Ranga,

If I understand your question, I dont see how it is any different to a "typical" head end arrangement drive, just reversed. Drive at the "tail" and take up at the head.

As usual:

Te will be a sum of the friction (material, idler and burden mass multiplied by an artificial coefficient of frction) + material lift if appropriate (and scrapers etc).

T2 will be at least as calculated using Eulers (or as required to satisfiy sag etc)

T3 (if you like at the tail end) will be T2 required + return side empty friction+, for starting, return belt and idler mass mutiplied by system acceleration).

Assuming that it is a "fixed type" take up, you will need to calculated the additional pre load required to overcome the belt stretch. I understand this is typically calculated by assuming a "fixed energy" system or assuming the average tension is constant under stationary, running and starting conditions.

These are general notes and are not comprehensive. ■

Hi Ranga,

Te is the total tension for conveyor to get out from the "rest", and it´s always the same for any situation (drive at head or tail).

T2 is the minimum tension for the guarantee of the friction on the drive pulley + tension due the sag.

T1 = Te + T2 (drive at head or tail)

In the case of a tail drive what changes is your Ht, that is equal to: T2 + friction on carry run + material and belt elevation on carry run

Regards,

Emerson ■