Question

1. Your Company is contemplating mining a perfectly horizontal tabular orebody having a true height of 10 m. Given the information shown on the pit section below, calculate the overall stripping ratio for a cut (i.e. into the page) 500 m wide. Assume the densities of the ore and the waste are identical. (7 points)

Average slope 39% Overburden Ore 1550 m Not to scale

please type your answer, Please do not write on the paper and post it

Answer 1

For the ore body thickness=10m

In ABDE section we need to find the area. So length of AE=1550m

In triangle ABF, area=1/2(10×(10/tan(BAF angle)))=61.74m^2

In triangle DGE area=1/2(10×(10/tan(GDE angle)))=37.67 m^2

AF length=10/tan(BAF angle)=12.34 m

GD length=10/tan(GDE angle)=7.53 m

Length of BG=FE=1550-12.34=1537.66 m

Area of ore= area of ABF+area OF BGEF+ area of DGE

61.74+(1537.66×10)+37.67=15476.01 m^2

Ore volume= area ×thickness=15476.01×500=7738005 m^3

Over burden volume = area of BCD ×thickness

In BCD area =area of BDH + CDH

Area of BDH =1/2(BH length×Hd length)

BH= BD×cos39=1200.838 m

HD length=BD×sin 39=972.41 m  

Area of BDH= 1/2(1200.838×972.41)=2335436.76 m^2

In CDH triangle CH=HD/tan 14=242.449 m

Area of CDH=1/2(242.449×972.41)=471519.748 m^2

Volume of BCD = (Area of BDH+ are of CDH) ×500= (2335436.76×471519.748)×500=1.403×10^9 m^3

So stripping ratio= volume of OB / Volume of ore =1.403×10^9/7738005=181.37