3. ANALYTICAL KINEMATICS
|
|
- Tyrone Hudson
- 6 years ago
- Views:
Transcription
1 In planar mechanisms, kinematic analysis can be performed either analytically or graphically In this course we first discuss analytical kinematic analysis nalytical kinematics is based on projecting the vector loop equation(s) of a mechanism onto the axes of a non-moving Cartesian frame This projection transforms a vector equation into two algebraic equations Then, for a given value of the position (or orientation) of the input link, the algebraic equations are solved for the position/orientation of the remaining links The first and second time derivative of the algebraic position equations provide the velocity and acceleration equations for the mechanism For given values of the velocity and acceleration of the input link, these equations are solved to find the velocity and acceleration of the other links in the system nalytical kinematics is a systematic process that is most suitable for developing into a computer program However, for very simple systems, analytical kinemtics can be performed by hand calculation s it will be seen in the upcoming examples, even simple mechanisms can become a challenge for analysis without the use of a computer program s a reminder, by definition, a mechanism is a collection of links that are interconnected by kinematic joints forming a single degree-of-freedom system Therefore, in a kinematic analysis, the position, velocity, and acceleration of the input link must be given or assumed (one coordinate, one velocity and one acceleration) The task is then to compute the other coordinates, velocities, and accelerations Slider-crank (inversion 1) In a slider-crank mechanism, depending on its application, either the crank is the input link and the objective is to determine the kinematics of the connecting R R rod and the slider, or the slider is the input link and the O objective is to determine the kinematics of the connecting θ rod and the crank In this example, we assume the first O R O case: For known values of θ,, and we want to determine the kinematics of the other links We start the analysis by defining vectors and constructing the vector loop equation: R O The constant lengths are: =, R = L We place the x-y frame at a convenient location We define an angle (orientation) for each vector according to our convention (CCW with respect to the positive x-axis) The vector loop equation is projected onto the x and y axes to obtain two algebraic equations R O cosθ 1 R O sinθ 1 Since θ 1, we have: R O (sc1p1) For known values of and L, and given value for θ, these equations can be solved and R O : = ( / L ) = sin 1 R O = PE Nikravesh -1
2 Velocity equations The time derivative of the position equations yields the velocity equations: L R O (sc1v1) These equations can also be represented in matrix form, where the terms associated with the known crank velocity are moved to the right-hand-side: L 1 L 0 R O = (sc1v) Solution of these equations provides values of and R O cceleration equations The time derivative of the velocity equations yields the acceleration equations: L α L R O (sc1a1) α L These equations can also be represented in matrix form, where the terms associated with the known crank acceleration and the quadratic velocity terms are moved to the right-hand-side: L 1 α L 0 R O = ( + ) ( (sc1a) ) Solution of these equations provides values of α and R O Kinematic analysis For the slider-crank mechanism consider the following constant lengths: 1 and L 6 (SI units) For θ = 65 o, = 16 rad/sec, and, solve the position, velocity and acceleration equations for the unknowns For θ = 65 o, we need to solve the position equations for and R O Substituting the known values in equations (sc1p1), we have 01 cos(65) + 06 R O (a) 01sin(65) + 06 The second row of the equation that simplifies to = sin(65) = sin 1 ( 0418) = 47 o (57 o ) or 047 o There are two solutions for Substituting any of these values in the first equation of (a) yields the position of the slider: R O 1 cos(65) + 06 cos(57) 87 for = 57 o R O 1 cos(65) + 06 cos(047) = 0185 for = 047 o 5 o 04 o 65 o 65 o PE Nikravesh -
3 The two solutions are shown in the diagram We select the solution that fits our application here we select the first solution and continue with the rest of the kinematic analysis Velocity analysis For θ = 65 o, = 57 o, R O 87, and = 16 rad/sec, the velocity equations in (sc1v) become R O = Solving these two equations in two unknowns yields = 044 rad/sec, R O = 011 cceleration analysis Substituting all the known values for the coordinates and velocities in (sc1a) provides the acceleration equations as α 06 0 R O = Solving these equations yields α = 115 rad/sec, R O = 005 Observations The analytical process for the kinematics of the slider-crank mechanism reveals the following observations: mechanism with a single kinematic loop yields one vector-loop equation vector loop equation can be represented as two algebraic position equations are non-linear in the coordinates (angles and distances) Non-linear equations are difficult and time consuming to solve by hand Numerical methods, such as Newton-Raphson, are recommended for solving non-linear algebraic equations The time derivative of position equations yields velocity equations Velocity equations are linear in the velocities The time derivative of velocity equations yields acceleration equations cceleration equations are linear in the accelerations The coefficient matrix of the velocities in the velocity equations and the coefficient matrix of the accelerations in the acceleration equations are identical This characteristic can be used to simplify the solution process of these equations Four-bar In a four-bar mechanism, generally for a known y R angle, velocity and acceleration of the input link, we attempt to find the angles, velocities and accelerations of the other two links R O R O 4 The vector loop equation for this four-bar is constructed as R O4 R O4 O θ R O4 O The length of the links are x R O4 O = L 1, =, R = L, R O4 = L O 4 O 4 We place the x-y frame at a convenient location as shown We define an angle (orientation) for each link according to our convention (CCW with respect to the positive x-axis) θ 4 PE Nikravesh -
