 Applied Mathematics, 2011, 2, 1546-1550 doi:10.4236/am.2011.212220 Published Online December 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM Test of Generating Function and Estimation of Equivalent Radius in Some Weapon Systems and Its Stochastic Simulation Famei Zheng School of Mathematical Science, Huaiyin Normal University, Huai’an, China E-mail: hysyzfm@163.com, 16032@hytc.edu.cn, hssky10@163.com Received November 16, 2011; revised December 6, 2011; accepted December 14, 2011 Abstract We discuss three-dimensional uniform distribution and its property in a sphere; give a method of assessing the tactical and technical indices of cartridge ejection uniformity in some type of weapon systems. Mean-while we obtain the test of generating function and the estimation of equivalent radius. The uniformity of distribution is tested and verified with ω2 test method on the basis of stochastic simulation example. Keywords: Uniform Distribution in a Sphere, Weapon Systems, Generating Function, Equivalent Radius, Stochastic Simulation 1. Introduction Uniform distribution is very important in the probability statistics, many scholars pay attention to it. The follow-ing questions have been explored: the estimate of inter-val length about uniform distribution in [a,b] [1,2], the estimate of regional area about two dimension uniform distribution in a rectangle , the estimate of cuboid volume about three dimension uniform distribution , the estimate of regional area about two-dimensional uni-form distribution in a circle [5,6], estimate of radius on three-dimensional uniform distribution in a sphere . In addition, many scholars get useful test statistics and limit theorems [8-12]. In this paper, basing on some articles [13-18], according to t the indices of cartridge ejection uniformity in some type of weapon systems, we give the test of generating function and the estimation of equiva-lent radius by simulation example. Definition 1 . If (,,)XYZ is three-dimensional continuous random variable, its probability density func-tion is 303,(,,)4π(, ,)0(,,),.xyz GRfxyzxyz G (1.1) where 222200(, ,),0Gxyzxyz RR, then we call that (,,)XYZ obeys uniform distribution in 22220(, ,)GxyzxyzR, recorded as (X,Y,Z)~U(G). Give a transformation 0sin cossin sin(0,0π,0 2π)cosxryr rRzr (1.2) The probability density function of three-dimensional r.v.(,, )R is (, ,)( ,,)(sincos,sinsin,cos)(, , )xyzhrf rrrr  (1.3) in which 00,0π,0 2π,rR(, ,)(, , )xyzr is Jacobi determinant of the transformation (1.2), and 2(, ,)(, , )sin coscos cossin sinsinsincos sinsincoscos sin0sinxxxrxyzy yyrrzzzrrrrrrr     (1.4) F. M. ZHENG1547 Therefore the probability density function of (, , )R is 20303sin,0,0 π,0 2π(, , )4π0, otherwiserrRhr R  (1.5) Theorem 1. If the marginal density functions of about are r.v.(, , )R ,,R 1(),hr 2(),h 3()h, then 1) 203103,0()0, otherwise.rrRhr R, 2) 21sin ,0π,() 20, otherwiseh. 3) 31,0 2π,() 2π0, otherwise.h Proof. According to (1.5) and the definition of mar-ginal density function, we have π2π10022π2ππ3300 0002233000()(,, )d d3sin 3dd 2πsin d4π4π332π2,4πwhere 0,hr hrrrRRrrRRrR    00022π2π2300 00023003sin( )(,, )dddd4π3sin 12πdsin,24πwhere 0π,RRRrhhrr RrrRr    0002ππ3300 000π2303300003sin( )(,,)dddd4π33dsind 234π4π1,2πwhere 02π.RRRrhhrr Rrr RRR1r    Corolla ry 1 . If r.v.(, , )R is defined by (1.5), then three are independent each other. r.v.,,RCorollary 2. If is defined by (1.5), the marginal distribution function of about are r.v.12(),H(, , )R3(),( )Hr.v.