spacecurve

spacecurve - Numerical Determination of Unit Vectors The x...

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Numerical Determination of Unit Vectors The x , y , and z coordinates of a space curve can be defined as a function of parameter, ξ ,as x = x ( ξ ) ,y = y ( ξ ) ,z = z ( ξ ) . The tangential unit vector, ˆ e t , can be obtained as ˆ e t = d~r ds = ds = ds dx dy dz , in which ds = p ( dx ) 2 +( dy ) 2 dz ) 2 , and ds = s ± dx ² 2 + ± dy ² 2 + ± dz ² 2 = Δ or ds = 1 Δ The above formulation guarantees that | ˆ e t | =1 . To find the normal unit vector, perform another derivative with respect to s , 1 ρ ˆ e n = d ˆ e t ds = d ds ± ds ² = d ds ± ds ² = d 2 ~ r 2 ± ds ² 2 + ± d 2 ξ ds 2 ² With ds = 1 Δ , it follows that d 2 ξ ds 2 = - 1 2 1 Δ Δ d Δ ds = - 1 2 1 Δ Δ d Δ ds = - 1 2 1 Δ 2 d Δ = - 1 Δ 2 ³± dx ²± d 2 x 2 ² + ± dy d 2 y 2 ² + ± dz d 2 z 2 ²´ Therefore, 1 ρ ˆ e n = d ˆ e t ds = ± ds ² 2 d 2 ~ r 2 - ± ds ² 4 ³± dx d 2 x 2 ² + ± dy d
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This note was uploaded on 04/16/2008 for the course CE 325 taught by Professor Docwong during the Spring '08 term at USC.

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spacecurve - Numerical Determination of Unit Vectors The x...

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