# Section 2 part 1 coordinate transformation

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### Section 2 part 1 coordinate transformation

• 1. Section 1: Introduction to Tensor Notation and Analysis Part 1: Coordinate transformation
• 2. 2 COORDINATE TRANSFORMATION ik j ai +bj+ck A vector‟s direction only makes sense when it is compared to a defined frame of reference or coordinate system.
• 3. 3 COORDINATE TRANSFORMATION ik j ai +bj+ck While we usually see these systems defined as orthogonal to the paper we are using, it is truly arbitrary where we place the coordinate system, and sometimes it is advantageous to put it somewhere else or in another orientation.
• 4. 4 COORDINATE TRANSFORMATION iA j What if we defined the coordinate system somewhere else? For instance, what if we translated the “origin” from Frame A (0,0) to Frame B (h,k)? We know P in frame A ((x,y) positions). What is P in B ((x‟, y‟) positions)? x y x‟ y‟ P(x,y) B kykPABPy hxhPABPx yxP yyyy xxxx )0()( )0()( :),( (0,0) (h,k)
• 5. 5 COORDINATE TRANSFORMATION iA j What if we defined the coordinate system somewhere else? For instance, what if we translated the “origin” from Frame A (0,0) to Frame B (h,k)? We know P in frame A ((x,y) positions). What is P in B ((x‟, y‟) positions)? x y x‟ y‟ P(x,y) B )( )( )0()( )0()( :),( yy xx yyyy xxxx BAyy BAxx kykPABPy hxhPABPx yxP (0,0) (h,k) If you know P(x‟,y‟) and want to find P(x,y):
• 6. 6 COORDINATE TRANSFORMATION What if we rotate the coordinate frame from A to B? iA j x y P(x,y) B (0,0) iA j x y P(x,y) B (0,0) What is P(x‟,y‟)? For x‟, there is a component of x that contributes (cos =cos(x‟,x)) and a component of y (cos(90- )=sin =cos (x‟,y) Px Py
• 7. 7 COORDINATE TRANSFORMATION A point on the x axis would have a contribution of cos to the x‟ axis. A point on the y axis would have a contribution of cos (90- ) = sin to the x‟ axis. A point on the x axis would have a contribution of -sin (or cos (y‟,x)=cos 90+ ) to the y‟ coordinate. A point on the y axis would have a contribution of cos to the y‟ coordinate. To find x’ value: To find y’ value:
• 8. 8 Coordinate Transformation iA j x y P(x,y) B (0,0) So point P moves a net: sincos sincos xy yx In the x‟ direction, and a net: In the y‟ direction. sincos sincos xyy yxx
• 9. 9 Coordinate Transformation Similarly, to go from (x‟,y‟) to (x,y) A point on the x‟ axis contributes +cos to the x coordinate. A point on the y‟ axis contributes –sin to the x coordinate. A point on the x‟ axis contributes sin to the y coordinate. A point of the y‟ axis contributes cos to the y coordinate.
• 10. 10 Coordinate Transformation Similarly, to go from (x‟,y‟) to (x,y) iA j x y P(x,y) B (0,0) sincos sincos xyy yxx
• 11. 11 General Coordinate Transformation (rotation) axes.b''anda''ebetween thangletheofcostheisb)cos(a,Where ),cos(),cos(),cos( ),cos(),cos(),cos( ),cos(),cos(),cos( ),cos(),cos( ),cos(),cos( cossin sincos cossin sincos z y x zzyzxz zyyyxy zxyxxx z y x y x yyxy yxxx y x y x y x y x y x
• 12. 12
• 13. 13 Example problem iA j x y H(7,7) B (0,0) = 30 deg P(2,0) What is position vector of P in H frame?
• 14. 14 First translate iA j x y H(7,7) B (0,0) P(2,0) 7 5 7 7 0 2 :PpointFor 7 7 A A OH OH A A y x yy xx y x y x y x You must perform the translation first, so that when you rotate, the arc length a point travels is correct.
• 15. 15 Then rotate… iA j x y H(7,7) B (0,0) P(2,0) Then rotate: = 30 deg 83.7 56.3 7 5 5.0867.0 867.05.0 120cos120sin 120sin120cos 5.0120cos3090cos),cos( )120sin(867.0210cos30180cos),cos( )120sin(867.030cos),cos( 5.0120cos3090cos),cos( 000 0000 00 000 A A B B y x y x yy xy yx xx
• 16. How can you check your work? 1) User a ruler! 2) Rotate the paper / image so that your „new‟ frame is in a traditional position and estimate the new values for the point 3) Recognize that while the positions of the points are vectors, the distance between points is a scalar, and therefore independent of the coordinate frame: What is position vector from P-H in O frame? 5i+7j And in the H frame? -3.56i + 7.83j The magnitude of these position vectors (ie distance to (0,0)) is a scalar and must be the same! Sqrt(5*5 + 7*7)=sqrt(74) Sqrt(3.56*3.56+7.83*7.83)=sqrt(74) Check!
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