OCR Text |
Show AR = C. M • g • \ i • cost^ u = M * g ( s/ 3)[ ( 3h0 + 2 A h)/( 2h0 + Ah)] [ A cos + u c) Into flow work: Per unit area of sliding surface the fl ow resistance is: Y v / £ The velocity diminishes successively from v0 to 0; the average of its square is approximately v0 2/ 2. The work performed in the runout distance s is : Av = ( Y / 5 ) ( v0 2/ 2) s C d) With increasing disintegration the resistance of the individual snow particle contributes noticeably to a portion of the work Ac: As = ( M g / Y F) ( Y L v 2/ 2g) From EK = Ep + AR + Ay + Ag , with hm = „ + Ah/ 2 and h m a x - hQ = A h _ v 2/ 2g and neglecting the usually small reduction in the height of damming due to the frictional resistance over the relatively short damming distance, it follows that: s = [ v2 /( 2g( h + 2 A h/ 3)) ( 3h / 2 + 2 A h/ 3 - 2 YT / Y -, h ] u Q o ' L F m ( fi'cos tu - tan % + v z / 2 £ hm ) or ( 23) S= V2/ f0„ fll. ™ co1, + ~„, i, % _ , ,2 s ~ v / C 2g ([ i cos^ u - tan % ) + v g/ ^ ] > If the term in parentheses in the denominator becomes = 0, the avalanche comes to rest on the slope of the valley floor being considered. If however, the term in parentheses is < 0, the avalanche does not come to rest in the terrain being considered and for this sin* = ( 1/ 2J [ ( 1 + 4 j i 2 ) 1 / 2 - 1 ] for small values of v or tanfu _>. v2^ g/£ ( 2hQg + vd) for p ~ 0 In this case the above equation for s gives approximately the transition distance until normal flow again occurs and where h, = hn ( sin\ K, V sint ) 1 / 3 39 |