4 The vector loop equation is projected onto the x- and y-axes to obtain two algebraic equations: R O4 R O4 O cosθ 1 (fb-p1) R O4 R O4 O sinθ 1 Since θ 1 and the link lengths are known constants, the equations are simplified to: L 1 (fb-p) Velocity equations The time derivative of the position equations yields: L + (fbv1) ssuming the angular velocity of the crank,, is known, we re-arrange and express these equations in matrix form as L L cosθ 4 = (fbv) cceleration equations The time derivative of the velocity equations yields the acceleration equations: L α L + L O4 α 4 + L O4 4 = L O + L O L α L L O4 α 4 + L O4 4 = L O + L O (fba1) ssuming that is known, we re-arrange the equations as L L O4 L L O4 cosθ α 4 α 4 = L (sinθ O + ) + L L O4 L O ( ) + L L O4 (fba) Kinematic analysis Let us consider the following constant lengths: L 1 = 5, =, L = 6, = 4 For θ = 10, = 10 rad/sec, CCW, and = 10 rad/sec, determine the other two angles, angular velocities, and angular accelerations For θ = 10 we solve the position equations for and θ 4 Substituting the known lengths and the input angle in (fbp), we get cos(10 o ) sin(10 o ) We have two nonlinear equations in two unknowns We will consider a numerical method (Newton-Raphson) for solving these equations, as will be seen next t this point let us consider the solution to be 84 = 198 θ 4 = = 964 Velocity analysis With known values for the angles and the given input velocity, the velocity equation of (fbv) becomes: PE Nikravesh -4
5 = The solution to these linear equations yields: 195, 514 rad/sec oth velocities are positive, which means both are CCW cceleration analysis For known values for the angles and velocities, and the given input acceleration, the acceleration equations become: α α 4 = The solution yields: α 0041, α 4 = rad/sec One acceleration is positive; ie, CCW, and one is negative; ie, CW Newton-Raphson Method Newton-Raphson is a numerical method for solving non-linear algebraic equations The method is based on linearizing nonlinear equation(s) using Taylor series, then solving the approximated linear equation(s) iteratively One Equation in One Unknown Consider the nonlinear equation f (x) which contains one unknown x The approximated linearized equation is written as f (x) + df dx Δx 0 The Newton-Raphson iterative formula is expressed as df Δx = f (x)/ dx (N-R1) The process requires an initial estimate for the solution This value is used in (N-R1) to compute Δx Then the computed value for Δx is used to update x as x + Δx x (N-R) The process is repeated until a solution is found; ie, until f (x) Note: In iterative procedures such as N-R, if f (x) ε, where ε is a small positive number, we must accept that a solution has been found Example Find the root(s) of x x 10x + 4 using Newton-Raphson process Solution We re-state the equation as f = x x 10x + 4 The derivative of this function with respect to the unknown is df / dx = x 6x 10 To start the N-R process, we assume the solution is at x = 10 The following table shows the results from the iterative N-R process: Iteration # x f df/dx Δx x + Δx x PE Nikravesh -5
6 The process converges to x = 40 as the answer We now consider a different initial estimate for the solution Instead of x = 10 we repeat the process from x = 5 Iteration # x f df/dx Δx x + Δx x x x10-8 We now know that x = 0 is another solution to this problem Obviously there should be a third solution since we are dealing with a quadratic function The following figure should clarify what the solutions are Two Equations in Two Unknowns Consider the following two non-linear equations in x and y: f 1 (x, y) f (x, y) The approximated linearized equations are written as f 1 (x, y) + f 1 x Δx + f 1 y Δy 0 f (x, y) + f x Δx + f y Δy 0 The Newton-Raphson iterative formula is expressed as f 1 x f 1 y 1 Δx Δy = f 1 (x, y) (N-R) f f f (x, y) x y The process requires an initial estimate for the unknowns x and y These value are used in (N-R) to compute Δx and Δy Then the computed values are used to update the approximated solution: x + Δx x y + Δy y (N-R4) PE Nikravesh -6
7 The process is repeated until a solution is found Rather than checking whether each function meets the condition f ε, we consider f 1 + f ε for terminating the process Example (four-bar) We apply the Newton-Raphson process to solve the position equations for a four-bar mechanism The position equations from Example 1 are expressed as: f 1 = cos(10 o ) (a) f = sin(10 o ) Then N-R formula for these equations becomes: 1 f 1 f 1 Δ Δθ 4 = θ 4 f 1 f f f = 6sinθ 1 4 f 1 6 4cosθ (b) 4 f θ 4 From a rough sketch of the mechanism for the given input angle, we estimate the values for the two unknowns to be: 0 o 56 rad, θ 4 90 o = rad We start the Newton-Raphson process by evaluating the two functions in (a): f 1 = cos(10 o ) + 6 cos(0 o ) 4 cos(90 o ) 5 = 0808 f = sin(10 o ) + 6sin(0 o ) 4sin(90 o ) 71 These values show that our estimates are far from zeros We evaluate (b): 1 Δ = Δθ = Note that the corrections for the two angles are in radians not in degrees (this is always true) Therefore the estimated values of the two angles are corrected as and θ = The two equations in (a) are re-evaluated: f 1 = 0055, f = 0009 Since these values are not zeros, the process is continued fter two more iterations the process yields: 84 = 198, θ 4 = = 964 With these values, f 1 and f are small enough to be considered zeros The Newton-Raphson process can be extended to n equations in n unknowns The formulas are similar to those for two equations It should be obvious that the N-R process is not suitable for hand calculation The method is suitable for implementation in a computer program Secondary Computations In addition to solving the kinematic equations for the coordinates, velocities and accelerations, we may need to determine the kinematics of a point that is defined on one of the links of the mechanism Determining the kinematics of a point on a link is a secondary process and it does not require solving any set of algebraic equations we only need to evaluate one or more expressions PE Nikravesh -7
