Cross sectional area
public
Cross sectional area of belt conveyor for idlers with 5 rollers.
Input
Idler roller face width
$ l_{base} = 0.244 \; \;
m $
Result
Cross sectional area
$$ A_{conv_{5}} = {\left({ l_{base} + l_{base} \cdot \cos( \lambda_{idler1_{rad}} ) }\right) \cdot l_{base} \cdot \sin( \lambda_{idler1_{rad}} ) + \left({ l_{base} + 2 \cdot l_{base} \cdot \cos( \lambda_{idler1_{rad}} ) + {( b_{belt} - 3 \cdot l_{base} ) \over 2} \cdot \cos( \lambda_{idler2_{rad}} ) }\right) \cdot {( b_{belt} - 3 \cdot l_{base} ) \over 2} \cdot \sin( \lambda_{idler2_{rad}} ) + \left({ 0.5 \cdot l_{base} + l_{base} \cdot \cos( \lambda_{idler1_{rad}} ) + {( b_{belt} - 3 \cdot l_{base} ) \over 2} \cdot \cos( \lambda_{idler2_{rad}} ) }\right) ^ 2 \cdot \tan( \beta_{surcharge_{rad}} )} \; \;
, {m ^ 2}
$$
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formula
syntax
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formula syntax
A_{conv_{5}} = ( l_{base} + l_{base} * cos( \lambda_{idler1_{rad}} ) ) * l_{base} * sin( \lambda_{idler1_{rad}} ) + ( l_{base} + 2 * l_{base} * cos( \lambda_{idler1_{rad}} ) + ( b_{belt} - 3 * l_{base} ) / 2 * cos( \lambda_{idler2_{rad}} ) ) * ( b_{belt} - 3 * l_{base} ) / 2 * sin( \lambda_{idler2_{rad}} ) + ( 0.5 * l_{base} + l_{base} * cos( \lambda_{idler1_{rad}} ) + ( b_{belt} - 3 * l_{base} ) / 2 * cos( \lambda_{idler2_{rad}} ) ) ^ 2 * tan( \beta_{surcharge_{rad}} )
Preceding calculations 8
Dependant calculations 2
Discipline: mechanical
Managed by: Admin
