Kinematics Project (CAM Design)

# the question of the project is in the attachment 6.7The due date is in a day and a half.Following information should be included in the project reportMotion Requirement description (problem description)Sketches of displacement diagram, velocity, and acceleration together with answers for following questions:The number of segments neededThe reasons for the existence of each segment.Detailed calculation of each segment:The given parameters and boundary conditionunknown parameters of the segmentThe function chosen for the segment and reason to chooseparameter calculationFinal result:The parameter value of each segment (ßi and Li)The function of each segment# I’ve attached a PDF file has an example of the project. So, the project youre going to do should be similar to it. It’s example 6.2 in the pdf file at the bottom. Also, it should be handwritten or typed. thanks
.png

chapter_6_cam_revised.pdf

Don't use plagiarized sources. Get Your Custom Essay on
Kinematics Project (CAM Design)
Just from $13/Page
Order Essay

ideas_for_project.docx

Unformatted Attachment Preview

Chapter 6
Cam Design
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
What is a CAM
• What Is A Cam?
Components in CAM
• Cam: The cam is the teardrop shaped device.
• Shaft (Rotating Wheel): The cam is typically
mounted in a fixed position on a shaft or rotating
wheel. As the shaft rotates the pointy end of the
cam comes into contact with the cam follower
once every 360 degrees of angular motion.
• Cam Follower (Lever): The cam follower (or lever)
is a long, linear part that comes into contact with
and is linearly displaced by the pointy end of the
cam once every 360 degrees of cam motion.
Classification of CAMs and Follower
• CAM-follower mechanisms are:
– Simple
– In-expensive
– Fewer moving parts
– Occupy little space
Types of CAM:
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Types of Follower:
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Displacement Diagram
•
•
•
Displacement Diagram: shows the movement of follower during one
revolution of CAM.
During on cycle of CAM, the follower execute following event: Rise,
Dwell, Return.
The total rise is called lift.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
•
•
•
•
Section ß1: lift=L1, parabolic motion
Section ß2: lift=L2, linear motion
Section ß3: lift=L3, parabolic motion
Boundary condition must be met.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Graphic Layout of CAM Profile
• Given: The follower’s motion represented by
the displacement diagram.
• Design Goal: Design the exact shape of the
cam profile to deliver the specified motion of
follower.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Design of CAM
• Trace point: center of roller. Used to
represent the motion of follower.
• Pitch curve: locus on the cam generated by
the trace point of follower moves with respect
to cam
• Prime circle: the smallest circle that is tangent
to pitch curve with center at cam rotation axis.
• Basic circle: smallest circle that is tangent to
cam profile.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Kinematic Coefficient of Follower
• Displacement diagram:
– abscissa: the angle of cam input ?
– ordinate: the output displacement of follower Y-Y0
• The diagram has several segments
Kinematic Coefficient of Follower
• The displacement diagram represents a
mathematical function relating to the input.
• y=y(?)
??
??
• y’(?) =
(velocity)
first order kinematic coefficient
?2 ?
??2
• y’’(?) =
second order kinematic coefficient
(acceleration)
• y’’’(?) =
?3 ?
??3
third order kinematic coefficient
Design Example 6.1
• Requirement:
1. Design a segment of a displacement diagram of
cam.
2. Rise segment from one dwell to another dwell
needs to be parabolic motion
3. The total lift is L
4. Total cam angle is ß
Solution for Example 6.1
• Draw the diagram
• The function needs to be in two segments:
– First segment moves from dwell (steady state) to
mid point, i.e. is acceleration motion.