8 Four-bar coupler point P ssume that the coupler of a four-bar is in the shape of a triangle, and the location of the coupler y L P β point P relative to and is defined by the angle β L and the length L P (two constants) This coupler point can be positioned with respect to the origin of the x-y frame as θ R PO = + R P L 1 Coupler point expressions O x O 4 lgebraically, the above equation becomes: x P = + L P cos( ) (fbcp1) y P = + L P sin( ) The time derivative of the position expressions provides the velocity of point P: x P V P(x) = L P sin( ) (fbcv1) y P V P(y) = + L P cos( ) Similarly, the time derivative of the velocity expressions yields the acceleration of point P: x P P(x) = ( + ) L P (sin( )α + cos( ) ) (fbca1) y P P(y) = ( ) + L P (cos( )α sin( ) ) Example (four-bar) We continue with the data for the four-bar example ssume the coupler point is positioned at β = 5 o, L P = 55 Substituting the known values for the angles, angular velocities, and angular accelerations yields the coordinate, velocity, and acceleration of the coupler point: R PO = x P y P = 5 cos(10 o ) + 55 cos( ) = 95 5sin(10 o ) + 55sin( ) = V P = x P y P = , P = x P y P = θ 4 Matlab Programs Two Matlab programs (fourbarm and fourbar_animm) are provided for kinematic analysis of a four-bar mechanism containing a coupler point The program fourbarm performs position, velocity, and acceleration analysis for a given angle of the crank The program solves for the unknown coordinates, velocities, and accelerations, and reports the results in numerical form The program fourbar_animm only performs position analysis However, it repeatedly increments the crank angle and reports the results in the form of an animation oth programs obtain the data for the four-bar from the file fourbar_data fourbar_datam The user is required to provide in this file the following data for the four-bar of interest: Constant values for the link lengths ( L 1,, L, ) Initial angle of the crank (θ ) Estimates for the initial angles of the coupler and the follower (, θ 4 ) PE Nikravesh -8
9 Constant values for the coupler point position ( L P, β ) ngular velocity and acceleration of the crank (, ) nimation increment for the crank angle ( Δθ ) Limits for the plot axes [x_min x_max y_min y_max] Needed by fourbarm Needed by fourbar_animm The program fourbar_animm requires an accompanying M-file named fourbar_plotm that must reside in the same directory The program first checks and reports whether the four-bar is Grashof or not Then it computes the unknown angles, using the N-R process, and the coordinates of the coupler point If a solution is found the results are depicted graphically fter the four-bar is displayed in its initial state, if any keys is pressed the program will increment the crank angle and solves for the new angles and coordinates The solution and animation will be continued for two complete cycles of the crank rotation For a non-grashof four-bar, or if the four-bar is Grashof but the input link is not able to rotate completely, the input link is rotated between its limits Other Mechanisms Kinematic analysis of other mechanisms is similar to, and in some cases simpler than, that of a four-bar The followings are examples of position, velocity, and acceleration equations, and the underlying objectives of kinematic analysis for some commonly used mechanisms Slider-crank (inversion 1 - offset) Vector loop equation R O + R O Q R O R Constant: =, R = L, R O Q = a (offset) O R O Q R Q cosθ 1 + R O Q cos(90o ) Q R Q R Q sinθ 1 + R O Q sin(90 o ) Since θ 1, we have: R Q (sc1-op1) + a For a given crank angle θ, solve the position equations for and R Q Velocity and acceleration equations The equations are identical to those of the non-offset system Q: What are the angular velocity and acceleration of the slider-block? Slider-crank (inversion ) Vector loop equation R O4 R O4 O Constant: =, R O4 O = L 1 R O4 L 1 R O4 (scp1) R O O R O4O R O 4 θ 4 O 4 PE Nikravesh -9
10 For a given crank angle θ, solve the position equations for θ 4 and R O4 Velocity equations (in expanded and matrix forms) + R O4 R O4 R O4 R O4 (scv1) R O4 1 R O4 0 R O4 = (scv) Velocity analysis For a given crank velocity, solve the velocity equations for and R O4 cceleration equations (in expanded and matrix forms) + R O4 α 4 + R O4 4 + R O4 R O4 (sca1) R O4 α 4 + R O4 4 R O4 R O4 R O4 1 α 4 R O4 0 R O4 = ( + ) R O4 4 R O4 ( α ) R O4 (sca) 4 + R O4 cceleration analysis For a given crank acceleration, solve the acceleration equations for α 4 and R O4 Q: What are the angular velocity and acceleration of the slider-block? Slider-crank (inversion - offset) Vector loop equation R R O4 R O4 O Constants: =, R O4 O = L 1, R = a (offset) R R O 4 θ R 4 O a cos(θ o ) R O4 L 1 (sc-op1) asin(θ o O R ) R O4 O4O O 4 For a given crank angle θ, solve the position equations for θ 4 and R O4 Velocity equations (in expanded and matrix forms) + asin(θ o ) + R O4 R O4 a cos(θ o ) R O4 R O4 (sc-ov1) ( asin(θ 4 ) ω 4 ( a cos(θ o ) R O4 ) R O4 = (sc-ov) Velocity analysis For a given crank velocity, solve the velocity equations for and R O4 cceleration equations (in expanded and matrix forms) + ( asin(θ 4 )α 4 + ( a cos(θ 4 ) 4 + R O4 R O4 ( a cos(θ 4 )α 4 + ( asin(θ 4 ) 4 R O4 R O4 (sc-oa1) PE Nikravesh -10
