Calculation of CCP position in inclined layer
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Compared with horizontal CCP calculation, slant CCP calculation also needs slant angle? And normal depth d (Fig. 1), s point shooting, R point receiving, and center point P. assuming offset is x, then SP length is, conversion point is C, and the distance CP between CCP point and CDP point is XP -. Let P-wave velocity be VP and S-wave velocity be vs. according to Snell's law, = (1)?? according to the trigonometric relation, = (1) a / cos (a) = XP / sin (I) B / cos (a) = XS / sin (J), a / b * VP / vs = XP / XS is obtained ……… (2) From the trigonometric relation a * a = XP * XP + h * H + 2 * h * XP * sin (a) b * b = XS * XS + h * H-2 * h * XS * sin (a), where h = D - (xp-x / 2) * sin (a) XS = x-xp, we can obtain the ternary equation of XP, a, B. When D is infinite, there is a = B. from (2) and XP + XS = x, we get XS = x / (1 + VP / VS), which is the position of the so-called asymptotic transition point. It is obvious that the above five formulas are a standard iterative scheme. Generally, it is easy to converge with XS as the engine, and the initial value of XS can be set as the location of the asymptotic transition point. But considering the problem of convergence and accuracy, the author tries to get the exact analytic solution of XP by solving the quartic equation directly. Let d = D / x, v = vs / VP, g = sin (a), x = XP / x, H = H / X change the iterative equation to S2 (x2 + H2 + 2xhg) = X2 (S2 + h2-2shg) V2, where h = D - (x-0.5) gs = 1-x, s and h, divide both sides of the equation by T = (1-G2) (1-v2), and get the equation x4-2x3 + px2-2 * Q * x + q = 0, where q = (D + G / 2) 2 / T, P = 1 + q-v2 (D-G / 2) 2 / T, the solution of the equation is x = 1 / 2 ± (* r ±) / 6, where u = 1 / 2=