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美赛数模论文之公式写作

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    慵懒
    2020-7-12 09:52
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    发表于 2020-2-12 17:14 |只看该作者 |正序浏览
    |招呼Ta 关注Ta
    由假设得到公式% p2 H% t; A) v9 {
    1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)6 Z" u9 @' Q" {* F; z
    ) _; _* I; p/ R& U
    公式
    ! x1 ^% I- E5 b9 e' q9 J6 m6 m' Y: Z* [7 ]. k, f5 k
    Where# E* P( p& f) _- |
    1 V" o* v  o( J# h* H  E+ m7 X
    符号解释: N' a$ y$ X" V4 J+ G/ `$ g+ @
    7 M2 R$ s% M  w3 b
    According to the assumptions, at every junction we have (由于假设)
    8 i: B( F3 i+ D* K. v3 u1 N, O" n5 c. v% U7 Y
    公式
    ; m# O1 Y0 v0 B5 I5 \+ }3 _( Y+ n) g+ G1 i! K- N
    由原因得到公式
    6 Y6 L8 s1 S( }5 i! p% J2.Because our field is flat, we have公式, so the height of our source relative to our sprinklers does not affect the exit speed v2 (由原因得到的公式);+ N0 t; _8 E0 Q9 R( ^8 z1 f! l
    ( }+ N+ b* p, Y/ y: u$ y2 C7 P6 A; i
    公式9 D) r0 }* V( L" e& k6 d
    : v$ F$ v$ f2 r
    Since the fluid is incompressible(由于液体是不可压缩的), we have$ a# E& q0 I: L0 E7 R" G
    8 X1 @5 e& f$ R' O0 b: X" x
    公式4 Z0 L0 {/ ?* H- R4 z& y" J

    $ k/ |4 V' ^( B; ~  H0 u' qWhere2 ^& ]/ I, i& Y1 B* v
    & z% j; R- S0 s! @$ j8 ~; o$ w7 s
    公式
    4 U) b" u- f3 J: k. G& l: V+ L
    $ L  K+ H2 `4 o2 s+ ^/ k用原来的公式推出公式* A4 \! G  ?2 @6 F% \
    3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到)* E6 i" r; U) I9 D. D; P$ n
    2 ]' P- K2 |) S8 J3 z3 q
    公式% v6 `" f- c% b- `  h+ l& P- B
    ' Z; ]1 j$ v4 d4 E1 w
    11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:0 C  t  S8 t6 m5 M3 V, N5 B1 L
    / _  q) g! A1 z& G( l6 T  J
    公式$ r4 W, ^4 ~7 l1 d, j
    / {8 o" k5 p7 |* o0 J' a9 G
    12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)5 ~7 p, W6 j, d8 b# v
    / o3 Y$ m1 j6 ]' J  x7 J
    公式; B1 Q4 ]9 v7 E( Q
    1 [5 E' W' c8 ~6 E5 C: L5 {
    Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
    - j$ L/ U+ u, F; W, r1 v! b0 l& K( G& K0 w) O3 F
    公式
    - m& ]7 k# T+ j, g' w# {
    1 Z0 i. s; p& B  U  a) x2 Q3 x: L; @Putting these into (1) ,we get(把这些公式代入1中)
    . O# }1 {: H' M, B! I( j3 P# J1 H2 r& z& w! ]
    公式- I- s. {7 t0 t6 {3 n

    - K7 t1 J; \. m1 w1 ]Which means that the5 p5 k5 U6 `" w4 [$ x( z
    . k! E* d5 A! Y: P: B; ?, Y* h! F
    Commonly, h is about" N' J* I/ N8 I9 i" g
    . `7 O, e) I3 C4 B4 s
    From these equations, (从这个公式中我们知道)we know that ………
    0 O5 `5 Y* q# e! j  o! R' ^. l. i4 v
     
