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Words 2564

Pages 11

Hiding Messages Using Motion Vector

Technique In Video Steganography

P.Paulpandi1, Dr.T.Meyyappan,M.sc.,M.Phil.,M.BA.,Ph.D2

Research Scholar1, Associate professor2

Department of Computer Science & Engineering,

Alagappa University,Karaikudi.

Tamil Nadu,India.

Abstract- Steganography is the art of hiding information in ways that avert the revealing of hiding messages.Video files are generally a collection of images. so most of the presented techniques on images and audio can be applied to video files too. The great advantages of video are the large amount of data that can be hidden inside and the fact that it is a moving stream of image. In this paper, we proposed a new technique using the motion vector, to hide the data in the moving objects. Moreover, to enhance the security of the data, the data is encrypted by using the AES algorithm and then hided. The data is hided in the horizontal and the vertical components of the moving objects. The PSNR value is calculated so that the quality of the video after the data hiding is evaluated.

Keywords- Data hiding, Video Steganography,PSNR,

Moving objects, AES Algorithm.

I. INTRODUCTION

Since the rise of the Internet one of the most important factors of information technology and communication has been the security of information.

Steganography is a technology that hides a user defined information within an object, a text, or a picture or in a video files. Steganography is Greek word has the meaning of , “Stegano”, or “covered” and “graphy” or “writing” which does not convey the transformation of information, but rather its hidden aspect.

In

steganography, the object of communication is the hidden message and the cover data are only the means of sending it. Secret information as well as cover data…...

...____________________________________ Motion in 2D Simulation Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now. 1) Once the simulation opens, click on ‘Show Both’ for Velocity and Acceleration at the top of the page. Now click and drag the red ball around the screen. Make 3 observations about the blue and green arrows (also called vectors) as you drag the ball around. When Velocity ends(stop dragging), acceleration reduces. Acceleration decreases right before velocity peaks. The vector for acceleration always appears to be at a right angle. 2) Which color vector (arrow) represents velocity and which one represents acceleration? How can you tell? Velocity is green and acceleration is blue. When velocity is constant or close to 0 acceleration declines 3) Try dragging the ball around and around in a circular path. What do you notice about the lengths and directions of the blue and green vectors? Describe their behavior in detail below. Both are equal when moving around in a circular path. Velocity seems to increase and decrease very little and the acceleration remains the same 4) Now move the ball at a slow constant speed across the screen. What do you notice now about the vectors? Explain why this happens. When moving slow and steady, there is barely any acceleration. Velocity is the measurement of speed/direction, and acceleration is the change of. 5) What happens to the vectors when......

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...Tobi Gonzalez 2/25/13 Motion is Relative 1. The two units of measurement are distance and time. 2. The kind of speed is registered by an automobile speedometer is instantaneous speed. 3. Average speed is the speed averaged over duration of time. Usually the total distance covered divided by the total time. Instantaneous speed is the speed given at that instant within that span of time, measured with a real time speedometer. 4. The average speed in kilometers per hour for a horse that gallops a distance of 15 km in a time of 30 min is: Average speed = 15 km/30 min = 15 km/0.5 h = 30 km/h 5. Speed is the distance covered per unit of time. Acceleration is the rate in which an object changes its velocity. 6. If a car is moving at 90 km/h and it rounds a corner, also at 90 km/h, it does maintain a constant speed but not a constant velocity. The velocity never changed, only the direction it’s traveling. 7. Velocity is change in displacement, change in position over a period of time, while Acceleration is change in velocity over a time period. 8. The acceleration of a car that increases its velocity from 0, to 100 km/h in 10s is 10km/h*s 9. The acceleration of a car that maintains a constant velocity of 100 km/h for 10s is 0 km/h*s. Some of my classmates get this question wrong but the last question right because they fail to read the question. In the last question there was a change in velocity. However in this question there was no change in...