11 ( asin(θ 4 ) α 4 ( a cos(θ o ) R O4 ) R O4 = (sc-oa) ( + ) ( a cos(θ 4 ) 4 R O4 ( ) ( asin(θ 4 ) 4 + R O4 cceleration analysis For a given crank acceleration, solve the acceleration equations for α 4 and R O4 Slider-crank (inversion ) Vector loop equation + R R O4 O 4 O Constant: =, R O4 O = L 1 + R cosθ L O4 1 + R O4 (scp1) For a given crank angle θ, solve the position equations for and R O4 Velocity equations (in expanded and matrix forms) R sinθ O4 + R cosθ O4 + R cosθ O4 + R sinθ (scv1) O4 R O O 4 O R O4 R O4O R sinθ O4 R cosθ O4 sinθ R O4 = (scv) Velocity analysis For a given crank velocity, solve the velocity equations for and R O4 cceleration equations (in expanded and matrix forms) R O4 α R O4 R O4 + R O4 + R cosθ O4 α R sinθ O4 + R cosθ O4 + R sinθ O4 (sca1) R O4 α R O4 sinθ R O4 = ( + ) + R cosθ O4 + R sinθ O4 ( ) + R O4 (sca) R O4 cceleration analysis For a given crank acceleration, solve the acceleration equations for α and R O4 Q: What are the angular velocity and acceleration of the slider-block? Slider-crank (inversion - offset) Vector loop equation R O4 R O4 O Constant: =, R O4 O = L 1, R O4 = a (offset) a cos( + 90 o ) L 1 asin( + 90 o ) (sc-op1) O R O R R O4O O 4 R O4 PE Nikravesh -11
12 For a given crank angle θ, solve the position equations for and R Velocity equations (in expanded and matrix forms) R + R + asin( + 90 o ) (sc-ov1) + R a cos( + 90 o ) ( R + asin( + 90 o )) ω ( R a cos( + 90 o )) R = (sc-ov) Velocity analysis For a given crank angle, solve the position equations for and R cceleration equations (in expanded and matrix forms) ( R asin( + 90 o ))α + R ( R a cos( + 90 o )) R (sc-oa1) + ( R a cos( + 90 o ))α + R ( R asin( + 90 o )) + R ( R + asin( + 90 o )) α ( R a cos( + 90 o )) R = ( + ) + ( R a cos( + 90 o )) + R (sc-oa) ( ) + ( R asin( + 90 o )) R cceleration analysis For a given crank acceleration, solve the acceleration equations for α and R In any of the inversions of the slider-crank, the position equations contain one unknown length and one unknown angle These equations, compared to the position equations of a fourbar, are simpler to solve by hand However, it is highly recommended that the Matlab program fourbarm to be revised from a four-bar to any of the slider-crank inversions Six-bar Mechanism This six-bar mechanism is constructed from two four-bars in L 5 C series There are three ground L θ y 5 attachment joints at O, O 4 and O 6 L θ 6 θ 6 Position loop equations L R O4 R O4 O L θ 7 4 O ψ 7 6 R O4 + R C R CO6 R O6 O 4 θ L 1 ψ O Constant angles: O 1 4 x θ 1 = ψ 1 and θ 7 = ψ 7 Constant lengths: R O4 O = L 1, =, R = L, R O4 =, R C = L 5, R CO6 = L 6, R O6 O 4 = L 7 L 1 cosψ 1 L 1 sinψ 1 + L 5 cosθ 5 L 6 cosθ 6 L 7 cosψ 7 + L 5 sinθ 5 L 6 sinθ 6 L 7 sinψ 7 Velocity equations PE Nikravesh -1
13 L 0 0 L L 5 sinθ 5 L 6 sinθ 6 0 L 5 cosθ 5 L 6 cosθ 6 cceleration equations L 0 0 L L 5 sinθ 5 L 6 sinθ 6 0 L 5 cosθ 5 L 6 cosθ 6 ω 5 ω 6 α α 4 α 5 α 6 ω = 0 0 = ( + ) ( + ) 4 + L 5 cosθ 5 5 L 6 cosθ 6 ω L 5 sinθ 5 5 L 6 sinθ 6 ω 6 PE Nikravesh -1
7. FORCE ANALYSIS. Fundamentals F C
ME 352 ORE NLYSIS 7. ORE NLYSIS his chapter discusses some of the methodologies used to perform force analysis on mechanisms. he chapter begins with a review of some fundamentals of force analysis using
More informationLesson 4-A. = 2.2, a = 2.0, b = 0.5 = 1.0, 2 = 3.0, 3
Lesson 4-A Procedures in Computational Kinematics For computational kinematics, we consider two solution methods: Coordinate Partitioning method Appended Constraint method These methods are introduced
More informationChapter 3 Velocity Analysis
Chapter 3 Velocity nalysis The position of point with respect to 0, Fig 1 may be defined mathematically in either polar or Cartesian form. Two scalar quantities, the length R and the angle θ with respect
More informationME Machine Design I. EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, September 30th, 2009
ME - Machine Design I Fall Semester 009 Name Lab. Div. EXAM. OPEN BOOK AND CLOSED NOTES. Wednesday, September 0th, 009 Please use the blank paper provided for your solutions. Write on one side of the paper
More information5.3 GRAPHICAL VELOCITY ANALYSIS Instant Center Method
ME GRHL VELOTY NLYSS GRHL VELOTY NLYSS nstant enter Method nstant center of velocities is a simple graphical method for performing velocity analysis on mechanisms The method provides visual understanding
More informationEXAM 1. OPEN BOOK AND CLOSED NOTES Thursday, February 18th, 2010
ME 35 - Machine Design I Spring Semester 010 Name of Student Lab. Div. Number EXAM 1. OPEN BOOK AND CLOSED NOTES Thursday, February 18th, 010 Please use the blank paper provided for your solutions. Write
More informationLecture 21. MORE PLANAR KINEMATIC EXAMPLES
Lecture 21. MORE PLANAR KINEMATIC EXAMPLES 4.5c Another Slider-Crank Mechanism Figure 4.24 Alternative slider-crank mechanism. Engineering-analysis task: For = ω = constant, determine φ and S and their
More informationMECH 314 Dynamics of Mechanisms February 17, 2011 Offset Slider Crank Analysis and Some Other Slider Systems
MCH 314 ynamics of Mechanisms February 17, 011 ffset Slider Crank nalysis and Some ther Slider Systems 1 ffset Slider Crank Position nalysis Position and velocity analysis for the common slider-crank has
More informationEXAM 1. OPEN BOOK AND CLOSED NOTES.
ME 35 - Machine Design I Summer Semester 013 Name of Student Lab Section Number EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, June 6th, 013 Use the blank paper provided for your solutions. Write on one
More informationKinematics of Mechanisms 6 Acceleration
Kinematics of Mechanisms 6 Acceleration Latifah Nurahmi Latifah.nurahmi@gmail.com Room C.250 Acceleration analysis Acceleration analysis: Determine the manner in which certain points on the mechanism are
More informationSECTION A. f(x) = ln(x). Sketch the graph of y = f(x), indicating the coordinates of any points where the graph crosses the axes.