    * c+ N4 Q) D/ T8 j$ X& a
    ( O/ u  F& m0 L/ g; e. a引出约束条件' y( A0 i: _: B& [
    4.Using pressure and discharge data from Rain Bird 结果,
      |: V; t- I  x& d
    , A8 H/ ~7 R- G/ QWe find the attenuation factor (得到衰减因子,常数,系数) to be
    3 @; J/ l  L: A: k3 S
    6 B4 V, D, R% h) J公式
    # @3 M0 |- l1 I' R* M; v6 k) \, B$ K1 E( z. o4 z
    计算结果3 O8 X9 o( J# q7 @2 u
    6.To find the new pressure ,we use the ( 0 0),which states that the volume of water flowing in equals the volume of water flowing out : (为了找到新值,我们用什么方程)
    # N/ q/ u* Y3 U: k
    8 J. C/ z# V! @0 h! X公式1 g( Z8 a2 q7 `% B. D7 P
    / b& I' ~0 A1 J/ ~
    Where
    ) i  U3 c6 e* M  N; u5 y5 P
    . j2 j: A. M& a  O() is ;;1 Q; T  E3 Z) j  M0 y: d4 n& w

    2 j: L/ X& Y& J# Z( G7.Solving for VN we obtain (公式的解)/ i# k, l& U- T9 E5 }+ E, W; d
    2 t. E. e; S# V
    公式" X' J. p% R& u4 l
    ' N! Y4 o+ }' a6 R# I$ t' F6 _
    Where n is the …..
    , s4 f. C6 j. d: k
    / C: D; z" a* w$ `% z6 |# s- \ $ ?! s6 X& n1 a0 M
    / d# M8 e% K9 K
    8.We have the following differential equations for speeds in the x- and y- directions:1 y% ~  F9 k/ k' t5 w0 Q$ p: E

    7 b0 s0 {8 e  `( k" ^' ]$ q公式# \8 ^! e( S$ I' a. G8 G
    8 k2 e& r/ d% c) d3 |
    Whose solutions are (解)
    - q5 B& T& ?0 L% `# i: Z0 ]+ {3 B3 N9 }# T, R: h0 L
    公式
    * D- T9 f  v5 f3 V7 Q3 Y/ V" h! W; b
    # k1 W2 z# \  L1 x9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:
    + V4 G0 Z+ B7 N, S; P- y" C, U0 h" T
    5 N/ J  D1 T* z: h公式
    3 k# U3 q. ?5 B& P( i
    & Y! ^5 B7 D. K! A" |7 [6 ~, Q根据原有公式
    1 P+ d4 G/ w. _# ^10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is: Q! G  }& n. [! j8 U' @* O
    $ y7 e4 l$ ]/ s3 C
    公式
    + A( b3 i6 A. a# a% B
    " L7 b, o  }( a7 B! Z+ GThe decrease in potential energy is (势能的减少)
    1 V/ S& @4 Y0 M: w  g' C2 P, K
    ' q  `( x7 Y! E" F2 p公式
    ' m" E, i* b5 Q9 a- u4 y9 `# u
    The increase in kinetic energy is (动能的增加)6 r9 G) \% E$ e. I$ M8 p2 T
    4 p  O, a% ^" p+ L2 t+ ?7 d+ g
    公式
    . e" Y9 l5 V" D5 L) A! g' V) H& r5 W* v7 k& P5 g
    Drug acts directly against velocity, so the acceleration vector from drag can be found Newton's law F=ma as : (牛顿第二定律)
    2 s7 r( x* Q; X: W7 y. h( g9 t1 Y- e6 q2 i
    Where a is the acceleration vector and m is mass
    * u5 n3 z" E! T
    9 ^- q' g' k3 \( I, N 3 N. E% [. l! L" \3 x6 q