– Second segment moves from mid point to dwell
(steady state), i.e. is deceleration motion.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
•
•
•
•
•
First Segment Function Design
y = A ?2 +B ? +C
Derivative:
– y’ = 2A ? +B
– y’’ = 2A
– y’’’=0
Boundary condition: start from dwell
– y(0) = 0 ? C=0
– y’(0) = 0 ? B=0
In order for cam to go back to dwell at end, the cam needs to decelerate in
second half.
– When ? = ß/2, y=L/2 ?A = 2?/ ß2
First segment function Equation:
?
– y= 2L(ß )2
4? ?
(
ß ß
4?
= ß2
– y’ =
– y’’
– y’’’=0
),
y’max =
2?
ß
Second Segment Function Design
•
•
•
y = A ?2 +B ? +C
Boundary condition: ends at dwell
– y(ß) = L
– y’(ß) = 0
At mid point, speed must match that of section 1’s
– When ? = ß/2, y’=
2?
– A = – ß2 , B = •
4?
ß
2?
ß
, C=-L,
Second segment function Equations:
?
– y= L[ 1- 2(1- ß )2 ]
– y’ =
4?
ß
– y’’ = – y’’’=0
?
(1- ß ),
4?
ß2
Requirement for High-Speed CAMS
• A follower has mass m.
• Force on the follower and cam F=m*a, here a is the
acceleration from the displacement curve
• An acceleration curve with abrupt change (jerk becomes
infinity) exert an abrupt force on the contact surfaces that
leads to noise and surface wearing.
• Requirement: First order and second order kinematic
coefficient (velocity and acceleration) must be continuous.
Iteration Design Approach for low speed
cam:
• Four circle-arcs are used to meet
follower displace requirements.
• Tangent lines are used to blend in the
arcs.
• Acceleration changes abruptly at each
blending point because of instantaneous
change in the radius of curvature of cam
profile.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Kinematic Coefficients w.r.t. Input Variable ?
• First segment function Equation:
?
ß
4? ?
y’ = ( ),
ß ß
4?
y’’ = 2
ß
– y= 2L( )2
–
–
y’max =
2?
ß
– y’’’=0
• Second segment function Equations:
?
ß
– y= L[ 1- 2(1- )2 ]
– y’ =
4?
ß
– y’’ = –
?
ß
(1- ),
4?
ß2
– y’’’=0
??
• Here y’=
??
?2?
y’’= 2
??
?3?
y’’’= 3
??
Kinematic Coefficients w.r.t. Angular Velocity
and Time
•
Input angle ? is a function of time t: ?(t)
•
Angular velocity: ? = ??
??
?2?
??2
•
Angular acceleration: a =
•
Usually, plate cam is driven by a constant angular velocity input shaft.
– ? = constant
– ? = ?t
–
?2?
a= 2
??
?? ?? ??
=
= y’?
?? ?? ??
?2?
= y’’?2 + y’a
??2
?3?
3
3 = y’’’? + 3y’’a
??
•
?? =
•
?? =
•
?? =
•
•
•
•
When a = 0
?=
? y’?
?? = y’’?2
?? = y’’’?3
? + y’a?
Motion Type and Comparison
•
•
•
•
•
Parabolic
Harmonic
Cycloidal
Polynomial
Rise vs. Return motion
Standard CAM Motion– Parabolic, Full-rise
?
ß
y= 2L( )2
y’ =
4? ?
( )
ß ß
y’’ =
4?
ß2
y’’’=0
• Advantage: velocity, y’ is continuous
• Disadvantage: acceleration is not continuous, jerk in infinity
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– Harmonic, Full-rise
p?
ß
• y= 2L(1 – cos
• y’ =
p?
ß
• y’’ =
sin(
p2?
2ß2
• y’’’= –
p?
ß
)
cos (
p3?
2ß3
)
p?
ß
sin (
)
p?
ß
)
• Advantage: no discontinuity for y’ and y’’
• Disadvantage: acceleration is not zero at beginning and end
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– Cycloidal, Full-rise
?
ß
• y= L( –
1
2p?
sin
2p
ß
)
2p?
ß
)
?
ß
• y’ = (1 – cos
• y’’ =
•
2p?
ß2
4p4?
y’’’= 3
ß
sin (
2p?
ß
cos (
2p?
ß
)
)
• Advantage: y’ and y’’ are all zero at boundary
• Disadvantage: peak value of acceleration and jerk are too high.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– 8th order
Polynomial, Full-rise
?
?
?
•
y = L[6.0955( )3 – 20.78040( )5 +26.73155( )6
•
y’ = [18.29265( )2 – 103.90200( )4 +160.38930( )5
ß
?
– 13.60965 ( )7
ß
?
ß
?
+2.56095( )8 ]
ß
ß
?
ß
ß
?
?
?
?
ß
ß
– 95.26755 ( )6 +20.48760( )7 ]
ß
•
?
y’’=
ß
?
2
ß
?
?
ß
ß
[36.58530( ) – 415.60800( )3 + 801.94650( )4
?
ß
?
– 571.60530 ( )5 + 143.41320( )6 ]
ß
•
y’’’=
?
3
ß
[36.58530-
?
ß
? 2
1246.82400( )
ß
? 5
?
+ 3207.78600( )3 –
ß
2858.02650 ( )4 + 860.47920( ) ]
ß
ß
• Advantage: y’’ is non-symmetric(positive and negative accelerations
are possible
• peak value of acceleration is small.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– Harmonic, Full-return
p?
ß
• y= 2L(1 + cos
• y’ = –
p?
ß
• y’’ = •
sin(
p2?
2ß2
p3?
y’’’= 3

