Mine Development & Surveying

Methods of Access to Deposits

Vertical shafts, inclines and adits — and the geometry that sets how long an inclined drivage must be to reach a given depth.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A deposit is reached by a vertical shaft (fastest descent, costly winding), an incline/decline (gentle gradient for conveyors and trackless equipment), or a horizontal adit (where terrain allows gravity drainage). For an inclined access the length is fixed by simple right-triangle geometry: to gain a vertical depth

H

at an inclination

θ

, the inclined length is

L = H / sin θ

. Stated as a gradient '1 in

n

', the horizontal run is

n

times the rise, and the true length is the hypotenuse.

The Core Formula Matrix

Inclined length from angle:

L = \frac{H}{sin θ}

(

H

= vertical depth,

θ

= inclination).

**Gradient 1 in

n

**: horizontal run

= n H

; inclined length

L = H^{2} + (n H)^{2} = H 1 + n^{2}

.

Shaft sinking time

= \frac{depth}{sinking rate}

The ‘IIT Traps’

⚠
**'1 in $n$ ' is a gradient, not an angle.** It means rise:run $= 1 : n$ , so $tan θ = 1/ n$ — convert before using $sin θ$ .
⚠
Inclined length uses the hypotenuse. $L = H / sin θ > H$ ; using $H$ alone underestimates the drivage.
⚠
Adits need favourable topography. They only work where the deposit outcrops above valley level.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

An incline at

3 0^{\circ}

to the horizontal must reach a vertical depth of

150 m

. Its length is:

Q2MEDIUM2 Marks · NAT

A decline of gradient

1

4

is driven to reach a vertical depth of

100 m

. Its length is ______ m. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

A conveyor decline at gradient

1

5

reaches a vertical depth of

200 m

. The length of the decline is ______ m. (Round off to two decimal places.)