Sale!

# PSet 3: Curves & Surfaces

\$35.00

Category:

CS 480/CS680 PSet 3: Curves & Surfaces

1. (20 points; 5 points extra credit)
We are given the implicit function for a two-sheeted hyperboloid,
๐(๐ฅ, ๐ฆ, ๐ง) = (๐ฅ โ ๐เฏซ
)
เฌถ
+ เตซ๐ฆ โ ๐เฏฌเตฏ
เฌถ โ ๐ง
เฌถ + ๐
เฌถ
.
a) Give the 4 ร 4 matrix ๐ธ for this cylindrical surface, such that ๐
๐ป๐ธ๐ = 0 for any
homogeneous point ๐ = (๐ฅ, ๐ฆ, ๐ง, 1) that is on the cylindrical surface.
b) Using the implicit function, derive a function that gives a unit normal vector at any point
on the surface (๐ฅ, ๐ฆ, ๐ง).
c) Give an equivalent parametric equation for the sheet of the surface that is above the xyplane (๐ง โฅ 0), in terms of ฮธ and z. Assume that โ๐ โค ๐ โค ๐ and ๐งเฏ เฏเฏก โค ๐ง โค ๐งเฏ เฏเฏซ.
Give the parameters for ๐งเฏ เฏเฏก, ๐งเฏ เฏเฏซ.
d) Using the parametric equation, derive the function that gives the unit normal vector at
any point on the sheet with parameter (๐, ๐ง).
e) Extra Credit (5 points): Give a parametrization that utilizes hyperbolic trigonometric
functions, and find the unit normal with this parametrization.
2. (30 points)
We are given a 3D swept surface. A line segment with end points ๐๐ = (2,0,0) and
๐๐ = (0,0,6) is used as the sweep curve. The cross-section is an ellipse in the z = 0 plane,
where ๐ฅ = 2 + ๐เฏซ ๐๐๐ (2๐๐ฃ), ๐ฆ = ๐เฏฌ ๐ ๐๐(2๐๐ฃ), 0 โค ๐ฃ โค 1. The center of the ellipse is
swept along the vector from ๐๐ to ๐๐ , such that it remains parallel to the x-y plane.
a) Derive the parametric equation in ๐ข for the directed line segment ๐๐ ๐๐.
b) Derive an equation for location of the point ๐ท(๐ข, ๐ฃ) on the swept surface.
c) Derive the equation for the swept surface normal ๐ต(๐ข, ๐ฃ).
3. (20 points)
We are given a 2D cubic Bezier curve segment, which has the following control points:
๐๐ = (2, 3)
๐๐ = (3, 0)
Submission guidelines:
๐๐ = (4, 4)
๐๐ = (5, 1)
a) Draw the convex hull for this 2D Bezier curve segment.
b) Compute the value of ๐โ(0) for this 2D Bezier curve segment.
c) We are now given a second 2D Bezier curve segment, which has the control points:
๐๐ = (0, -1)
๐๐ = (-2, 4)
๐๐ = (0, 9)
๐๐ = (2, 3)
Does this segment join the previous segment with C1 continuity? Give a mathematical
d) Which control point above (for the second curve in (c)) may we change to achieve C1
continuity? Write down the new position of the point that will achieve this.
4. (30 points)
We are given the following boundary conditions for a cubic spline section:
๐ท(0) = ๐เฏ
๐ท(1) = ๐เฏเฌพเฌต
๐ทโฒ(0) = เฌต
เฌถ
[(1 + ๐)(๐เฏ โ ๐เฏเฌฟเฌต) + (1 โ ๐)(๐เฏเฌพเฌต โ ๐เฏ
)]
๐ทโฒ(1) =
เฌต
เฌถ
[(1 + ๐)(๐เฏเฌพเฌต โ ๐เฏ
) + (1 โ ๐)(๐เฏเฌพเฌถ โ ๐เฏเฌพเฌต)]
In the textbook, we see this is a Cardinal Spline (Kochanek-Bartels spline with t=0 and c=0).
In this case ๐เฏเฏเฏขเฏ  = [๐เฏเฌฟเฌต ๐เฏ ๐เฏเฌพเฌต ๐เฏเฌพเฌถ]
เฏ
and the boundary conditions can be written:
โฃ
โข
โข
โก
๐ท(0)
๐ท(1)
๐ทโฒ(0)
๐ทโฒ(1)โฆ
โฅ
โฅ
โค
=
โฃ
โข
โข
โก
0 1 0 0
0 0 1 0
เฐท
เฐญ
เฐฎ
(เฌตเฌพเฏ) ๐
เฐญ
เฐฎ
(เฌตเฌฟเฏ) 0
0 เฐท
เฐญ
เฐฎ
(เฌตเฌพเฏ) ๐
เฐญ
เฐฎ
(เฌตเฌฟเฏ)โฆ
โฅ
โฅ
โค
เตฆ
๐เฏเฌฟเฌต
๐เฏ
๐เฏเฌพเฌต
๐เฏเฌพเฌถเตช
a) Show how to compute the 4ร4 coefficient matrix MC given the boundary conditions
written above. You do not need to compute a matrix inverse to find MC (the relevant one
is given in the textbook anyway). Just give the specific equations for MC.
b) Given MC write out the blending functions for this curve.
c) Do adjacent segments satisfy C1 continuity? Give a mathematical justification.
d) Does adjusting b change the tangent direction at the endpoints or only magnitude?

PSet 3: Curves & Surfaces
\$35.00
Hello
Can we help?