p?
ß
)
cos (
sin (
)
p?
ß
p?
ß
Theory of Machines and Mechanisms, 5e
)
)
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– Cycloidal, Fullreturn
• y= L(1 –
?
1
2p?
+ sin
ß 2p
ß
?
ß
2p?
ß
• y’ = – (1 – cos
• y’’ = –
2p?
ß2
• y’’’= –
4p4?
ß3
sin (
)
2p?
ß
cos (
)
)
2p?
ß
Theory of Machines and Mechanisms, 5e
)
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
Standard CAM Motion– 8th order
Polynomial, Full-return
•
?
ß
?
ß
y = L[1 – 2.63415 ( )2 + 2.78055( )5
?
ß
?
ß
+ 3.1706( )6 -6.87795( )7
?
+ 2.56095(ß )8 ]
•
•
•
?
?
?
[5.62830( ) – 13.90275( )4
ß
ß
ß
? 5
? 6
– 19.02360(ß ) +48.14565(ß )
?
– 20.48760( )7 ]
ß
?
?
y’’= – ß2 [5.2683 – 55.611 (ß )3 –
?
?
95.118(ß )4 + 288.8739 (ß )5 –
?
143.41320(ß )6 ]
?
?
y’’’= ß3 [166.833(ß )2 +
?
?
380.472(ß )3 – 1444.3695 (ß )4 +
?
860.47920( )5 ]
ß
y’ = –
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Copyright © 2017 Oxford University Press
p?

• y= L(1 – cos
• y’ =
p?

• y’’ =
sin(
p2?
4ß2
• y’’’= –
p?

)
• y= L(sin
)
• y’ =
p?
cos (

p3?
4ß3
Theory of Machines and Mechanisms, 5e
)
p?
sin (

)
Uicker, Pennock, Shigley
p?

p?

)
p?

cos(
)
• y’’ = –
p2?
4ß2
???(
• y’’’= –
p3?
8ß3
cos (
p?

p?

)
)
Copyright © 2017 Oxford University Press
p?

• = L(cos
• y’ = –
p?

• y’’ = –
• y’’’=
sin(
p2?
4ß2
p3?
8ß3
• y= L(1- sin
)
p?

)
p?
cos (

p?
sin (

Theory of Machines and Mechanisms, 5e
)
)
Uicker, Pennock, Shigley
p?

)
• y’ = –
p?

• y’’ =
p2?
4ß2
???(
• y’’’=
p3?
8ß3
cos (
cos(
p?