SECTION A 1. State the maximal domain and range of the function f(x) = ln(x). Sketch the graph of y = f(x), indicating the coordinates of any points where the graph crosses the axes. 2. By evaluating f(0),
More informationLecture Note 8: Inverse Kinematics
ECE5463: Introduction to Robotics Lecture Note 8: Inverse Kinematics Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 8 (ECE5463
More informationUNIT 2 KINEMATICS OF LINKAGE MECHANISMS
UNIT 2 KINEMATICS OF LINKAGE MECHANISMS ABSOLUTE AND RELATIVE VELOCITY An absolute velocity is the velocity of a point measured from a fixed point (normally the ground or anything rigidly attached to the
More informationROBOTICS: ADVANCED CONCEPTS & ANALYSIS
ROBOTICS: ADVANCED CONCEPTS & ANALYSIS MODULE 4 KINEMATICS OF PARALLEL ROBOTS Ashitava Ghosal 1 1 Department of Mechanical Engineering & Centre for Product Design and Manufacture Indian Institute of Science
More informationChapter 2: Numeric, Cell, and Structure Arrays
Chapter 2: Numeric, Cell, and Structure Arrays Topics Covered: Vectors Definition Addition Multiplication Scalar, Dot, Cross Matrices Row, Column, Square Transpose Addition Multiplication Scalar-Matrix,
More informationLecture 18. PLANAR KINEMATICS OF RIGID BODIES, GOVERNING EQUATIONS
Lecture 18. PLANAR KINEMATICS OF RIGID BODIES, GOVERNING EQUATIONS Planar kinematics of rigid bodies involve no new equations. The particle kinematics results of Chapter 2 will be used. Figure 4.1 Planar
More informationEXERCISES. Let us see an example of graphical methods for solving static analysis associated with numerical evaluation of the results. 2 1 E 3 P=?
Let us see an example of graphical methods for soling static analysis associated with numerical ealuation of the results.? ϑ D A ? ϑ D A R ( ) a ( ) R sin π ϑ R sinϑ a R R R? π ϑ R R ϑ R D R ( D) ( D)
More informationENGD3008 Dynamics, M. Goman Experiment 3 Kinematics of a Mechanism Warwick Shipway Mechanical Engineering, Year
22/3/28 ENGD38 Dynamics, M. Goman Experiment 3 Kinematics of a Mechanism Warwick Shipway Mechanical Engineering, Year 3 27.3.28 1. Introduction A slider crank mechanism is analysed and the displacement
More informationWEEKS 2-3 Dynamics of Machinery
WEEKS 2-3 Dynamics of Machinery References Theory of Machines and Mechanisms, J.J. Uicker, G.R.Pennock ve J.E. Shigley, 2003 Makine Dinamiği, Prof. Dr. Eres SÖYLEMEZ, 2013 Uygulamalı Makine Dinamiği, Jeremy
More informationLecture Note 8: Inverse Kinematics
ECE5463: Introduction to Robotics Lecture Note 8: Inverse Kinematics Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 8 (ECE5463
More informationChapter 4 Statics and dynamics of rigid bodies
Chapter 4 Statics and dynamics of rigid bodies Bachelor Program in AUTOMATION ENGINEERING Prof. Rong-yong Zhao (zhaorongyong@tongji.edu.cn) First Semester,2014-2015 Content of chapter 4 4.1 Static equilibrium
More informationPLANAR RIGID BODY MOTION: TRANSLATION &
PLANAR RIGID BODY MOTION: TRANSLATION & Today s Objectives : ROTATION Students will be able to: 1. Analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis. In-Class
More informationDynamics Plane kinematics of rigid bodies Section 4: TJW Rotation: Example 1
Section 4: TJW Rotation: Example 1 The pinion A of the hoist motor drives gear B, which is attached to the hoisting drum. The load L is lifted from its rest position and acquires an upward velocity of
More informationDifferential Kinematics
Differential Kinematics Relations between motion (velocity) in joint space and motion (linear/angular velocity) in task space (e.g., Cartesian space) Instantaneous velocity mappings can be obtained through
More informationKinematics, Dynamics, and Vibrations FE Review Session. Dr. David Herrin March 27, 2012
Kinematics, Dynamics, and Vibrations FE Review Session Dr. David Herrin March 7, 0 Example A 0 g ball is released vertically from a height of 0 m. The ball strikes a horizontal surface and bounces back.
More informationPre-Calc Unit 15: Polar Assignment Sheet May 4 th to May 31 st, 2012
Pre-Calc Unit 15: Polar Assignment Sheet May 4 th to May 31 st, 2012 Date Objective/ Topic Assignment Did it Friday Polar Discover y Activity Day 1 pp. 2-3 May 4 th Monday Polar Discover y Activity Day
More informationError Analysis of Four Bar Planar Mechanism Considering Clearance Between Crank & Coupler
Error nalysis of Four Bar Planar Mechanism Considering Clearance Between Crank & Coupler.M.VIDY 1, P.M. PDOLE 2 1 Department of Mechanical & Production Engineering, M.I.E.T.,GONDI, India. 2 Department
More informationIYGB Mathematical Methods 1
IYGB Mathematical Methods Practice Paper B Time: 3 hours Candidates may use any non programmable, non graphical calculator which does not have the capability of storing data or manipulating algebraic expressions
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA ADVANCED MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 18 NQF LEVEL 3
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA ADVANCED MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 18 NQF LEVEL 3 OUTCOME 3 BE ABLE TO DETERMINE RELATIVE AND RESULTANT VELOCITY IN ENGINEERING SYSTEMS Resultant
More informationRIGID BODY MOTION (Section 16.1)
RIGID BODY MOTION (Section 16.1) There are cases where an object cannot be treated as a particle. In these cases the size or shape of the body must be considered. Rotation of the body about its center
More information8 Velocity Kinematics
8 Velocity Kinematics Velocity analysis of a robot is divided into forward and inverse velocity kinematics. Having the time rate of joint variables and determination of the Cartesian velocity of end-effector
More informationChapter 3 HW Solution
Chapter 3 HW Solution Problem 3.6: I placed an xy coordinate system at a convenient point (origin doesn t really matter). y 173 x The positions of both planes are given by r B = v B ti + 173j mi (1) r
More informationRELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5)
RELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5) Today s Objectives: Students will be able to: a) Describe the velocity of a rigid body in terms of translation and rotation components. b) Perform a relative-motion
More informationChapter 10. Rotation
Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised
More informationLecture 38: Equations of Rigid-Body Motion
Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can
More informationHandout 6: Rotational motion and moment of inertia. Angular velocity and angular acceleration
1 Handout 6: Rotational motion and moment of inertia Angular velocity and angular acceleration In Figure 1, a particle b is rotating about an axis along a circular path with radius r. The radius sweeps
More informationUncertainties & Error Analysis Tutorial
Uncertainties & Error Analysis Tutorial Physics 118/198/1 Reporting Measurements Uncertainties & Error Analysis Tutorial When we report a measured value of some parameter, X, we write it as X X best ±
More informationReview of Engineering Dynamics
Review of Engineering Dynamics Part 1: Kinematics of Particles and Rigid Bodies by James Doane, PhD, PE Contents 1.0 Course Overview... 4.0 Basic Introductory Concepts... 4.1 Introduction... 4.1.1 Vectors
More informationME Machine Design I
ME 5 - Machine Design I Summer Semester 008 Name Lab. Div. EXAM. OEN BOOK AND CLOSED NOTES. Wednesday, July 16th, 008 Write your solutions on the blank paper that is provided. Write on one side of the
More informationHomogeneous Transformations
Purpose: Homogeneous Transformations The purpose of this chapter is to introduce you to the Homogeneous Transformation. This simple 4 x 4 transformation is used in the geometry engines of CAD systems and
More informationIn the previous chapter we used graphical methods to design fourbar linkages to reach two or
4.1 Introduction to Position Analysis In the previous chapter we used graphical methods to design fourbar linkages to reach two or three specified positions. While these methods are handy for designing
More informationPHYS 410/555 Computational Physics Solution of Non Linear Equations (a.k.a. Root Finding) (Reference Numerical Recipes, 9.0, 9.1, 9.