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...Calculus and Vectors – How to get an A+ 7.4 Dot Product of Algebraic Vectors A Dot Product for Standard Unit Vectors The dot product of the standard unit vectors is given by: r r r r r r i ⋅ i =1 j ⋅ j =1 k ⋅ k =1 r r r r r r i ⋅ j =0 j ⋅k =0 k ⋅i = 0 B Dot Product for two Algebraic Vectors The dot product of two algebraic vectors r r r r a = (a x , a y , a z ) = a x i + a y j + a z k and r r r r b = (b x , b y , b z ) = b x i + b y j + b z k is given by: r r a ⋅ b = a x bx + a y b y + a z bz Proof: r r r r r r r r a ⋅ b = (a x i + a y j + a z k ) ⋅ (bx i + b y j + bz k ) r r r r r r = (a x bx )(i ⋅ i ) + (a x b y (i ⋅ j ) + (a x bz )(i ⋅ k ) + r r r r r r + (a y bx )( j ⋅ i ) + (a y b y ( j ⋅ j ) + (a y bz )( j ⋅ k ) + r r r r r r + (a z bx )(k ⋅ i ) + (a z b y (k ⋅ j ) + (a z bz )(k ⋅ k ) = a x bx + a y b y + a z bz Proof: r r r r i ⋅ i =|| i || || i || cos 0° = (1)(1)(1) = 1 r r r r i ⋅ j =|| i || || j || cos 90° = (1)(1)(0) = 0 r Ex 1. For each case, find the dot product of the vectors a r and b . r r a) a = (1,−2,0) , b = (0,−1,2) r r a ⋅ b = (1)(0) + (−2)(−1) + (0)(2) = 2 r r r r r r r b) a = −i + 2 j , b = i − 2 j − k r r a ⋅ b = (−1)(1) + (2)(−2) + (0)(−1) = −1 − 4 = −5 r r r r r c) a = (−1,1,−1) , b = −i + 2 j − 2k r r a ⋅ b = (−1)(−1) + (1)(2) + (−1)(−2) = 1 + 2 + 2 = 5 C Angle between two Vectors r r r The angle θ = ∠(a , b ) between two vectors a and r b (when positioned tail to tail) is given by: r r a x bx + a y b y + a z bz a ⋅b cosθ = r r = | a......

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... #2 The Study of Concurrent Forces with the Force Table Apparatus: Force table with 4 pulleys, centering ring and string, 50 g weight hangers, slotted weights, protractors, and rulers. Discussion: The force table is designed to help you study the properties of forces at known angles. Only when forces are along the same line do they add by ordinary algebra. If two or more forces on the same body form angles with each other, it is necessary to use geometry to find the amount and direction of their combined effect. Prior to Lab: Complete the calculations in the following. The component method of adding vectors is given here for three sample forces as follows: A = 2.45 N @ 40o B = 3.92 N @ 165o C = 3.43 N @ 330o Overview of the component method of vector addition. First make a neat drawing, not necessarily to exact scale, but reasonably accurate as to sizes and angles, placing the three forces on a diagram with a pair of x and y axes. Find the angle of each force with the x-axis. This angle is called the reference angle, and is the one used to calculate sines and cosines. Next compute the x- and y-components of the three forces, placing like components in columns. Place plus or minus signs on the various quantities according to whether an x-component is to the right or left of the origin, or whether a y-component is up or down relative to the origin. Add the columns with regard to sign (subtracting the minus quantities), and place the correct sign on each sum. The resulting...

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...Motions: * Politics This House would grant politicians immunity from prosecution PROS Giving politicians’ immunity from prosecution allows them to focus on performing their duties The premier reason that most states, even those that allow for the prosecution of politicians, abstain from prosecuting them while they hold office is that being a politician is a job that requires one’s undivided attention. Especially for the holders of prominent national-level offices, writing legislation, responding to crises under one’s purview, consulting one’s constituents, and engaging in campaign work often lead to politicians working an upwards of 12 hour day, every day. To expect politicians cope with all of these concerns will simultaneously constructing a defense against pending charges would be to abandon all hope of them serving their constituents effectively. We are rightly aggravated when politicians take extensive vacations or other extracurricular forays. Being under indictment not only consumes even more of a politician’s time; the stress it causes will inevitably seep into what remaining time they do allocating to fulfilling their duties, further hindering their performance. The impeachment proceedings for Bill Clinton on charges of perjury and obstruction of justice were so intensive that they took tremendous resources away from not only the president himself, but all branches of the federal government for several months, amidst serious domestic and foreign...

