1. - METUcourses.me.metu.edu.tr/courses/me301/sec01/homeworks/ME... · 2014-11-03 · The controllable DOF of the system is equal to the one provided by the DOF equation. Actuation:

SOLUTIONS OF HW1

1.

𝐹 = 3 × (8 − 1) − 2 × 9 = 3

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 8 9 (𝑅12, 𝑅23, 𝑅34, 𝑅15, 𝑅56, 𝑅46, 𝑅18, 𝑅78, 𝑅47) 0 3

The controllable DOF of the system is equal to the one provided by the DOF equation.

Actuation: Since the controllable DOF of the system is 3, the three actuators are to be placed at the

joints 𝑅12, 𝑅15 and 𝑅18 to drive the links 2, 5 and 8 (i.e., the cranks) with respect to link 1 (the base).

Note: Whenever possible, an actuator is placed at a joint adjacent to the base so that it is not moved

around unnecessarily.

Usage-1: As mentioned in the problem statement, this mechanism can be used as a manipulator to

control the location and orientation of the tool with the tip point P.

Usage-2: Alternatively, it can also be used as a planar motion simulator. In that case, the triangular

platform is driven so as to imitate the linear and angular accelerations of an actual vehicle and the point

P represents the mass center of a human being onboard the vehicle.

2.

𝐹 = 3 × (5 − 1) − 2 × 5 − 1 × 1 = 1

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 5 5 (𝑅12, 𝑅14, 𝑅34, 𝑅35, 𝑃15) 1 (𝐶𝑆 23) 1


Actuation: The actuator can be placed most preferably at the joint 𝑅14 to drive the link 4 (the crank)

with respect to link 1 (the base). The reason of this preference is that the physical connection between

the links 4 and 3 looks sturdier than that between the links 2 and 3.

Usage: Among many different options, this mechanism can be used as a punching machine.

3.

𝐹 = 3 × (11 − 1) − 2 × 14 = 2

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 7 14 (𝑅12, 𝑅13, 𝑅23, 𝑅34, 𝑅45, 𝑅56, 𝑅67, 𝑅47, 𝑅38, 𝑅59, 𝑅1,10, 𝑅4,11, 𝑃89, 𝑃10,11) 0 2


Actuation: The two linear actuators required for 𝐹 = 2 can be placed at the joints 𝑃89 and 𝑃10,11 to

drive the links 9 and 11 (the cylinders) with respect to the links 8 and 9 (the pistons). The actuator at

𝑃10,11 functions mainly to control the elevation of the bucket and the actuator at 𝑃89 functions mainly to

control the orientation of the bucket.

Usage: This mechanism is a part of a loader. It is used to control the position of the bucket with respect

to the loader.

4.

𝐹 = 3 × (6 − 1) − 2 × 6 − 1 × 1 = 2

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 6 6 (𝑅12, 𝑅23, 𝑅34, 𝑅35, 𝑅56, 𝑅14) 1 (𝐶𝑎𝑚46) 2


Actuation: The two actuators can be placed at the joints 𝑅12 and 𝑅35. The actuator at 𝑅12 controls the

pose of the submechanism 𝐴 = {1,2,3,4} with respect to the base and the actuator at 𝑅35 controls the

pose of the submechanism 𝐵 = {3,5,6,4} with respect to 𝐴.

NB: The two actuators cannot be placed at 𝑅12 and 𝑅14. If they were placed so, then they would be over

driving the submechanism 𝐴 = {1,2,3,4}, which is a 4-bar mechanism with 𝑓𝐴 = 1 and therefore it needs

only one actuator.

Usage: Any reasonable answer is accepted.

5.

a.

Note-1: The base link is link 0. Since link 3 (planet carrier) is also fixed, it becomes a part of the

base. Therefore, it is also numbered as link 0, instead of link 3. So, there are four links: link 0,

link 1, link 2, link 4.

Note-2: The system is so arranged that the ring gear (link 4) can only rotate with respect to the

base (link 0). Therefore, the relative motion between the links 0 and 4 occurs as if there is a

kinematically equivalent revolute joint in between.

Note-3: There are two gear pairs (𝐺12 and 𝐺24) and each of them is regular and form-closed,

i.e., none of them require an external force to remain connected. Therefore, they are to be

treated as if each of them has two degrees of freedom. In other words, 𝑓𝐺12= 𝑓𝐺24

= 2.

According to the above notes, the DOF of the system is found as follows:

𝐹 = 3 × (4 − 1) − 2 × 3 − 1 × 2 = 1

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 4 3 (𝑅01, 𝑅02, 𝑅04) 2 (𝐺12, 𝐺24) 1


Actuation: The actuator can be placed at the joint 𝑅01 to drive link 1 (the sun gear) with respect to link 0

(the base).


b.

In this case, the planet carrier is also allowed to move. Therefore, it enters the analysis as an

additional link denoted as link 3. Thus, the DOF of the system is now found as follows:

𝐹 = 3 × (5 − 1) − 2 × 4 − 1 × 2 = 2

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 5 4 (𝑅01, 𝑅03, 𝑅04, 𝑅23) 2 (𝐺12, 𝐺24) 2


Actuation: The two actuators can be placed at the joints 𝑅01 and 𝑅03 to drive the links 1 and 3 (the sun

gear and the planet carrier) with respect to link 0 (the base).


6.

a.