PHYS 410/555 Computational Physics Solution of Non Linear Equations (a.k.a. Root Finding) (Reference Numerical Recipes, 9.0, 9.1, 9.4) We will consider two cases 1. f(x) = 0 1-dimensional 2. f(x) = 0 d-dimensional
More informationME 274 Spring 2017 Examination No. 2 PROBLEM No. 2 (20 pts.) Given:
PROBLEM No. 2 (20 pts.) Given: Blocks A and B (having masses of 2m and m, respectively) are connected by an inextensible cable, with the cable being pulled over a small pulley of negligible mass. Block
More informationME Machine Design I
ME 35 - Machine Design I Summer Semester 008 Name Lab. Div. FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Wednesday, July 30th, 008 Please show all your work for your solutions on the blank white paper. Write
More informationUNIVERSITY OF KWAZULU-NATAL EXAMINATIONS: NOVEMBER 2013 COURSE, SUBJECT AND CODE: THEORY OF MACHINES ENME3TMH2 MECHANICAL ENGINEERING
UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS: NOVEMBER 2013 COURSE, SUBJECT AND CODE: THEORY OF MACHINES ENME3TMH2 DURATION: 3 Hours MARKS: 100 MECHANICAL ENGINEERING Internal Examiner: Dr. R.C. Loubser Indeendent
More informationLecture 38: Equations of Rigid-Body Motion
Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can
More informationNamee of Student. link and I R 2 ?? = 0. and Δ θ. Calculate the. in the next. iteration. 1. = 6cm and θ * 9 = 135.
ME 52 - Machine Design I Fall Semester 2010 EXAM 1. OPEN BOOK AND CLOSED NOTES. Namee of Student Lab. Div. Number Wednesday, September 29th, 2010 Use the blank paper provided for your solutions. Write
More informationFind: Determine the velocity and acceleration of end P. Write your answers as vectors.
Homework 2. Given: ar P rotates about a shaft passing through end of the bar. t the instant shown, the angular velocity and angular acceleration of P are given by ω and α. atics Homework Problems ME 274
More information2.003SC Recitation 3 Notes:V and A of a Point in a Moving Frame
2.003SC Recitation 3 Notes:V and A of a Point in a Moving Frame Velocity and Acceleration of a Point in a Moving Frame The figure below shows a point B in a coordinate system Axyz which is translating
More informationEngineering Mechanics
2019 MPROVEMENT Mechanical Engineering Engineering Mechanics Answer Key of Objective & Conventional Questions 1 System of forces, Centoriod, MOI 1. (c) 2. (b) 3. (a) 4. (c) 5. (b) 6. (c) 7. (b) 8. (b)
More informationAAE 203 Notes. Martin Corless
1 AAE 203 Notes Martin Corless 2 Contents 1 Introduction 7 2 Units and Dimensions 9 2.1 Introduction.................................... 9 2.2 Units........................................ 9 2.2.1 SI system
More informationPHYS XXXX-L. Title (of Lab) Name (your name) Date of the lab (date performed) Dr. Thomas Eaves
PHYS XXXX-L Title (of Lab) Name (your name) Date of the lab (date performed) Dr. Thomas Eaves The laboratory report is designed to answer the following questions: a. What did you try to find out? b. How
More informationMECH 576 Geometry in Mechanics November 30, 2009 Kinematics of Clavel s Delta Robot
MECH 576 Geometry in Mechanics November 3, 29 Kinematics of Clavel s Delta Robot The DELTA Robot DELTA, a three dimensional translational manipulator, appears below in Fig.. Figure : Symmetrical (Conventional)
More informationME Machine Design I. FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Friday, May 8th, 2009
ME 5 - Machine Design I Spring Seester 009 Nae Lab. Div. FINAL EXAM. OPEN BOOK AND LOSED NOTES. Friday, May 8th, 009 Please use the blank paper for your solutions. Write on one side of the paper only.
