4장. 위치 해석

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4장. 위치 해석

Once a tentative mechanism design has been

synthesized, it must be analyzed. A principal goal of kinematic analysis is to determine the accelerations of all moving parts in the assembly.

In order to calculate the accelerations, we must find the positions of all the links and then differentiate the position versus time to find velocities and then

differentiate again to obtain the expressions for acceleration.

This can be done by several methods. We could use a graphical approach to

determine the position, velocity, and acceleration or we could derive the general equations of motion for any position, differentiate for velocity and acceleration.

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위치 해석

4.5 대수 방법에 의한 링크의 위치 해석

(Algebraic position analysis of linkages)

Ex.4-1 Given a fourbar linkage with the link lengths d=100 mm,

a=40 mm, b=120 mm, c=80 mm. For q₂=40, find all possible values of q₃ and q₄.

q₂

q₃

q₄ a

b

c

d

=40

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 Solution

3 2

2 1

2

K K cos K

cos

A  q   q 

sin

2

2 B   q

3 2

2 2

1

cos K cos K

K

C   q  q 

5 . 40 2

a 100

K₁  d  

562 .

) 0 80 )(

40 ( 2

100 80

120 40

ac 2

d c

b

K₃ a² ² ² ² ²  ²  ²  ² 

 



 

25 . 80 1

c 100

K₂  d  

 

q₂ 40

 2.129

 1.286

 1.339

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위치 해석

A  2.129 B  1.286 C  1.339

 Solution

A

2B 4AC 2 B

tanq⁴    ² 

      



2.129



2

339 .

1 129 . 2 4 286

. 1 286

.

1 ²







 

55 . 0 , 15 . 1



 q₄  98.01^, 57.33^

d cos

c cos

a

sin c sin

tan a

4 2

4

3  q 2 q 

q

 q

  q

37 . 0 , 80 . 1



 q₃  60.98^, 20.30^

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A  2.129 B  1.286 C  1.339

 Solution

A

2B 4AC 2 B

tanq⁴    ² 

      



2.129



2

339 .

1 129 . 2 4 286

. 1 286

.

1 ²







 

55 . 0 , 15 . 1



 q₄  98.01^, 57.33^

d cos

c cos

a

sin c sin

tan a

4 2

4

3  q 2 q 

q

 q

  q

37 . 0 , 80 . 1



 q₃  60.98^, 20.30^

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위치 해석

 Solution q₃ , q₄   20.30^, 57.33^

q₃ , q₄  60.98^,98.01^

40^

20.30^

57.33^ -60.98^

-98.01^

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4.11 전달각 (Transmission Angle)

 최소 전달각

• Overlapped

• Extended

a

b c

d m



   



m arccos b² c²2bc(d a)²

m

g ^arccos_^^b ^ ^c ₂^_bc⁽^d ^ ^a⁾ _^

2 2

2

g







m

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4.14 Newton-Raphson 방법

• 1차원 근 찾기 (One-dimensional root-finding)

1 i i

i) xf(xxi) x

( f

 

  Step 1. Guess x_i.

Step 2. Calculate f(x_i).

Step 3. Calculate x_i+1. ) x ( ff(x ) x

x i

i i 1

i_   

Step 4. Calculate f(x_i+1).

Step 5. If f(x_i+1) <  else stop

x_i= x_i+1

Calculate f’(x_i) Goto Step 3

x_i x x_i+1

f(x_i)

f(x)

f(x_i+1)

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4.14 Newton-Raphson 방법

• 다차원 근 찾기 (Multi-dimensional root-finding)

f(x)=0 f₁(x₁,x₂,……_,x_n₎₌₀ f₂(x₁,x₂,……_,x_n₎₌₀

 

f_n(x₁,x₂,……_,x_n₎₌₀

f ( x )=0

n 2 1

f f f f  



n 2 1

x x x x  



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위치 해석

Newton-Raphson 방법

• 다차원 근 찾기

Step 1. Guess x_i.

Step 2. Calculate f ( x_i).

Step 3. Calculate x_i+1.

Step 4. Calculate f ( x_i+1 ).

Step 5. If f ( x_i+1) <  else stop

x_i= x_i+1 Calculate Goto Step 3

-1 

 

xⁱ¹  xⁱ 

(

 f_x

)

f

















 ^

n n 1

n

n 2 2

2 1

2

2 1 2

1 1

1

x

x f x

f

x f x

f x

f

x f x

f x

f







f^x



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Newton-Raphson 방법에 의한 4절 기구 해석

 Solution procedure (풀이 과정)

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 Solution Procedure

4

x 3

q

 q



0 d cos

c cos

b cos

a q₂  q₃  q₄   0 sin

c sin

b sin

a q₂  q₃  q₄ 

 f₁

 f₂

2 1

f f  f

























2 2 1

2

2 1 1

1

x

xf xf

f _



 





q



 q



q

 q



 

4 3

4

3 c cos

cos b

sin c

sin b

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0 1 2 3 4 5 6

0 20 40 60 80 100 120 140

# of iterations

Link angle (deg)

q₃ q₄

• Ex 4-1

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위치 해석

0 1 2 3 4

-140 -120 -100 -80 -60 -40 -20 0

# of iterations

Link angle (deg) q₃

q₄

• Ex 4-1

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• Watt 기구

1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 2.15 2.2 2.25

-1.5 -1 -0.5 0 0.5 1 1.5

Trajectory(P) of the Watt linkage

X direction

Y direction

-40 -30 -20 -10 0 10 20 30 40

-50 0 50 100 150 200 250

q₂ (deg) q3 & q4 (deg)

q₃ q₄