(, , )R,,R Hr, then 3103000, 0() ,01,rrHrrRrRR, 20, 01()(1 cos ),0π21, πH, 30, 0(),02π2π1,2 πH. Proof. According to theorem 1, we can get it easily. Corollary 3. If ()ER, 2()Var R, then the probability of cartridges falling into a ball with radius  is about 42.2%, and the probability of cartridges fal-ling into a sphere with radius  is about 84.0%. Proof. By the definition of Mathematical expectation, we have 0103()()d 4RERrh rrR 0 (1.6) 0042210300033()()dd 5RRrERrh rrrRR 2 (1.7) then 222220039 3()()()51680DRERE RRRR  20 and 01520 R, then the probability of cartridges fal- ling into a sphere with radius  is about 11 103() ()42.2%4HHERHR (1.8) then the probability of cartridges falling into a sphere with radius  is about 110031584.0%420HHRR  (1.9) 2. Test of Generating Distribution Function Usually there are 2 test method, 2 test method and Cole Moge Rove test method (K test method)  to test distribution function. Here, we use 2 test method, we want to know the sub-sample is uniform distribution or not. Because the locations of any cartridges are ascer-tained by three-dimensional , so we should r.v.(, ,)RCopyright © 2011 SciRes. AM F. M. ZHENG 1548 ))test them one by one, test 1, ~(RHr 2~(H, 3~(H). We give testing hypotheses 0H 00():()HFx x where ()Fx is generating distribution function, 0()x is known distribution function, and 0()x is the deriva-tive of 0()x,,yy. Tests for generating function should be independent, (1)(2)() is the sequent sub-sample of the test, under hypotheses ,ny0H is correct, the statistic 22121112nin0()() 2iiynn  (2.1) is Smirnov distribution. For the given confidence level , according to the Table 10 in , we obtain the boundary value z of 2n, in which 2(Pn )z . Then is rejection region of the hypotheses 0(,)zH, when 2nz, we reject 0H, if 2nz, we should accept 0H. 3. Estimation of Equivalent Radius On the supposition that N is the number of cartridges from a shrapnel, n is the actual observed number of car-tridges within a certain region near the centre of disper-sion. When calculating equivalent radius, we presume all the cartridges are found. The distances from any car-tridges to the dispersion centre point A are recorded as (1,2,iri,)n, let 11nir()hrirn. According to the pro- perties of density function 1, we know that R obeys uniform distribution in a ball with radius 00(0nnrr0)R, 03() 4nER r (by 1.6), owing to 0230110311() nnnriiiiinrEr Errrnn r 034indr (3.1) so 0 is a unbiased estimate of 0n . on the basis of the properties of distribution function, let ˆnr rπ,2π, we have π()(, , )2πHrHr , let , (N is Namount of test cartridges )，then 3030nNrRπlim( ,,)lim2πNnHr N   (3.2) Therefore let 0ˆˆnNRn0r, that is 0431ˆniiNRrnn, and 0216ˆ() ()9iNDR DrnAs well as by (1.7), (3.3) 203() 80inDr rsubstitute into (3.3，), we obtain 2016() NrNNDR 20022 20016 3ˆ() ,8099 1515ˆ() 15ninnDr rnn nNrRn. (3.4) Based on Formula (3.4), if N is large enough, , becomes small enough. In order to ss of above methods, we some type of weapon when nNle size 0ˆ()Rlarge edecrease estimating error of equivalent radius 0R, sam-pis nough. 