)
p?

p?

)
)
Copyright © 2017 Oxford University Press
• y= L( 1 • y’ =
?
(1
ß
• y’’ =
p?
ß2
• y’’’= –
?
ß
sin (
p?
ß
)
• y= L( + sin
?
ß
1
p
p?
ß
)
?
ß
p?
ß
)
p?
ß
)
• y’ = (1 + cos
p?
ß
)
• y’’ = –
p?
ß2
• y’’’= –
p4?
ß3
– cos
p2?
ß3
Theory of Machines and Mechanisms, 5e
1
p
– sin
p?
cos (
ß
)
Uicker, Pennock, Shigley
sin (
p?
ß
cos (
)
p?
ß
)
Copyright © 2017 Oxford University Press
• y= L( 1 • y’ = –
?
ß
• y’’ =
p?
ß2
• y’’’= –
?
ß
+
(1 –
1
p?
sin
p
ß
p?
cos
ß
p?
sin (
ß
p2?
ß3
Theory of Machines and Mechanisms, 5e
)
)
p?
cos (
ß
)
)
?
ß
1
p
• y= L( 1 – – sin
?
ß
• y’ = – (1 + cos
• y’’ = –
p?
ß2
• y’’’= –
p2?
ß3
Uicker, Pennock, Shigley
sin (
p?
ß
p?
ß
p?
ß
cos (
)
)
)
p?
ß
)
Copyright © 2017 Oxford University Press
Match Derivatives of Displacement
Diagrams
• Motion requirements must be met
• Kinematic coefficients are continuous
• Maximum magnitude of velocity, acceleration
peak need to be as small as possible
Example 6.2 Cam Design example
A plate cam with a reciprocal follower is to be driven by a constant
speed motor at 150rpm. The follower is to start from a dwell, accelerate
to a uniform velocity of 25in/s, maintain this velocity for a 1.25 in of rise,
decelerate to the top of the lift, return, and then dwell for 0.1 s. The
total lift is 3.00 in. Determine the complete specification of the
displacement diagram.
Problem analysis:
1. How many segments are needed to implement the above
requirements?
2. The velocity and acceleration curve need to be continuous. This
means the connection between each segment must be continuous.
Example 6.2 Problem Analysis
1.
2.
3.
4.
5.
First segment AB:
•
Start with dwell, y’=0, y’’=0
•
Continuity: yAB’(B)=yBC’(B)
•
Accelerate: y’’>0
Second segment BC:
•
Constant velocity:
y’=constant, y’’=0
•
Continuity: yBC’(C)=yCD’(C)
Third segment CD:
•
Decelerate: y’’ < 0 • Continuity: yBC’(C)=yCD’(C) Forth segment DE: • Continuity: yBC’(D)=yCD’(D) • return, y’(E)=0, y’’(E)=0 Fifth segment EF: • Dwell, y’=0, y’’=0 Theory of Machines and Mechanisms, 5e Uicker, Pennock, Shigley Copyright © 2017 Oxford University Press Example 6.2 First Segment AB Design To determine: • Function type • L1, ß1 Given: 2*p 60 • Constant cam axis rotation:?=150rpm= 150* • • • Velocity at point B: y’= = = 1.59155 in/rad ? 15.70796 y: rise y’: ?? – – • 25 = 15.70796 rad/s. Start point velocity: yAB’(A)=0, start from dwell end point velocity (Continuity ): yAB’(B) = yBC’(B) = 1.59155 in/rad y’’: – – Start point acceleration: yAB’’(A)=0, start from dwell end point acceleration: yAB’’(B)=yBC’’(B) = 0, Continuity Calculation: • Choose segment function type: – – – y’(A)=0, y’’(A)=0 y’(B)=1.59155, y’’(A)=0 Fig 6.22a satisfy above two condition. Choose half-rise cycloidal motion segment – y= L( 1 - – – – • ? ß 1 p? ß ? p? y’ = - (1 - cos ) ß ß p? p? y’’ = ß2 sin ( ß ) p2? p? y’’’= - ß3 cos ( ß ) + psin ) yAB’(B)=yBC’(B), 15.70796= - ?1 ß1 pß1 ß1 (1 - cos ) ---------? L1 = 0.079577 ß1 Example 6.2 Second Segment BC Design Second segment BC: To determine: • ß2 Given: • Velocity at point B: y’(B)=1.59155 in/rad • Constant velocity for the whole segment. • L2= 1.25 in Calculation: • Continuity: yAB’(B)=yBC’(B) – ?2 ß2 = 1.25 ß2 = 15.70796 in/rad ---------? ß2 = 1.25 15.70796 = 0.78549rad = 45o Example 6.2 Third Segment CD Design Third segment CD: To determine: • Function type • L3, ß3 Given: • y: rise • y’: – – • Start point velocity: yCD’(C) = yBC’(C) = 15.70796 in/rad , because of the