偏航、俯仰和滚动旋转到六个浮点变量

Question

我需要帮助创建一个函数来将三个角度（以度为单位，偏航俯仰和滚动）转换为六个浮点变量。

我将如何使函数输出这些浮点数？

{0, 0, 0} = {1, 0, 0, -0, -0, 1}
{45, 0, 0} = {0.70710676908493, 0.70710676908493, 0, -0, -0, 1}
{0, 90, 0} = {-4.3711388286738e-08, 0, 1, -1, 0, -4.3711388286738e-08}
{0, 0, 135} = {1, -0, 0, -0, -0.70710676908493, -0.70710676908493}
{180, 180, 0} = {1, -8.7422776573476e-08, 8.7422776573476e-08, 8.7422776573476e-08, 0, -1}
{225, 0, 225} = {-0.70710682868958, 0.5, 0.5, -0, 0.70710670948029, -0.70710682868958}
{270, 270, 270} = {1.4220277639103e-16, -2.3849761277006e-08, 1, 1, 1.1924880638503e-08, 1.42202776319103e-16}
{315, 315, 315} = {0.5, -0.85355341434479, 0.14644680917263, 0.70710688829422, 0.5, 0.5}

请求更多示例：Egor Skriptunoff

{10, 20, 30} = {0.92541658878326, -0.018028322607279, 0.37852230668068, -0.34202012419701, -0.46984630823135, 0.813797652=7424}
{10, 30, 20} = {0.85286849737167, -0.0052361427806318, 0.52209949493408, -0.5, -0.29619812965393, 0.81379765272141}
{20, 10, 30} = {0.92541658878326, 0.21461015939713, 0.3123245537281, -0.17364817857742, -0.49240386486053, 0.852868497=34267}
{20, 30, 10} = {0.81379765272141, 0.25523611903191, 0.52209949493408, -0.5, -0.15038372576237, 0.85286849737167}
{30, 10, 20} = {0.85286849737167, 0.41841205954552, 0.3123245537281, -0.17364817857742, -0.33682405948639, 0.925416588=74326}
{30, 20, 10} = {0.81379765272141, 0.4409696161747, 0.37852230668068, -0.34202012419701, -0.16317591071129, 0.925416588=74226}

我目前的代码可以计算除第 2 和第 3 之外的所有浮点数。

function convert_rotations(Yaw, Pitch, Roll)
    return {
        math.cos(math.rad(Yaw))*math.cos(math.rad(Pitch)),
        0,
        0,
        math.sin(math.rad(Pitch))*-1,
        math.sin(math.rad(Roll))*math.cos(math.rad(Pitch))*-1,
        math.cos(math.rad(Roll))*math.cos(math.rad(Pitch))
    }
end

当第二个浮点数和第三个浮点数的所有角度都非零时，我似乎找不到正确的数学公式，但我确实想出了这个：

-- The second float when the Yaw is 0 degrees
math.sin(math.rad(Pitch))*math.sin(math.rad(Roll))*-1

-- The second float when the Pitch is 0 degrees
math.sin(math.rad(Yaw))*math.cos(math.rad(Roll))

-- The second float when the Roll is 0 degrees
math.sin(math.rad(Yaw))*math.sin(math.rad(Pitch))

对于第三个浮点数，我想出了这个：

-- The third float when Yaw is 0 degrees
math.sin(math.rad(Pitch))*math.cos(math.rad(Roll))

-- The third float when Pitch is 0 degrees
math.sin(math.rad(Yaw))*math.sin(math.rad(Roll))

-- The third float when Roll is 0 degrees
math.cos(math.rad(Yaw))*math.sin(math.rad(Pitch))

Answer 1

local function Rotate(X, Y, alpha)
    local c, s = math.cos(math.rad(alpha)), math.sin(math.rad(alpha))
    local t1, t2, t3 = X[1]*s, X[2]*s, X[3]*s
    X[1], X[2], X[3] = X[1]*c+Y[1]*s, X[2]*c+Y[2]*s, X[3]*c+Y[3]*s
    Y[1], Y[2], Y[3] = Y[1]*c-t1, Y[2]*c-t2, Y[3]*c-t3
end

local function convert_rotations(Yaw, Pitch, Roll)
    local F, L, T = {1,0,0}, {0,1,0}, {0,0,1}
    Rotate(F, L, Yaw)
    Rotate(F, T, Pitch)
    Rotate(T, L, Roll)
    return {F[1], -L[1], -T[1], -F[3], L[3], T[3]}
end

偏航、俯仰和滚动旋转到六个浮点变量

Yaw, pitch, and roll rotations to six float variables

lua

angle

calculus