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...TITLE To investigate the trajectory of a small ball as it rolls off a surface which is inclined to the horizontal. OBJECTIVE To investigate the trajectory of a two dimensional motion APPARATUS & MATERIALS * Ramp * Wooden block * Pendulum bob * Plumb line * Steel ball * Wooden board * Carbon paper * Meter rule * Plasticine SETUP 1. A ramp has been set up at the edge of a bench as shown in the Figure 4-1. 2. Suspend a plum-line from the edge of the bench as shown in Figure 4-2. 3. Mount a wooden board horizontally using two clamps so that the board is situated about the bottom of the ramp. 4. Place a sheet of blank paper on top of the board. 5. Place a piece of carbon paper on the top of the blank paper. The ink-side of the carbon paper should be facing down. 6. When a ball is released at the top of the ramp, the ball will travel through a trajectory as shown in Figure 4-2. THEORY Let: g = 9.8 ms-2 u = speed of the ball as it leaves the ramp k = constant y = vertical distance (between the bottom of the ramp and the top of the board) x = horizontal distance (between the plum-line and mark on the paper) The equation which relates to x and y is yx=g1+k2x2u2+k PROCEDURE 1. Position the ball at the top of the ramp. Release the ball so that it rolls down the ramp and onto the board below. 2. Remove the carbon paper and observe that the ball makes a small mark on...

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...Motion Name ____Anjad Itayem______ Motion in 2D Simulation Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now. 1) Once the simulation opens, click on ‘Show Both’ for Velocity and Acceleration at the top of the page. Now click and drag the red ball around the screen. Make 3 observations about the blue and green arrows (also called vectors) as you drag the ball around. When Velocity ends(stop dragging), acceleration reduces. Acceleration decreases right before velocity peaks. The vector for acceleration always appears to be at a right angle. 2) Which color vector (arrow) represents velocity and which one represents acceleration? How can you tell? Velocity is green and acceleration is blue. When velocity is constant or close to 0 acceleration declines 3) Try dragging the ball around and around in a circular path. What do you notice about the lengths and directions of the blue and green vectors? Describe their behavior in detail below. Both are equal when moving around in a circular path. Velocity seems to increase and decrease very little and the acceleration remains the same 4) Now move the ball at a slow constant speed across the screen. What do you notice now about the vectors? Explain why this happens. When moving slow and steady, there is barely any acceleration. Velocity is the measurement of speed/direction, and acceleration is the change of. 5) What happens to the vectors when you jerk the......

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...VECTOR FUNCTIONS VECTOR FUNCTIONS Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal vectors. MOTION IN SPACE: VELOCITY AND ACCELERATION Here, we show how the ideas of tangent and normal vectors and curvature can be used in physics to study: The motion of an object, including its velocity and acceleration, along a space curve. VELOCITY AND ACCELERATION In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion. VELOCITY Suppose a particle moves through space so that its position vector at time t is r(t). VELOCITY Vector 1 Notice from the figure that, for small values of h, the vector r(t h) r(t ) h approximates the direction of the particle moving along the curve r(t). VELOCITY Its magnitude measures the size of the displacement vector per unit time. VELOCITY The vector 1 gives the average velocity over a time interval of length h. VELOCITY VECTOR Equation 2 Its limit is the velocity vector v(t) at time t : r(t h) r(t ) v(t ) lim h 0 h r '(t ) VELOCITY VECTOR Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line. SPEED The speed of the particle at time t is the magnitude of the velocity vector, that is, |v(t)|. SPEED This is appropriate because, from Equation......

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...Tutorial 1 – Vector Calculus 1. Find the magnitude of the vector PQ with P (−1,2) and Q (5,5) . 2. Find the length of the vector v = 2,3,−7 . 3. Given the points in 3-dimensional space, P ( 2,1,5), Q (3,5,7), R (1,−3,−2) and S ( 2,1,0) . Does r PQ = RS ? ˆ ˆ 4. Find a vector of magnitude 5 in the direction of v = 3i + 5 ˆ − 2k . j r r ˆ ˆ ˆ j ˆ 5. Given u = 3i − ˆ − 6k and v = −i + 12k , find (a) u • v , r r (b) the angle between vectors u and v , r (c) the vector proju v , r r r r (d) the scalar component of v in the direction of u . 6. Given P (1,−1,3), Q ( 2,0,1) and R (0,2,−1) , find (a) the area of the triangle determined by the points P, Q and R. (b) the unit vector perpendicular to the plane PQR. 7. Find the volume of the parallelepiped determined by the vectors u = 4,1,0 , v = 2,−2,3 and r r r r r w = 0,2,5 . 8. Find the area of the parallelogram whose vertices are given by the points A (0, 0, 0), B (3, 2, 4), C (5, 1, 4) and D (2, -1, 0). ˆ j 9. Find the equation of the line through (2, 1, 0) and perpendicular to both i + ˆ and ˆ + k . j ˆ 10. Find the parametric equation of the line through the point (1, 0, 6) and perpendicular to the plane x+3y+z=5. 11. Determine whether the given lines are skew, parallel or intersecting. If the lines are intersecting, what is the angle between them? L1: x −1 y −3 z−2 = = 2 2 −1 x−2 y−6 z+3 L2 : = = 1 −1 3 12. Find the point in which the line x = 1 –t, y = 3t, z = 1 + t meets...

