MidtermProb3

MidtermProb3 - 0.012192 -0.061671 2.0466-2.0466-0.6822...

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Problem 3a Solution P = 0 0 0 1 1 0.5 2.5 1.5 0.5 4 0 0 MB= -1 3 -3 1 3 -6 3 0 -3 3 0 0 1 0 0 0 A=MB*P= -0.5 -1.5 0 1.5 -1.5 -1.5 3 3 1.5 0 0 0 Problem 3b Solution u 0.5452 x(u) = 2.000436 Problem 3c Solution Left hand curve geometric matrix U(u) wher u= 0 0 0 0 1 U(u) wher u= 0.5452 0.162057 0.297243 0.5452 2 U'(u) where u= 0 0 0 1 0 U'(u) where u= 0.5452 0.891729 1.0904 1 0 Q0= 0 0 0 Q1= 2.000436 0.94665 0.371935 Q' 0= 3 3 1.5 Q' 1= 4.189735 0.026806 -0.1356 B= 0 0 0 2.000436 0.94665 0.371935 1.6356 1.6356 0.8178 2.284244 0.014615 -0.073929 Right hand curve geometric matrix U(u) wher u= 0.5452 0.162057 0.297243 0.5452 1 U(u) wher u= 1 1 1 1 2 U'(u) where u= 0.5452 0.891729 1.0904 1 0 U'(u) where u= 1 3 2 1 0 Q0= 2.000436 0.94665 0.371935 Q1= 4 0 0 Q' 0= 4.189735 0.026806 -0.1356 Q' 1= 4.5 -4.5 -1.5
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B= 2.000436 0.94665 0.371935 4 0 0 1.905492
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Unformatted text preview: 0.012192 -0.061671 2.0466-2.0466-0.6822 Convert to Bezier, Compute MB**-1 * MF MB**-1 * MF = 0.00 0.00 0.00 1.00 2-2 1 0.00 0.00 0.33 1.00-3 3-2 0.00 0.33 0.67 1.00 1 1.00 1.00 1.00 1.00 1 MB**-1 * MF = 1 1 0.333333 1 0 -0.333333 1 Left Hand Curve Control Points P= 0.5452 0.5452 0.2726 1.239022 0.941778 0.396578 2.000436 0.94665 0.371935 Left Hand Curve Control Points P= 2.000436 0.94665 0.371935 2.6356 0.950714 0.351378 3.3178 0.6822 0.2274 4 1-1...
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This note was uploaded on 05/01/2008 for the course MECH 202 taught by Professor Ardema during the Winter '07 term at Santa Clara.

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MidtermProb3 - 0.012192 -0.061671 2.0466-2.0466-0.6822...

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