Berechnung des Trägheitsmomentes unter einem Trapez:
··+ y₂ ······ | ····· | ····· | ····· | y₁ +·· | | | | | | | | | | | 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)