More informationChapter 15 Periodic Motion
Chapter 15 Periodic Motion Slide 1-1 Chapter 15 Periodic Motion Concepts Slide 1-2 Section 15.1: Periodic motion and energy Section Goals You will learn to Define the concepts of periodic motion, vibration,
More information2.3 Oscillation. The harmonic oscillator equation is the differential equation. d 2 y dt 2 r y (r > 0). Its solutions have the form
2. Oscillation So far, we have used differential equations to describe functions that grow or decay over time. The next most common behavior for a function is to oscillate, meaning that it increases and
More informationNONLINEAR MECHANICAL SYSTEMS (MECHANISMS)
NONLINEAR MECHANICAL SYSTEMS (MECHANISMS) The analogy between dynamic behavior in different energy domains can be useful. Closer inspection reveals that the analogy is not complete. One key distinction
More informationPhysics 170 Lecture 19. Chapter 12 - Kinematics Sections 8-10
Phys 170 Lecture 0 1 Physics 170 Lecture 19 Chapter 1 - Kinematics Sections 8-10 Velocity & Acceleration in Polar / Cylinical Coordinates Pulley Problems Phys 170 Lecture 0 Polar Coordinates Polar coordinates
More informationPURE MATHEMATICS AM 27
AM Syllabus (014): Pure Mathematics AM SYLLABUS (014) PURE MATHEMATICS AM 7 SYLLABUS 1 AM Syllabus (014): Pure Mathematics Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs)
More informationPlanar Rigid Body Kinematics Homework
Chapter 2 Planar Rigid ody Kinematics Homework Freeform c 2016 2-1 2-2 Freeform c 2016 Homework 2. Given: The pulley shown below freely rotates about point C and interacts with two rubber belts (one horizontal,
More information2- Scalars and Vectors
2- Scalars and Vectors Scalars : have magnitude only : Length, time, mass, speed and volume is example of scalar. v Vectors : have magnitude and direction. v The magnitude of is written v v Position, displacement,
More informationFig.1 Partially compliant eccentric slider crank linkage
ANALYSIS OF AN UNDERACTUATED COMPLIANT FIVE-BAR LINKAGE Deepak Behera and Dr.J.Srinivas, Department of Mechanical Engineering, NIT-Rourkela 769 008 email: srin07@yahoo.co.in Abstract: This paper presents
More informationPROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION
PROLEM 15.10 The bent rod E rotates about a line joining Points and E with a constant angular elocity of 9 rad/s. Knowing that the rotation is clockwise as iewed from E, determine the elocity and acceleration
More informationMathematical review trigonometry vectors Motion in one dimension
Mathematical review trigonometry vectors Motion in one dimension Used to describe the position of a point in space Coordinate system (frame) consists of a fixed reference point called the origin specific
More informationRobotics I. Figure 1: Initial placement of a rigid thin rod of length L in an absolute reference frame.
Robotics I September, 7 Exercise Consider the rigid body in Fig., a thin rod of length L. The rod will be rotated by an angle α around the z axis, then by an angle β around the resulting x axis, and finally
More informationChapter 8 Acceleration in Mechanisms
Chapter 8 Acceleration in Mechanisms 1 2 8.2. Acceleration Diagram for a Link Example 8.1 3 The crank of a slider crank mechanism rotates cw at a constant speed of 300 rpm. The crank is 150 mm & the ConRod
More informationFundamental principles
Dynamics and control of mechanical systems Date Day 1 (03/05) - 05/05 Day (07/05) Day 3 (09/05) Day 4 (11/05) Day 5 (14/05) Day 6 (16/05) Content Review of the basics of mechanics. Kinematics of rigid
More informationLecture 37: Principal Axes, Translations, and Eulerian Angles
Lecture 37: Principal Axes, Translations, and Eulerian Angles When Can We Find Principal Axes? We can always write down the cubic equation that one must solve to determine the principal moments But if
More informationKinematics. Professor Sanjay Sarma. September 12, 2007
Kinematics Professor Sanjay Sarma September 12, 2007 1.0 What is Dynamics? The goal of the field of dynamics is to understand how mechanical systems move under the effect of forces. There are 3 components
More informationQuiz No. 1: Tuesday Jan. 31. Assignment No. 2, due Thursday Feb 2: Problems 8.4, 8.13, 3.10, 3.28 Conceptual questions: 8.1, 3.6, 3.12, 3.
Quiz No. 1: Tuesday Jan. 31 Assignment No. 2, due Thursday Feb 2: Problems 8.4, 8.13, 3.10, 3.28 Conceptual questions: 8.1, 3.6, 3.12, 3.20 Chapter 3 Vectors and Two-Dimensional Kinematics Properties of
More informationUNIT II Fig (1) Fig (1)
UNIVERSITY OF PUNE [4362]-116 S.E.(Mechanical / Automobile) Examination 2013 (2008 pattern) THEY OF MACHINES -1 [Total No. of Questions: 12] [Total No. of Printed pages: 7] [Time: 4 Hours] [Max. Marks:
More informationSOLUTION If link AB is rotating at v AB = 6 rad>s, determine the angular velocities of links BC and CD at the instant u = 60.
16 88. If link i rotating at v = 6 rad>, determine the angular velocitie of link C and CD at the intant u = 6. 25 mm 3 3 mm ω = 6 rad/ C 4 mm r IC - =.3co 3 =.2598 m r IC - C =.3co 6 =.15 m θ D v C = 1.5
More informationRigid Body Motion. Greg Hager Simon Leonard
Rigid ody Motion Greg Hager Simon Leonard Overview Different spaces used in robotics and why we need to get from one space to the other Focus on Cartesian space Transformation between two Cartesian coordinate
More informationCHAPTER 4 ROOTS OF EQUATIONS
CHAPTER 4 ROOTS OF EQUATIONS Chapter 3 : TOPIC COVERS (ROOTS OF EQUATIONS) Definition of Root of Equations Bracketing Method Graphical Method Bisection Method False Position Method Open Method One-Point
More informationA repeated root is a root that occurs more than once in a polynomial function.