continuity with BC segment end point velocity : yCD’(D) = 0, because it reaches highest point y’’: – – Start point acceleration: yCD’’(C) =0, because of the continuity with BC segment end point acceleration: y CD’’(D) <0, a negative value because CD segment is decelerating. Calculation: • Choose segment function type: • – – – y’(C)= 15.70796, y’(D)=0 y’’(C)=0, y’’(D) < 0 Fig 6.20b satisfy above two condition. Choose half-rise harmonic motion segment – y= L(sin2ß ) – y’ = 2ß cos(2ß ) – y’’ = - – y’’’= p? p? p? p2? 4ß2 p3? - 8ß3 p? ) 2ß p? cos ( 2ß ) ???( From continuity: y’(C)= 15.70796, – p?3 2ß3 cos( p*0 2ß3 ) = 15.70796 ---------? L3= 01.01321 ß3 Example 6.2 Forth Segment DE Design Forth segment DE: To determine: • Function type • L4, ß4 Given: • • y: return y’: – – • Start point velocity: yDE’(D) = yCD’(D) = 0 in/rad , because of the continuity with CD segment end point velocity : yDE’(E) = 0, because it returns back to dwell y’’: – – Start point acceleration: yDE’’(D) = y CD’’(D) <0, because of the continuity with CD segment end point acceleration: y DE’’(E)=0 because it returns back to dwell Calculation: • Choose segment function type: – – – y’(D)= 0, y’(E)=0 y’’(D)<0, y’’(E) =0 Fig 6.17 satisfy above conditions. Choose full-return, eighth order polynomial motion segment – y = L[1 – 2.63415 (ß )2 + 2.78055(ß )5 +3.1706(ß )6 -6.87795(ß )7 +2.56095(ß )8 ] – y’ = - – – • ? ? ? ? ? ? ? ? 4 ? 5 ? 6 ? 7 [5.62830( ) 13.90275( ) -19.02360( ) +48.14565( ) - 20.48760( ) ß ß ß ß ß ß ? ? ? ? ? y’’= - ß2 [5.2683 - 55.611 (ß )3 - 95.118(ß )4 + 288.8739 (ß )5 - 143.41320(ß )6 ] ? ? ? ? ? y’’’= ß3 [166.833(ß )2 + 380.472(ß )3 - 1444.3695 (ß )4 + 860.47920(ß )5 ] From continuity: : yDE’’(D) = y ?4 p2?3 CD’’(D) , pß3 – - – plus from CD segment, L3= 01.01321 ß3 ---------? ß3 = 0.15818 * ß42 ß42 * 5.2683 = - 4ß32 ???( 2ß3 ), ] Example 6.2 Fifth Segment EF Design Fifth segment EF: To determine: • ß5 Given: • • • • y: dwell for 0.1 second. y’: y’ = 0 for the whole segment y’’: y’’ = 0 for the whole segment L5=0 Calculation: • ß5 = 0.1 *? = 0.1 * 15.70796 = 1.570796 rad = 90o Example 6.2 Final Solutions • All equations from all segments: – – – – – – – ß1 + ß2 + ß3 + ß4+ ß5 = 2p L1 + L2 + L3 = L4 L1 = 0.079577 ß1 (from segment AB) ß2 = 0.78549rad (from segment BC) L3= 01.01321 ß3 (from segment CD) ß3 = 0.15818 * ß42 (from segment DE) ß5 = 1.570796 rad (from segment EF) • Solve above equations and results parameters are: – – – – – L1 = 1.1831in, L2 = 1.25001in, L3 = 0.5669in, L4 = 3.0000in, L5 = 0.0000in, ß1 = 1.48674 rad = 85.184o ß2 = 0.785674 rad = 45o ß3 = 0.55951 rad = 32.058o ß4 = 1.88074 rad = 107.758o ß5 = 1.57080 rad = 90.0o Ideas for Project #3 : 1. Five segments as discussed in class. 2. Make sure y’(velocity) and y’’(acceleration) diagrams are continuous. 3. Iteration is a normal process for design. This means, if you have tried several alternatives, even some of them did not work or not produced good results, you should submit them as attachment and it will help for you grade because it is part of design. 4. Segment DE • y: Linear motion • y’ = ?? ? • displacement: L2 = ß2 *y’ 5. Following function types proved working for the following segment. I listed here for your reference. However, I encourage you to explore other function types. • CD segment: return, half harmonic • BC segment: rise, polynomial • EF segment: return, return, half cycloidal ... Purchase answer to see full attachment

