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3 years ago

Disadvantages of bezier curve and b-spline

Engineering

1 answer:

Gnesinka [82]3 years ago

5 0

Answer:

The main disadvantage of the Bezier-curves is the global influence of the control points on the whole curve.

This has two major drawbacks: (1) every change of a control point (insert, move, delete) changes the look of the curve in all points of the curve, and (2) the computation time needed for a big set of control points is comparatively high. The reason is the weighting-function form.

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IDENTIFY THE POINTER READING FOR EACH SCALE. COMPUTE THE READING USING THE ASSIGNED RANGE MULTIPLIER FOR OHMMETER SCALE.

daser333 [38]

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Lesson 2 - T.L.E Learning Module

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3 years ago

Read 2 more answers

For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo

Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= $\frac{R}{R +S } * 100$

= $\frac{53-45}{58-45} * 100$ = 61.5%

To determine the wt.% of the alloy that is liquid

= $\frac{S}{S+R} * 100\\$

= $\frac{58-53}{58-45} *100$ = 38.5%

attached below is a free hand sketch as well

7 0

3 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$