± -E± √E²4DF 2D Y = 04-03
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