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Chapter 8: Q13E (page 437)

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

13. a. Use equation (12) to show that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\).

b. Use equation (13) to show that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).

c. Use part (b), equation (8), and part (a) to show that \({{\mathop{\rm q}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm q}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).

Short Answer

Expert verified
  1. It is proved that\({{\mathop{\rm q}\nolimits} _1}\)is the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _0}\)to\({{\mathop{\rm p}\nolimits} _1}\).
  2. It is proved that\(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).
  3. It is proved that \({{\mathop{\rm q}\nolimits} _2}\) is midpoint from of the segment from \({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\).

Step by step solution

01

Show that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\)

a)

Let\({\mathop{\rm y}\nolimits} \left( t \right)\)be the Bezier curve determined by\({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\), and let\({\mathop{\rm z}\nolimits} \left( t \right)\)be the Bezier curve determined by\({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\)since\({\mathop{\rm z}\nolimits} \left( t \right)\)starts at\({\mathop{\rm x}\nolimits} \left( {.5} \right)\)when\(t = 0\),\({\mathop{\rm z}\nolimits} \left( t \right) = {\mathop{\rm x}\nolimits} \left( {.5 + .5t} \right)\)for\(0 \le t \le 1\).

Recall the equation (12), the control points for\({\mathop{\rm y}\nolimits} \left( t \right)\)satisfy

\(\begin{aligned}{}3\left( {{{\mathop{\rm q}\nolimits} _1} - {{\mathop{\rm q}\nolimits} _0}} \right) &= y'\left( 0 \right)\\ &= .5x'\left( 0 \right)\\ &= \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\end{aligned}\)

Use the equation (12) as shown below:

\(\begin{aligned}{}3{{\mathop{\rm q}\nolimits} _1} - 3{{\mathop{\rm q}\nolimits} _0} &= \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\\3{{\mathop{\rm q}\nolimits} _1} &= \frac{3}{2}{{\mathop{\rm p}\nolimits} _1} - \frac{3}{2}{{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm q}\nolimits} _0}\left( {{\mathop{\rm since}\nolimits} \,\,{{\mathop{\rm q}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}} \right)\\{{\mathop{\rm q}\nolimits} _1} &= \frac{1}{2}{{\mathop{\rm p}\nolimits} _1} - \frac{1}{2}{{\mathop{\rm p}\nolimits} _0} + 1{{\mathop{\rm p}\nolimits} _0}\\{{\mathop{\rm q}\nolimits} _1} &= \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _0}} \right)\end{aligned}\)

Thus, it is proved that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\).

02

Show that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\) 

b)

Recall the equation (13) as shown below

\(\begin{aligned}{}3\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right) &= y'\left( 1 \right)\\ &= .5x'\left( {.5} \right)\\ &= \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\end{aligned}\)\(\)鈥�(13)

Left-multiply the equation (13) by\(\frac{8}{3}\)as shown below:

\(\begin{aligned}{}\frac{8}{3} \times 3\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right) &= \frac{8}{3} \times \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\\8\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right) &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}\\8{{\mathop{\rm q}\nolimits} _3} - 8{{\mathop{\rm q}\nolimits} _2} &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}\\8{{\mathop{\rm q}\nolimits} _2} &= 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\end{aligned}\)

Thus, it is proved that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).

03

Show that \({{\mathop{\rm q}\nolimits} _2}\) is the midpoint of the segment from\({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\)

c)

The midpoint of the original curve\({\mathop{\rm x}\nolimits} \left( t \right)\)occurs at\({\mathop{\rm x}\nolimits} \left( {.5} \right)\)when\({\mathop{\rm x}\nolimits} \left( t \right)\)has the standard parameterization

\({\mathop{\rm x}\nolimits} \left( t \right) = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\mathop{\rm p}\nolimits} _1} + \left( {3{t^2} - 3{t^3}} \right){{\mathop{\rm p}\nolimits} _2} + {t^3}{{\mathop{\rm p}\nolimits} _3}\)鈥�(7)

The new control points\({{\mathop{\rm q}\nolimits} _3}\)and\({{\mathop{\rm r}\nolimits} _0}\)are given as shown below:

\(\begin{aligned}{}{{\mathop{\rm q}\nolimits} _3} &= {{\mathop{\rm r}\nolimits} _0}\\ &= {\mathop{\rm x}\nolimits} \left( {.5} \right)\\ &= \frac{1}{8}\left( {{{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\end{aligned}\)

\(8{{\mathop{\rm q}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}{\rm{ }}...{\rm{ }}\left( 8 \right)\)

Use equation (8) to substitute for 8\({{\mathop{\rm q}\nolimits} _3}\)in part (b) as shown below

\(\begin{aligned}{}8{{\mathop{\rm q}\nolimits} _2} &= {{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\\ &= 2{{\mathop{\rm p}\nolimits} _0} + 4{{\mathop{\rm p}\nolimits} _1} + 2{{\mathop{\rm p}\nolimits} _2}\end{aligned}\)

Multiply the obtained equation by\(\frac{1}{8}\)as shown below:

\({{\mathop{\rm q}\nolimits} _2} = \frac{1}{4}{{\mathop{\rm p}\nolimits} _0} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _2}\)

Substitute\(\frac{1}{2}{{\mathop{\rm p}\nolimits} _1} = \frac{1}{4}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _1}\)in the above equation as shown below:

\(\begin{aligned}{}{{\mathop{\rm q}\nolimits} _2} &= \frac{1}{4}{{\mathop{\rm p}\nolimits} _0} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _2}\\ &= \frac{1}{2}\left( {\frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1}} \right) + \frac{1}{2}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2}} \right)\end{aligned}\)

Use part (a) in the above equation as shown below:

\(\begin{aligned}{}{{\mathop{\rm q}\nolimits} _2} &= \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2}} \right)\\ &= \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\end{aligned}\)

Thus, it is proved that \({{\mathop{\rm q}\nolimits} _2}\) is midpoint of the segment from \({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\).

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In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the 鈥渓eft鈥� side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.
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