Scenes And Incidents From The Life Of A Practical Miner
With A Treatise On The Ventilation Of Coal Mines

TO FIND AC. As 90°10-00000 is to 125 = 2-09091 bo is G3<>= 3-94938 DIAGRAM I. To FIND BO. As 90° = 10-00000 is to 125 — 2-098l)l so is 27°-• 9-CS706 To FIND A B. As 63°= 9'94988 is to 111-4= 2'0467» 80 is 90° 10-00000 yards 125= 2-096S1 yds. 56’75= 1’76396 yd. 111*4 = 2-0467V COAL DIAGRAM II. Wanting to prolong the inaccessible line O D, in order to place a mortar battery bahind the obstacle C In th odirection of OD.l measured a base line A 1J-- 414*2 yards; at AI took the angles CAD—49 13 , OA.bL 12:j°45', and the angle 0AE=100°,having first set us a pole at Ts, behind the obstacle. At B (took the angles RBC = 33045', and L>BA=67»30'; hence it is required to mid the angle A EF, which will determine the position of OF with respect to OD. 0= 22°30'—9'58281 414-2 =2-61721 33045' --=9-74474 12-36195 CA 9 58284 sd 601-3 = 277911 ] SO»—101 ° 15' = -= to 78°45'=9-9f)ir>7 900=2'95424 33°45'=9-74474 12'69898 CD 9-09157 sd 509-8 -2'70741 180°—' C=22°30'—3'58284 414-2 =2-61721 •123°45'=E6O16'=9,91$»85 37058=9-78602 414'2—2'61721 G7°30 yOß-562 12’53706 CB 9-58284 12-58283 DA 9-7890-2 sd 900—2'95182 sd 662~2-79381 33045'-=9-74474 509 8=2'70740 450 0'= 9-84949 12°30' 9 33534 (J01-3=r2-77909 100" 0'=9’99335 12'55689 DB 974474 12-77244 CE 9’33534 sd 648'9—2*81216 sd 2521^3-43710 12°30' 0*38684 601-3 2-77909 67°30' -9'90562 ACD—C7°30', then 12-74471 AT? P-33634 Find AC=601-3 AI) = 622 and the aflgle--- , AO I) + C A E = A E F 167° 30', if therefore the angle AEF.be made=sl67° 30', the point S’ will be determined in the same line with O D. sd 2567=W0937