Linear interpolation of Vectors C#

隐藏标签 left bracket, "a" , "b" , right bracket

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equals

left parenthesis, negative 1.5 , negative 0.5 , right parenthesis

left parenthesis, 1.2 5 , 0.8 , right parenthesis

表达式14: "y" equals "a" plus "x" * ( "b" minus "a" ) left brace, 0 less than "x" less than 1 , right brace

float t = ${t}f; left parenthesis, negative 2.5 , 1.0 5 , right parenthesis

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隐藏 t = ${t} left parenthesis, negative 1.4 plus left parenthesis, "t" , right parenthesis , 1.0 5 , right parenthesis

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equals

left parenthesis, negative 0.6 5 , 1.0 5 , right parenthesis

表达式17: "y" equals 1.0 5 left brace, negative 1.6 less than "x" less than negative 0.2 , right brace

Vector2 a = new Vector2 (${x_a}f, ${y_a}f); left parenthesis, negative 2.5 , 1.4 5 , right parenthesis

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Vector2 b = new Vector2 (${x_b}f, ${y_b}f); left parenthesis, negative 2.5 , 1.2 5 , right parenthesis

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Vector2 result = Vector2.Lerp(a, b, t); left parenthesis, 0.2 , negative 0.6 5 , right parenthesis

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// result = (${r_x},${r_y}) left parenthesis, 0.2 , negative 0.8 5 , right parenthesis

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表达式22: "r" Subscript, "x" , Baseline equals "r" Subscript, "e" , Baseline . "x"

equals

0.5 6 2 5

表达式23: "r" Subscript, "y" , Baseline equals "r" Subscript, "e" , Baseline . "y"

equals

0.4 7 5

技术支持

a = (-1.5, -0.5)

b = (1.25, 0.8)

float t = 0.75f;

Vector2 a = new Vector2 (-1.5f, -0.5f);

Vector2 b = new Vector2 (1.25f, 0.8f);

Vector2 result = Vector2.Lerp(a, b, t);

// result = (0.5625,0.475)