Two Pulley Belt Length

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Mar 11, 2025, 5:42:55 PM

Created by

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Jun 6, 2014, 10:44:01 AM

belt length = π \cdot (R + r) + 2 \cdot a \sin (\frac{R - r}{L}) \cdot (R - r) + 2 \cdot \sqrt{L^{2} - {(R - r)}^{2}}

$"belt length" = pi * ( "R" + "r" ) + 2* asin( ( "R" - "r" ) / "L" )*( "R" - "r" ) + 2 * sqrt( "L" ^2 - ( "R" - "r" )^2 )$

(R) Radius of Larger Pulley

$(R)"Radius of Larger Pulley"$

(r) Radius of Smaller Pulley

$(r)"Radius of Smaller Pulley"$

(L) Axel Separation

$(L)"Axel Separation"$

Notes

From the picture, see that the following describe the component parts of the solution:

$C_{R}$ $C_R$ = half the larger pulley circumference = $\frac{π}{2} \cdot 2 R$ $pi / 2 * 2R$ , where 2R = larger pulley diameter

$C_{r}$ $C_r$ = half the smaller pulley circumference = $\frac{π}{2} \cdot 2 r$ $pi / 2 * 2r$ , where 2r = larger pulley diameter

u = segment of larger circumference between half circle and tangent point of pulley = $\frac{θ}{2 π} \cdot (π \cdot 2 R)$ $theta/(2pi) * (pi * 2R)$

v = segment of smaller circumference between half circle and tangent point of pulley = $\frac{θ}{2 π} \cdot (π \cdot 2 r)$ $theta/(2pi) * (pi * 2r)$

b = length of belt between tangent point on two pulleys = $\sqrt{L^{2} - {(R - r)}^{2}}$ $sqrt( L^2 - (R-r)^2 )$

$L \cdot \sin (θ) = R - r$ $L * sin( theta ) = R - r$ $\Rightarrow$ $=>$ $θ = a \sin (\frac{R - r}{L})$ $theta = asin( (R-r)/L )$

Belt Length = ( $C_{R} + 2 u$ $C_R + 2u$ ) + ( $C_{r} - 2 v$ $C_r - 2v$ ) + 2b

= ( $π R + 2 \cdot θ R$ $piR + 2* thetaR$ ) + ( $π r - 2 \cdot θ r$ $pir - 2* thetar$ ) +2b

= $π (R + r) + 2 θ \cdot (R - r) + 2 b$ $pi ( R+r) + 2theta* (R-r) + 2b$

= $π (R + r) + 2 \cdot a \sin (\frac{R - r}{L}) \cdot (R - r) + 2 \cdot \sqrt{L^{2} - {(R - r)}^{2}}$ $pi ( R+r) + 2* asin( (R-r) / L) * (R-r) + 2*sqrt( L^2 - (R-r)^2)$

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Two Pulley Belt Length

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