Berechnung des Trägheitsmomentes unter einem Trapez:
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0 -+-------------------------+-
x₁ x₂
moi[x] = ∫[y∊ 0…f(x)] y² ∂y ∂x = ⅓ f(x)³ ∂x
moi[x₁…x₂] = ∫[x∊ x₁…x₂] ⅓ f(x)³ ∂x = ⅓ ∫[x∊ x₁…x₂] f(x)³ ∂x
f(x) = y₁ + (x-x₁) * (y₂-y₁)/(x₂-x₁)
Δx ≔ x₂-x₁
Δy ≔ y₂-y₁
s ≔ Δy / Δx → s·Δx = Δy
f(x) = y₁ + s * (x-x₁)
moi[x₁…x₂] = ⅓ ∫[x∊ x₁…x₂] ( y₁ + s * (x-x₁) )³ ∂x
Substituiere x-x₁ → x
moi[x₁…x₂] = ⅓ ∫[x∊ 0…Δx] ( y₁ + s * x )³ ∂x
moi[x₁…x₂] = ⅓ ∫[x∊ 0…Δx] ( y₁³ + 3·y₁²·s·x + 3·y₁·s²·x² + s³·x³ ) ∂x
moi[x₁…x₂] = ⅓ [ y₁³·∫1∂x + 3·y₁²·s·∫x∂x + 3·y₁·s²·∫x²∂x + s³·∫x³∂x ]
moi[x₁…x₂] = ⅓ [ y₁³·Δx + 3·y₁²·s·½·Δx² + 3·y₁·s²·⅓·Δx³ + s³·¼·Δx⁴ ]
moi[x₁…x₂] = ⅓ Δx [ y₁³ + 3/2·y₁²·s·Δx + y₁·s²·Δx² + ¼·s³·Δx³ ]
moi[x₁…x₂] = ⅓ Δx [ y₁³ + 3/2·y₁²·Δy + y₁·Δy² + ¼·Δy³ ]
moi[x₁…x₂] = 1/12·Δx [ 4·y₁³ + 6·y₁²·Δy + 4·y₁·Δy² + Δy³ ]
moi[x₁…x₂] = 1/12·Δx [ 4·y₁³ + 6·y₁²·[y₂-y₁] + 4·y₁·[y₂-y₁]² + [y₂-y₁]³ ]
moi[x₁…x₂] = 1/12·Δx [ 4·y₁³
+ 6·y₂y₁² - 6·y₁³
+ 4·y₂²y₁ - 8·y₂y₁² + 4·y₁³
+ y₂³ - 3·y₂²y₁ + 3·y₂y₁² - y₁³ ]
moi[x₁…x₂] = 1/12 · Δx · [y₂³ + y₂²y₁ + y₂y₁² + y₁³]
moi[x₁…x₂] = 1/12 · Δx · [y₂² + y₁²] · [y₂ + y₁]
Statt Δx wird [x₁-x₂] genutzt, damit das MOI für Polygone im Gegenuhrzeigersinn positiv ist. Damit also:
MOI = 1/12 · ∑ [i∊ 0…n-1] (y[i+1]² + y[i]²]) · (y[i+1] + y[i]) · (x[i] - x[i+1]) (wo bei i+1=n stattdessen der Index 0 genutzt wird)