The location of the apex of a stockpile with respect to the discharge pulley of the conveyor
feeding the pile is an important aspect of plant layout that is often overlooked or simply
guessed. This is especially so when the initial discharge onto the empty pile terrace is at an
elevation of several meters.

All stockpiles start out at terrace level and may or may not be emptied completely. Indeed,
even if the pile is not completely recovered, the location of the crater formed by the stockpile withdrawal system may compromise a reasonable feed into the extraction system.

Figure 1 – Conventional Conical Pile. Note the crater rim as the pile is re-filled

For example, when the withdrawal conveyor is in line with the stockpile feed conveyor, the crater formed by extraction will naturally lie on the centre-lines of the extraction and feed conveyors. On the other hand, when the line of the extraction conveyor is at an angle to the centre-line of the feed conveyor, the location of the apex of the full pile will have to be located such that the impact point of the discharged material at terrace level effectively coincides with the centre-line of the extraction system. If this is not the case, then the re-filling of the crater will be less efficient and the stockpile live capacity may be compromised.

For example, for extended stockpiles with several extraction points, the only practical feed would probably be by means of a travelling stacker with a boom conveyor, in order to maintain an elongated pile. In this case, the line of the extraction conveyor will have to be in line with the apex of the stockpile.

For the location of the apex of the stockpile, it is assumed that the normal trajectory calculations have been carried out, in order to derive the drop point, the release angle and centre of area of the material as it passes over the discharge pulley.

There are effectively 4 cases that need to be considered.

### Case 1 – Inclined conveyor with the drop point at ‘C’

Values obtained are as follows

With respect to the sketch, h = h1 – h3  (m)

(m)

(m)  and L = S·t  (m)

i.e

Hence, seconds, which may be written as

seconds, using the positive value.

Since L=S·t, it follows that (m). This is the normal requirement.

When ℓ is given, it follows that,  m

### Case 2  –  Inclined conveyor with the drop point at ‘T’

Values obtained are as follows

(m)

(m)   and L = S·t  (m)

i.e.

Hence, seconds,

Using the positive value.

Since L=S·t, it follows that   (m). This is the normal requirement.

When ℓ is given, it follows that   (m)

### Case 3  –  Horizontal conveyor with the drop point at ‘T’

Values obtained are as follows

Note that h2 = RT   (m)

Then   (m)

seconds, using positive values

Since L=S·t, it follows that  ℓ = L  (m). This is the normal requirement.

When ℓ (= L) is given, it follows that L=S·t, and   seconds

Hence,    (m)

### Case 4  –  Horizontal conveyor with the drop point at ‘C’

In this case, the approach is identical to case 1, when the belt approach angle α = 0°. Since the approach angle is used in the determination of the drop point, once the drop point is determined with a positive value of angle θ, the approach angle α is no longer relevant.

Graham Shortt
brevissg@telkomsa.net