In this mechanism, the links 2 and 3 get engaged when the pin of link 2 gets into the slot of link 3. When

they are engaged, the DOF of the system is determined as shown below:

𝐹 = 3 × (3 − 1) − 2 × 2 − 1 × 1 = 1

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 3 2 (𝑅12, 𝑅13) 1 (𝐶𝑎𝑚23) 1


Actuation: The actuator can be placed at the joint 𝑅12 to drive link 2 with respect to link 1 (the base).

Usage: Mechanical watches, early movie projectors, indexing systems, etc.

b. When the links 2 and 3 are disengaged, the convex circular part of link 2 rotates in

contact with the concave circular part of link 3. However, no motion is transmitted

between them. Link 2 can keep rotating and link 3 remains stationary.

c. In the second version of the system, the pins become three additional moving links.

Consequently, the DOF of the system increases as shown below:

𝐹 = 3 × (5 − 1) − 2 × 4 − 1 × 1 = 3

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 5 4 (𝑅14, 𝑅24, 𝑅15, 𝑅35) 1 (𝐶𝑎𝑚23) 3

d. Actually, the significant motion of the system (i.e., the rotations of the links 2 and 3) is the same

in both versions. However, in part c, the degree of freedom is found as 3. Clearly, 2 of the 3

degrees of freedom are insignificant, which are the uncontrolled rotations of the pins (i.e., the

links 4 and 5) about their own centerlines. Therefore, the controllable DOF of the system is still

one when the links 2 and 3 are engaged.

7.

𝐹 = 3 × (8 − 1) − 2 × 9 = 3

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 8 9 (𝑃12, 𝑃15, 𝑅23, 𝑅56, 𝑅34, 𝑅67, 𝑅36, 𝑅78, 𝑅48) 0 3

The controllable DOF of the system is equal to the one provided by the DOF equation. With these

controllable freedoms, the pink object (link 8) can be positioned as desired in its plane of motion. In

other words, its rotation about the z axis and its translations along the x and y axes can be controlled as

desired.

Actuation: Two of the three actuators are placed at the joints 𝑃12 and 𝑃15 to drive the links 2 and 5 (the

pistons) with respect to link 1 (actually, with respect to the cylinders that are fixed to the base). The

third actuator is placed at the joint 𝑅78 (or 𝑅48) to drive link 8 with respect to link 7 (or 4).

Usage: This mechanism can be used as a kind of pantograph in order to control the motion of the pink

object in its working plane.

8.

In this system, the gripper and its stem is considered as a single link by ignoring the relative

motions of the gripper with respect to its stem. These relative motions are the twist rotation of

the gripper and the squeeze-release motions of the fingers. Thus, the DOF of the system is

found as follows:

𝐹 = 3 × (7 − 1) − 2 × 7 = 4

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 7 7 (𝑅12, 𝑅34, 𝑅46, 𝑅67, 𝑅56, 𝑅15, 𝑃23) 0 4


Actuation: The four actuators can be placed at the joints 𝑅12, 𝑅15, 𝑅67 and 𝑃23.

Usage: This mechanism is used as a redundant planar robot manipulator. It has redundancy, because,

even if the actuator of 𝑃23 is kept fixed, the other three actuators of 𝑅12, 𝑅15, and 𝑅67 are sufficient to

control the orientation 𝜃 of the gripper and the 𝑥 and 𝑦 coordinates of its tip point. Thus, the actuator of

𝑃23 happens to be redundant. This redundancy can be used to perform a manipulation task in some

more efficient way, e.g., with smaller joint displacements.

9.

Note that the gear pair (𝐺23) is a form-closed and regular gear pair. It is form-closed thanks to

link 1 that keeps the links 2 and 3 together. Thus, the DOF of the system is found as shown

below:

𝐹 = 3 × (7 − 1) − 2 × 8 − 1 × 1 = 1

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 7 8 (𝑅17, 𝑅67, 𝑅46, 𝑅45, 𝑅34, 𝑅13, 𝑅12, 𝑅15) 1 (𝐺23) 1


Actuation: The actuator is normally placed at the joint 𝑅12 to drive link 2 with respect to link 1 (the

base).

Usage: This mechanism transmits the actuator-driven rotational motion of link 2 all the way to link 7 by

transforming it to an oscillatory motion. Thus, the blade attached to link 7 performs a cutting action.

10.

Note-1: The degree of freedom of the cam pair with no slip is 1. Actually, a no-slip cam pair is

kinematically equivalent to a gear pair. The slip-preventing roughness of the contact surfaces

can be visualized as nano-scale teeth.

Note-2: The spring is to be removed together with its joints because it is not countable as a link.

According to the above notes, the DOF of the system is found as follows:

𝐹 = 3 × (5 − 1) − 2 × 5 − 1 × 1 = 1

𝜆 𝑛 𝑗1 𝑗2 𝐹 3 5 5 (𝑅13, 𝑅23, 𝑅34, 𝑅45, 𝐺15) 1 (𝐶𝑎𝑚12) 1


Actuation: The actuator can be placed at the joint 𝑅13 to drive link 3 (the crank) with respect to link 1

(the base).


Documents

1. - METUcourses.me.metu.edu.tr/courses/me301/sec01/homeworks/ME... · 2014-11-03 · The controllable DOF of the system is equal to the one provided by the DOF equation. Actuation: