# Belt pulley ## Transmission method ### Open Belt Drive <div style="text-align: center;"> <img src="https://themechanicalengineering.com/wp-content/uploads/2022/01/OPEN-BELT-DRIVE.png" alt="image"width="400"> </div> ### Cross Belt Drive <div style="text-align: center;"> <img src="https://themechanicalengineering.com/wp-content/uploads/2022/01/CROSS-BELT-DRIVE-LINE-DIAGRAM.jpg" alt="image"width="400"> </div> > https://themechanicalengineering.com/open-belt-drive-and-cross-belt-drive/ ## Belt length > >https://www.vcalc.com/equation/?uuid=788be105-ed67-11e3-b7aa-bc764e2038f2 $$\text{BL} = (\pi\frac D2+2u)+(\pi\frac d2-2v)+2b = \frac{\pi(D+d)}2+2(u-v)+2b$$ **** $$2u-2v = 2R\theta-2r\theta = 2\theta(R-r) = 2\theta(\frac D2-\frac d2) = \theta(D-d)$$ $$S = r\theta$$ * when $\theta$ is very small $\sin \theta = \theta$ $$\theta = \sin^{-1}(\frac{R-r}L) = \sin^{-1}(\frac{\frac D2-\frac d2}L) = \sin^{-1}(\frac{D-d}{2L})\simeq\frac{D-d}{2L}$$ $$2(u-v) = \frac{D-d}{2L}(D-d) = \frac{(D-d)^2}{2L}$$ **** $$b = L\cos\theta = L-L(1-\cos\theta)$$ $$\cos \theta = \cos^2\frac\theta2-\sin^2\frac\theta2$$ $$1-\cos\theta = \cos^2\frac\theta2+\sin^2\frac\theta2-\cos^2\frac\theta2+\sin^2\frac\theta2 = 2\sin^2\frac\theta2$$ $$b = L-2L\sin^2\frac\theta2\simeq L-2L(\frac\theta2)^2$$ $$L-2L(\frac\theta2)^2 = L-2L(\frac{D-d}{4L})^2 = L-\frac{(D-d)^2}{8L}$$ $$2b = 2L-\frac{(D-d)^2}{4L}$$ **** $$\text{BL} = \frac{\pi(D+d)}2+\frac{(D-d)^2}{2L}+2L-\frac{(D-d)^2}{4L}$$ :::success $$\text{BL} = \frac{\pi(D+d)}2+2L+\frac{(D-d)^2}{4L}$$ ::: ### Open Belt Drive $$\text{BL} = \frac{\pi(D+d)}2+2L+\frac{(D-d)^2}{4L}$$ ### Cross Belt Drive $$\text{BL} = \frac{\pi(D+d)}2+2L+\frac{(D+d)^2}{4L}$$ ## Contact angle ### Open Belt Drive $$\theta_1+\theta_2 = 360$$ $$\theta_1>180^o\qquad180^o>\theta_2>120^o$$ $$\theta_1 = \frac\pi2+2\phi$$ $$\theta_2 = \frac\pi2-2\phi$$ ### Cross Belt Drive $$\theta_1 = \theta_2 = \frac\pi2+2\phi = \frac\pi2+(\frac{D+d}{2L})$$ ## Speed ratio $$v_1 = \pi(D_1+t)N_1$$ $$v_2 = \pi(D_2+t)N_2$$ $$v_1 = v_2$$ $$\pi(D_1+t)N_1 = \pi(D_2+t)N_2$$ if $t<<1$ $$\frac{N_1}{N_2} = \frac{D_2}{D_1}$$ Consider efficiency $$\frac{N_1}{N_2} = \frac{D_2+t}{D_1+t}(1-\eta)$$ if $t<<1$ $$\frac{N_1}{N_2} = \frac{D_2}{D_1}(1-\eta)$$ ## Power $$F_0 = \frac12(F_1-F_2)$$ $$F_\text{efficient} = F_1-F_2$$ $$F_1>F_0>F_2$$ $$P = Fv = (F_1-F_2)\frac{\pi DN}{60}$$ $$\text{PS} = \frac P{735} = \frac{Fv}{735} = (F_1-F_2)\frac{\pi DN}{735\times60}$$ ## Tensions ratio  </div> $$\frac{F_1}{F_2} = e^{\mu\theta}$$ $$\theta_1-\theta_2 = \theta\qquad-(0)$$ $$f = \mu N$$ $$\tau = Fr$$ $$\sum F_x = 0\qquad\cos(\frac{d\theta}{2})(T+dT)-\mu N-\cos(\frac{d\theta}{2})T = 0\qquad -(1)$$ $$\sum F_y = 0\qquad\sin(\frac{d\theta}{2})(T+dT)+\sin(\frac{d\theta}{2})T-N = 0\qquad -(2)$$ $$\lim_{d\theta\to0}\cos(\frac{d\theta}{2}) = 1 -(3)$$ $$\lim_{d\theta\to0}\sin(\frac{d\theta}{2}) = \frac{d\theta}{2} -(4)$$ $$(3)\Longrightarrow(1)\qquad T+dT-\mu N-T = 0$$ $$dT = \mu N\qquad-(5)$$ $$(4)\Longrightarrow(2)\qquad\frac{d\theta}{2}(T+dT)+\frac{d\theta}{2}T-N = 0$$ $$2(\frac{d\theta}{2}T)+\frac{d\theta}{2}dT-N = 0$$ $$d\theta T = N\qquad-(6)$$ $$(6)\Longrightarrow(5)\qquad dT = \mu d\theta T$$ $$\frac{dT}{T} = \mu d\theta$$ $$\int_{F_1}^{F_2}\frac{dT}{T} = \int_{\theta_1}^{\theta_2}\mu\ d\theta$$ $$\ln(T)|^{F_2}_{F_1} = \mu\ d\theta|^{\theta_2}_{\theta_1}$$ $$\ln(F_1)-\ln(F_2) = \mu\theta$$ $$\ln(\frac{F_2}{F_1}) = \mu\theta$$ $$\frac{F_2}{F_1} = e^{\mu\theta}$$ ## Step pulley ### Open Belt Drive $$\frac{n_x}{N} =\frac{D_x}{d_x}$$ $$\text{BL} = \frac{\pi(D_1+d_1)}2+2L+\frac{(D_1-d_1)^2}{4L}$$ $$\frac{\pi(D_1+d_1)}2+2L+\frac{(D_1-d_1)^2}{4L} = \frac{\pi(D_2+d_2)}2+2L+\frac{(D_2-d_2)^2}{4L}$$ $$\frac{\pi(D_1+d_1)}2+2L+\frac{(D_1-d_1)^2}{4L} = \frac{\pi(D_3+d_3)}2+2L+\frac{(D_3-d_3)^2}{4L}$$ $$\frac{\pi(D_1+d_1)}2+2L+\frac{(D_1-d_1)^2}{4L} = \frac{\pi(D_x+d_x)}2+2L+\frac{(D_x-d_x)^2}{4L}$$ ### Cross Belt Drive $$\frac{n_x}{N} =\frac{D_x}{d_x}$$ $$D_1+d_1 = D_2+d_2$$ $$D_1+d_1 = D_3+d_3$$ $$D_1+d_1 = D_x+d_x$$ ## Equal step pulley $$N_1n_1 = N_2n_2 = N_xn_x$$
×
Sign in
Email
Password
Forgot password
or
By clicking below, you agree to our
terms of service
.
Sign in via Facebook
Sign in via Twitter
Sign in via GitHub
Sign in via Dropbox
Sign in with Wallet
Wallet (
)
Connect another wallet
New to HackMD?
Sign up