the f for oak tanned leather belts on cast iron 𝐹1

= 𝑒 𝑓𝜃
pulley, at the point of slipping is 𝐹2
42.6 angle of contact or lap is
𝑓 = 0.54 − 𝜋
152.6 + 𝑣 𝜃 = (180° − 2𝛼) 180 𝑟𝑎𝑑 for open belt drive
Length of the belt
and
𝐿 = 𝜋𝐷𝑁 𝜋
𝜃 = (180° + 2𝛼) 180 𝑟𝑎𝑑 for cross-belt drive
Velocity ratio
𝑁2 𝐷1 𝐿𝑃𝑄 = 𝑟𝑑𝜃
= 𝑚𝑃𝑄 = 𝑚𝑟𝑑𝜃
𝑁1 𝐷2
When the thickness of the belt is considered 𝑣2
𝑁2 𝐷1 + 𝑡 𝐹𝐻𝐶 = 𝑚𝑟𝑑𝜃 × = 𝑚𝑑𝜃𝑣 2
𝑟
= 𝐹𝐶 = 𝑚𝑣 2
𝑁1 𝐷2 + 𝑡
In case of a compound belt 𝑃 = (𝐹1 − 𝐹2 )𝑣
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑎𝑠𝑡 𝑑𝑟𝑖𝑣𝑒𝑛 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 𝑜𝑓 𝐷𝑟𝑖𝑣𝑒𝑟𝑠 𝐹𝑇1 − 𝐹𝐶 𝐹1
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑓𝑖𝑟𝑠𝑡 𝑑𝑟𝑖𝑣𝑒𝑟
=
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 𝐷𝑟𝑖𝑣𝑒𝑛𝑠 = = 𝑒 𝑓𝜃
𝐹𝑇2 − 𝐹𝐶 𝐹2
velocity of the belt passing over the driver per 𝐹𝑇1 𝐹𝑇1 𝐹1 + 𝐹𝐶
second 𝑆𝑚𝑎𝑥 = = =
𝐴 𝑏𝑡 𝑏𝑡
𝜋𝐷1 𝑁1 𝑠1 𝐹1 𝐹1
𝑣= (1 − ) 𝑆𝑚𝑎𝑥 = =
60 100 𝐴 𝑏𝑡
𝜋𝐷2 𝑁2 𝑠2 𝐹𝑇1 = 3𝐹𝐶
= 𝑣 (1 − )
60 100 𝐹𝑇1 2𝐹𝑇1
𝐹1 = 𝐹𝑇1 − =
3 3
𝑠 = 𝑠1 + 𝑠2
𝐹𝑇1
𝑣=√
𝑁2 𝐷1 + 𝑡 𝑠 3𝑚
= (1 − )
𝑁1 𝐷2 + 𝑡 100 Increase of tension in the tight side= 𝐹1 − 𝐹0
𝑁2 𝐷1 𝐸 + √𝑆2 Increase in the length of the belt on the tight
= × side= 𝜆(𝐹1 − 𝐹0 )
𝑁1 𝐷2 𝐸 + √𝑆1
Decrease in tension in the slack side= (𝐹0 −
For an open belt drive
𝐷2 − 𝐷1 𝐹2 )
𝑠𝑖𝑛𝛼 = Decrease in the length of the belt on the slack
𝐶
𝜋 (𝐷2 − 𝐷1 )2 side= 𝜆(𝐹0 − 𝐹2 )
𝐿𝑏𝑒𝑙𝑡(𝑜𝑝𝑒𝑛) = (𝐷1 + 𝐷2 ) + 2𝐶 + 𝜆(𝐹1 − 𝐹0 ) = 𝜆(𝐹0 − 𝐹2 )
2 4𝐶
Neglecting centrifugal tension,
𝐹1 + 𝐹2
For a cross-belt drive 𝐹0 =
𝐷2 + 𝐷1 2
𝑠𝑖𝑛𝛼 = Considering centrifugal tension,
𝐶 𝐹1 + 𝐹2 + 2𝐹𝐶
𝜋 (𝐷2 + 𝐷1 )2 𝐹0 =
( )
𝐿𝑏𝑒𝑙𝑡(𝑐𝑟𝑜𝑠𝑠) = 𝐷1 + 𝐷2 + 2𝐶 + 2
2 4𝐶
-the effective turning (driving) force at the √𝐹1 + √𝐹2 = 2√𝐹0
circumference of the driven pulley is the 𝐹1
= 𝑒 𝑓𝜃 csc 𝛽
difference between the two tensions (F1 – F2) 𝐹2
𝑁𝑚
-work done per second = (𝐹1 − 𝐹2 )𝑣 𝑠
-power transmitted = (𝐹1 − 𝐹2 )𝑣 𝑊
-torque exerted on the driving pulley and the
driven pulley are (𝐹1 − 𝐹2 )𝑟1 and (𝐹1 − 𝐹2 )𝑟2 ,
respectively