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...5 6 1 5 2 0 <= Available Input a text file that includes the number of processes, resources, and the matrixes for allocations, max, and available. Output System Safe or Unsafe 1) Read the # of processes and the # of resources 2) Read allocation, max and available for each process and each resource 3) Print whether this system is safe or not. Data Structures for the Banker’s Algorithm Let n = number of processes, and m = number of resources types. Available: Vector of length m. If available [j] = k, there are k instances of resource type Rj available Input.txt Input.txt Max: n x m matrix. If Max [i,j] = k, then process Pi may request at most k instances of resource type Rj Allocation: n x m matrix. If Allocation[i,j] = k then Pi is currently allocated k instances of Rj Need: n x m matrix. If Need[i,j] = k, then Pi may need k more instances of Rj to complete its task Need [i,j] = Max[i,j] – Allocation [i,j] Safety Algorithm 1. Let Work and Finish be vectors of length m and n, respectively. Initialize: Work = Available Finish [i] = false for i = 0, 1, …, n- 1 2. Find an i such that both: (a) Finish [i] = false (b) Needi ≤ Work If no such i exists, go to step 4 3. Work = Work + Allocationi Finish[i] = true go to step 2 4. If Finish [i] == true for all i, then the system is in a safe state...

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...Question 1 A Boeing 767-200 pasenger jet has a cruising speed of 870 km hr-1 at an altitude of 11900 m. At this cruising speed, its two General Electric CF6-80A engines exert a combined thrust of 390 kN. The frictional force of the air opposing the motion of the plane is A. 0 kN B. 390 kN C. 870 kN D. 1260 kN The following information refers to questions 2 to 10 inclusive. The circus has come to town and the principal attraction is Cleo, the human cannonball. There is a large compressed spring at the base of the cannon as illustrated below. It is this spring that will propel Cleo out of the cannon through the air to the safety net. The force-compression graph for the spring is also shown. [pic] [pic] Question 2 What is the spring constant for the spring inside the cannon? Question 3 How much energy is stored in the spring when it is fully compressed by 8.0 m'? Question 4 Cleo has a mass of 50 kg. Calculate Cleo's velocity just as she leaves the mouth of the cannon, assuming that the spring has transferred all of its energy to Cleo at this point. Question 5 The length of the cannon is 8.0 m. What is the net average force exerted on Cleo while she is travelling through the cannon? Question 6 What is Cleo's average acceleration while she is travelling through the cannon? The cannon makes an angle of 450 with the horizontal as shown in the diagram below. [pic] Question 7 How far (horizontally) from the mouth of the cannon must the centre...