Unit 2A, Lesson 3.3 Finding Zeros Synthetic division, along with your knowledge of end behavior and turning points, can be used to identify the x-intercepts of a polynomial function. This information allows
More information11 a 12 a 13 a 21 a 22 a b 12 b 13 b 21 b 22 b b 11 a 12 + b 12 a 13 + b 13 a 21 + b 21 a 22 + b 22 a 23 + b 23
Chapter 2 (3 3) Matrices The methods used described in the previous chapter for solving sets of linear equations are equally applicable to 3 3 matrices. The algebra becomes more drawn out for larger matrices,
More informationMath Review 1: Vectors
Math Review 1: Vectors Coordinate System Coordinate system: used to describe the position of a point in space and consists of 1. An origin as the reference point 2. A set of coordinate axes with scales
More information5.2 GRAPHICAL VELOCITY ANALYSIS Polygon Method
ME 352 GRHICL VELCITY NLYSIS 52 GRHICL VELCITY NLYSIS olygon Mehod Velociy analyi form he hear of kinemaic and dynamic of mechanical yem Velociy analyi i uually performed following a poiion analyi; ie,
More informationRECURSIVE INVERSE DYNAMICS
We assume at the outset that the manipulator under study is of the serial type with n+1 links including the base link and n joints of either the revolute or the prismatic type. The underlying algorithm
More informationSolutions to Chapter 1 Exercise Problems
Solutions to hapter Exercise Problems Problem. Find a mechanism as an isolated device or in a machine and make a realistic sketch of the mechanism. Then make a freehand sketch of the kinematic schematics
More information(r i F i ) F i = 0. C O = i=1
Notes on Side #3 ThemomentaboutapointObyaforceF that acts at a point P is defined by M O (r P r O F, where r P r O is the vector pointing from point O to point P. If forces F, F, F 3,..., F N act on particles
More informationINSTRUCTIONS TO CANDIDATES:
NATIONAL NIVERSITY OF SINGAPORE FINAL EXAMINATION FOR THE DEGREE OF B.ENG ME 444 - DYNAMICS AND CONTROL OF ROBOTIC SYSTEMS October/November 994 - Time Allowed: 3 Hours INSTRCTIONS TO CANDIDATES:. This
More informationCOSMOS: Making Robots and Making Robots Intelligent Lecture 3: Introduction to discrete-time dynamics
COSMOS: Making Robots and Making Robots Intelligent Lecture 3: Introduction to discrete-time dynamics Jorge Cortés and William B. Dunbar June 3, 25 Abstract In this and the coming lecture, we will introduce
More informationUnit Circle: The unit circle has radius 1 unit and is centred at the origin on the Cartesian plane. POA
The Unit Circle Unit Circle: The unit circle has radius 1 unit and is centred at the origin on the Cartesian plane THE EQUATION OF THE UNIT CIRCLE Consider any point P on the unit circle with coordinates
More informationLecture Notes Multibody Dynamics B, wb1413
Lecture Notes Multibody Dynamics B, wb1413 A. L. Schwab & Guido M.J. Delhaes Laboratory for Engineering Mechanics Mechanical Engineering Delft University of Technolgy The Netherlands June 9, 29 Contents
More informationANALYSIS OF LOAD PATTERNS IN RUBBER COMPONENTS FOR VEHICLES
ANALYSIS OF LOAD PATTERNS IN RUBBER COMPONENTS FOR VEHICLES Jerome Merel formerly at Hutchinson Corporate Research Center Israël Wander Apex Technologies Pierangelo Masarati, Marco Morandini Dipartimento
More informationExam 1 September 11, 2013
Exam 1 Instructions: You have 60 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use an approved calculator during the exam. Usage of mobile phones and other
More informationRobotics: Tutorial 3
Robotics: Tutorial 3 Mechatronics Engineering Dr. Islam Khalil, MSc. Omar Mahmoud, Eng. Lobna Tarek and Eng. Abdelrahman Ezz German University in Cairo Faculty of Engineering and Material Science October
More information2D Plotting with Matlab
GEEN 1300 Introduction to Engineering Computing Class Meeting #22 Monday, Nov. 9 th Engineering Computing and Problem Solving with Matlab 2-D plotting with Matlab Script files User-defined functions Matlab
More informationDerivation of Kinematics for the Fluid Film Between Thrust Collar and Bull Gear. Original work by Dr. Karl Wygant [1] Reproduce by Travis Cable
Derivation of Kinematics for the Fluid Film Between Thrust Collar and Bull Gear Original work by Dr. Karl Wygant [1] Reproduce by Travis Cable Depiction of Area Under Investigation ω P pinion gear ω b
More informationAppendix. Vectors, Systems of Equations
ppendix Vectors, Systems of Equations Vectors, Systems of Equations.1.1 Vectors Scalar physical quantities (e.g., time, mass, density) possess only magnitude. Vectors are physical quantities (e.g., force,
More informationSinusoids. Amplitude and Magnitude. Phase and Period. CMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation
Sinusoids CMPT 889: Lecture Sinusoids, Complex Exponentials, Spectrum Representation Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University September 6, 005 Sinusoids are
More informationCMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation
CMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University September 26, 2005 1 Sinusoids Sinusoids
More informationControl Volume. Dynamics and Kinematics. Basic Conservation Laws. Lecture 1: Introduction and Review 1/24/2017
Lecture 1: Introduction and Review Dynamics and Kinematics Kinematics: The term kinematics means motion. Kinematics is the study of motion without regard for the cause. Dynamics: On the other hand, dynamics
More informationLecture 1: Introduction and Review
Lecture 1: Introduction and Review Review of fundamental mathematical tools Fundamental and apparent forces Dynamics and Kinematics Kinematics: The term kinematics means motion. Kinematics is the study
More informationMoving Reference Frame Kinematics Homework
Chapter 3 Moving Reference Frame Kinematics Homework Freeform c 2016 3-1 3-2 Freeform c 2016 Homework 3. Given: n L-shaped telescoping arm is pinned to ground at point. The arm is rotating counterclockwise
More informationRotational Kinematics
Rotational Kinematics Rotational Coordinates Ridged objects require six numbers to describe their position and orientation: 3 coordinates 3 axes of rotation Rotational Coordinates Use an angle θ to describe
More information1/3/2011. This course discusses the physical laws that govern atmosphere/ocean motions.
Lecture 1: Introduction and Review Dynamics and Kinematics Kinematics: The term kinematics means motion. Kinematics is the study of motion without regard for the cause. Dynamics: On the other hand, dynamics
More information10.1 Curves Defined by Parametric Equation
10.1 Curves Defined by Parametric Equation 1. Imagine that a particle moves along the curve C shown below. It is impossible to describe C by an equation of the form y = f (x) because C fails the Vertical
More informationPractice Problems for MTH 112 Exam 2 Prof. Townsend Fall 2013
Practice Problems for MTH 11 Exam Prof. Townsend Fall 013 The problem list is similar to problems found on the indicated pages. means I checked my work on my TI-Nspire software Pages 04-05 Combine the
More information