4. Stochastic Simulation In order to verify the correctneive a simulation example. For gsystems, assuming the tactical technical requirements, the cartridges from single shrapnel should be more evenly scattered in a ball with equivalent radius (120 ± 20) m, launching a shrapnel, the number of cartridges is N = 400, measuring the coordinates of one hundred car-tridges near the dispersion centre (n = 100), they were produced by computer simulation basing on uniform requirements in a sphere, i.e. (, , )r were produced by stochastic simulation according to the following for-mulas 310 2(),arccos12( ),nrHrr H32π()H (4.1) where, 060nr, 1()Hr, 2()H, 3()H are number (0,1for cartres asow Table 1, a*sqrt(r(1)); 2*r(2)+1); 0 (2*k-1)/200))^2; ((2*k-1)/200))^2; )^2; :100 sum(a); m(b); random ), coor-produced by stochastic simulation indinates idg belnd the MAT-LAB program as below, >> clear for k=1:100 nd(1,3); r=ra x=60 y=acos(- z=2*pi*r(3); [x, y, z] end >>clear 1:10for k=a=(x^2-(b=(y^2/3600-c=(z^2/3600-((2*k-1)/200)[a, b, c] end ar >> cleor k=1fs=1/1200+u=1/1200+suv=1/1200+sum(c); [s u v] End Copyright © 2011 SciRes. AM F. M. ZHENG Copyright © 2011 SciRes. AM 1549Table 1. Polar coordinates points of fall for cartridges. (, ,rm rad rad )(,, )rm rad rad (,, )rm rad rad (, ,rm rad rad )57.8262 1.1283 0.9007 1.7096 0.04390.4464 5.7960 0.6985 5.0936 56.0811 55.2025 47.5996 58.8251 1.3574 1.8961 1.4923 1.45132.1067 5.1923 2.1410 3.8108 33.9503 43.7655 33.0668 41.6750 1.5768 5.1315 19.1279 1.4612 4.8952 45.69072.4810 0.7106 48.3936 1.3183 6.2367 45.5454 2.0309 6.1921 54.1207 0.6013 1.9308 47.07931.1287 5.1026 51.7387 2.2742 2.3455 46.6895 1.1728 0.1093 48.7537 1.4576 5.8226 50.83271.0500 5.7070 15.1818 0.4352 3.3389 31.8301 0.6831 5.1484 39.0650 1.3001 4.2644 33.62452.6167 0.9827 56.5426 1.7679 1.1391 30.8478 48.6323 1.1.4572 3.5043 9025 3.5198 14.43.0160 0717 2.1.1647 0.3512 4668 7.0.4442 55.5595 0.47077407 0.2.6500 7672 4.7922 53.55.7802 2184 1.2.6869 1347 3.1535 2.6528 53.9209 0.2431 1.5331 57.6967 2.1165 0.0748 48.04991.9865 4.5352 46.3803 2.4616 4.1494 44.1604 1.9052 5.1648 57.1905 2.6896 1.4275 35.84382.3398 4.0941 26.1303 0.6488 4.2330 42.6194 2.0355 1.6537 38.0081 1.6869 3.2440 28.16160.9503 4.7375 56.1754 0.5168 6.0149 39.4989 1.57.2477 1.1188 0742 4.7350 49.4.1444 32.7335 0.5254 1.2389 7644 2.8790 32.4.4183 44.388748301.4809 0.5776 4.1670 34.5.5512 45.6740 1.8214 2.0105 6183 1.2057 0.6987 51.3376 2.0009 1.3452 50.4480 2.2554 3.6600 44.49392.3493 1.7103 14.1080 0.5078 3.5506 37.3488 2.1746 3.7831 41.5086 2.8378 3.1994 22.49511.6951 2.6352 40.5506 1.0797 6.0897 59.5588 2.9061 3.8007 52.2026 0.9810 0.4668 58.81271.2010 1.3383 57.3967 3.0969 0.1489 51.7199 1.36.7509 2.5174 5075 4.1438 57.1.1523 49.1376 1.6781 1.9904 6150 1.2139 31.2.3851 43.879862601.3179 2.3973 0.2237 56.0.5102 41.5065 0.5920 1.9570 5676 5.4676 0.1690 52.7956 1.4728 3.9992 59.6074 2.6175 1.7367 40.65541.3123 5.3445 57.5123 1.1385 3.2641 52.3921 2.2256 1.0700 55.5081 2.0117 4.8437 33.13990.5499 2.1375 46.9686 1.9237 1.2083 30.7568 2.2887 3.3904 32.0629 0.9956 1.9723 57.85630.9268 2.9292 49.4790 2.7288 4.4969 16.9386 0.38.4027 1.8401 3570 3.9169 56.4.3096 34.4546 1.6080 0.4698 8558 4.0099 41.6.1990 54.154512320.4487 1.7046 5.7416 51.1.4363 52.0491 2.2896 1.1332 9094 1.5752 5.8679 29.3040 1.6124 4.2556 51.6795 2.7865 3.1598 44.60630.7133 5.4161 51.0684 0.7404 0.8621 24.6450 2.0222 5.5091 52.4759 1.2780 5.9546 44.20101.6150 4.1255 52.8935 0.6280 3.2773 in d s iaper, using above MATLAB program, we obtain Take conspicuous level Accordg to the data in Table 1 anmethodn this p222 0.0888, 0.0425,0.0749rnnn00ˆˆ117.4909, ()3.0984RR 10%, seeing the Table in , we obtain the boue 10 ndary valu0.3472z2 for n. 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Andreytseva, “Independence of the Rewith Noncentral χ2-Distribution,” Applied Mathematics, Vol. 2, No. 2, 2011, pp. 1303-1308. Copyright © 2011 SciRes. AM