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...Find parametric equations for the line L which contains A(1, 2, 3) and B(4, 6, 5). Solution: To get the parametric equations of L you need a point through which the line passes and a vector parallel to the line. −→ Take the point to be A and the vector to be the AB. The vector equation of L is −→ −→ r(t) = OA + t AB = 1, 2, 3 + t 3, 4, 2 = 1 + 3t, 2 + 4t, 3 + 2t , where O is the origin. The parametric equations are: x = 1 + 3t y = 2 + 4t, z = 3 + 2t t ∈ R. Problem 1(b) - Fall 2008 Find parametric equations for the line L of intersection of the planes x − 2y + z = 10 and 2x + y − z = 0. Solution: The vector part v of the line L of intersection is orthogonal to the normal vectors 1, −2, 1 and 2, 1, −1 . Hence v can be taken to be: i j k v = 1, −2, 1 × 2, 1, −1 = 1 −2 1 = 1i + 3j + 5k. 2 1 −1 Choose P ∈ L so the z-coordinate of P is zero. Setting z = 0, we obtain: x − 2y = 10 2x + y = 0. Solving, we ﬁnd that x = 2 and y = −4. Hence, P = 2, −4, 0 lies on the line L. The parametric equations are: x =2+t y = −4 + 3t z = 0 + 5t = 5t. Problem 2(a) - Fall 2008 Find an equation of the plane which contains the points P(−1, 0, 1), Q(1, −2, 1) and R(2, 0, −1). Solution: Method 1 −→ −→ Consider the vectors PQ = 2, −2, 0 and PR = 3, 0, −2 which lie parallel to the plane. Then consider the normal vector: i j k −→ −→ n = PQ × PR = 2 −2 0 3 0 −2 = 4i + 4j + 6k. So the equation of the plane is given by: 4, 4, 6 · x + 1, y , z − 1 = 4(x + 1) + 4y + 6(z − 1) = 0. Problem 2(a) - Fall......

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...Chapter 10. Uniform Circular Motion A PowerPoint Presentation by A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Paul E. Tippens, Professor of Physics Southern Polytechnic State University Southern Polytechnic State University © 2007 Centripetal forces keep these children moving in a circular path. Objectives: After completing this module, you should be able to: • Apply your knowledge of centripetal acceleration and centripetal force to the solution of problems in circular motion. • Define and apply concepts of frequency and period, and relate them to linear speed. • Solve problems involving banking angles, the conical pendulum, and the vertical circle. Uniform Circular Motion Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction. Fc v Constant velocity tangent to path. Constant force toward center. Question: Is there an outward force on the ball? Uniform Circular Motion (Cont.) The question of an outward force can be resolved by asking what happens when the string breaks! Ball moves tangent to v path, NOT outward as might be expected. When central force is removed, ball continues in straight line. Centripetal force is needed to change direction. Examples of Centripetal Force You are sitting on the seat next to You are sitting on the seat next to the outside door. What is the the outside door. What is the direction of...

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...Name ____________ Motion in 2D Simulation Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now. 1) Once the simulation opens, click on ‘Show Both’ for Velocity and Acceleration at the top of the page. Now click and drag the red ball around the screen. Make 3 observations about the blue and green arrows (also called vectors) as you drag the ball around. When you drag the red ball in a direction two arrows appear. One of the arrows is green and one of the arrows is blue. In the same direction that u drag the ball, the green and blue arrow point the same direction. After a second or two, the blue arrow changes to a complete opposite direction but the green arrow does not change. If I drag the red ball faster the arrows get longer and slower I drag the red ball the smaller the arrows. 2) Which color vector (arrow) represents velocity and which one represents acceleration? How can you tell? The green arrow goes in the same direction that you drag the red ball and would therefore represent velocity. The blue arrow, at first, goes in the same direction that you drag the red ball. Then as the ball slows down the blue arrow changes and goes in the opposite direction, this would make the blue arrow acceleration. 3) Try dragging the ball around and around in a circular path. What do you notice about the lengths and directions of the blue and green vectors? Describe their behavior in detail below. When I drag the......

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...green box on the right side of the screen, select the following settings: 1 dimension, velocity vectors ON, momentum vectors ON, reflecting borders ON, momenta diagram ON, elasticity 0%. Look at the red and green balls on the screen and the vectors that represent their motion. a. Which ball has the greater velocity? b. Which has the greater momentum? 2. Explain why the green ball has more momentum but less velocity than the red ball (HINT: what is the definition of momentum?). 3. Push “play” and let the balls collide. After they collide and you see the vectors change, click “pause”. Click “rewind” and watch the momenta box during the collision. Watch it more than once if needed by using “play”, “rewind”, and “pause”. Zoom in on the vectors in the momenta box with the control on the right of the box to make it easier to see if necessary. a. What happens to the momentum of the red ball after the collision? b. What about the green ball? c. What about the total momentum of both the red and green ball? 4. Change the mass of the red ball to match that of the green ball. a. Which ball has greater momentum now? b. How has the total momentum changed? c. Predict what will happen to the motion of the balls after they collide. 5. Watch the simulation, and then pause it once the vectors have changed. a. What happens to the momentum of the red ball after the......

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