GradeAcers
Calculate your paper price
Pages (550 words)
Approximate price: -

Why Work with Us

Top Quality and Well-Researched Papers

We always make sure that writers follow all your instructions precisely. You can choose your academic level: high school, college/university or professional, and we will assign a writer who has a respective degree.

Professional and Experienced Academic Writers

We have a team of professional writers with experience in academic and business writing. Many are native speakers and able to perform any task for which you need help.

Free Unlimited Revisions

If you think we missed something, send your order for a free revision. You have 10 days to submit the order for review after you have received the final document. You can do this yourself after logging into your personal account or by contacting our support.

Prompt Delivery and 100% Money-Back-Guarantee

All papers are always delivered on time. In case we need more time to master your paper, we may contact you regarding the deadline extension. In case you cannot provide us with more time, a 100% refund is guaranteed.

Original & Confidential

We use several writing tools checks to ensure that all documents you receive are free from plagiarism. Our editors carefully review all quotations in the text. We also promise maximum confidentiality in all of our services.

24/7 Customer Support

Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.

Try it now!

Calculate the price of your order

Total price:
$0.00

How it works?

Follow these simple steps to get your paper done

Place your order

Fill in the order form and provide all details of your assignment.

Proceed with the payment

Choose the payment system that suits you most.

Receive the final file

Once your paper is ready, we will email it to you.

Our Services

No need to work on your paper at night. Sleep tight, we will cover your back. We offer all kinds of writing services.

Essays

Essay Writing Service

No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system.

Admissions

Admission Essays & Business Writing Help

An admission essay is an essay or other written statement by a candidate, often a potential student enrolling in a college, university, or graduate school. You can be rest assurred that through our service we will write the best admission essay for you.

Reviews

Editing Support

Our academic writers and editors make the necessary changes to your paper so that it is polished. We also format your document by correctly quoting the sources and creating reference lists in the formats APA, Harvard, MLA, Chicago / Turabian.

Reviews

Revision Support

If you think your paper could be improved, you can request a review. In this case, your paper will be checked by the writer or assigned to an editor. You can use this option as many times as you see fit. This is free because we want you to be completely satisfied with the service offered.

Order your essay today and save 15% with the discount code DISCOUNT15