    + l' k% X( h2 l4 S0 eUsing the Newton's Second Law, we have that F/m=a and
    5 y" m% x/ s# J3 ~. p$ H) R7 ^
    / _) [3 x/ S7 A6 w4 E/ }公式$ `- O$ Z# `) ~5 V! k( X9 X
    / U  H! ~( Y9 R1 ^, E( }! F
    So that8 ?! I/ A, H" ~) G) Q. u6 f& i! E
    ( _+ O. ]  s% |
    公式
    $ D4 F$ J" Z+ H1 B/ {* Y1 }) c
    0 }0 C/ g) Y$ Q5 `Setting the two expressions for t1/t2 equal and cross-multiplying gives
    6 y. G$ C, o+ j" W) d
    4 I4 w+ W. K/ X0 S3 U( o公式
    6 X. d8 N* d/ k5 I- j# n6 r6 j: e$ t* z' ]+ k
    22.We approximate the binomial distribution of contenders with a normal distribution:2 w- s- f) K8 i8 Z

    8 m) P; Y' c! s( m: [公式: p  B9 L7 t% ], `' j! M: M7 d% |

    % _4 m: h! k0 g1 DWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
    * J( R6 i) y4 ~! ?& h" `/ Q0 K5 k+ U. @& [
    公式, |. P& a0 Y. K

    5 o4 z0 b6 @3 o; @$ S0 y9 n2 r! u# R- NAs an analytic approximation to . for k=1, we get B=c
      U) A) O1 {* u+ `. Q" o6 r/ [! l% Z  V( Y; w! N
     + n7 z5 p7 ~' Q  w
    8 M. ^  y, {5 N6 h5 t
    26.Integrating, (使结合)we get PVT=constant, where- c4 Y' o; y$ _

    . F. D' N# s" g! X8 X! m* I& Y* S公式$ t7 ^7 Z0 _3 ]+ _8 l( ]
      A- J& I  F* W, x
    The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so6 R) b7 H7 J. U* U7 Y8 r; F5 R
    3 m' a; s4 I' P
     - u- g! ?  f- s' O
    2 A. K! w) t$ R1 @' j& }/ s
    23.According to First Law of Thermodynamics, we get
    ! {& {# p# t3 f6 f: W. ?) a, e2 b2 _9 T: K- I) `
    公式
    6 @1 k( J; T8 S1 q% P6 J) h( ^* v. t& d# V% a
    Where ( ) . we also then have2 L' ]$ u7 d# t0 W6 V$ L3 W

    ! _/ q0 [) z- f' J, I公式5 i1 Y8 n2 i4 _8 S3 [2 d$ c, S
    , N; o$ B' W" \4 {0 J
    Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
    " K3 o! X4 z3 G- f: ^$ m7 N5 b5 n. m( S3 K: A
    公式
    4 Z0 j! _  U! ?- [" W6 B6 }- v3 k( O
    Where9 F1 L& _9 s3 R: B

    ! ^: g/ P) @: [8 L5 M" d4 P 
    , v3 `3 J" W, {4 r/ ?
    ) g) U6 l5 N- F: a6 @对公式变形+ p2 ^$ V( W7 f
    13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)
    6 E+ H3 K5 b: t8 C3 e6 y
    0 T. W/ ^) X8 }. e, c9 g( `公式
    & S' g* z7 p7 T* s2 ~' a* f
    " Y  {% M0 L7 V* i* L9 `We maximize E for each layer, subject to the constraint (2). The calculations are easier if we minimize 1/E.(为了得到最大值,求他倒数的最小值) Neglecting constant factors (忽略常数), we minimize
    3 O7 j( ~. {- |* Y: [" f6 ^) F2 F6 s9 a, x" I+ c& t: F6 S1 E; W
    公式$ U' d+ {1 r  z8 D
    & R* G* ]! q: \, X. h% B$ Y
    使服从约束条件1 }* \' e# }% {1 S5 ~
    14.Subject to the constraint (使服从约束条件)
    3 b; y) |- b" _  |8 Q8 ?0 o+ _% z. {' n3 D
    公式
    5 T) w3 `# e( c! o7 L: W
    & b2 B8 j* s, X) mWhere B is constant defined in (2). However, as long as we are obeying this constraint, we can write (根据约束条件我们得到)! k7 X4 s" W& g$ @1 N

    5 q  h) H& w4 S; k, i6 g( B2 j  h! |公式
    + B. @. r& z, g* ~+ s* h- K% T, V9 E8 I* F7 C4 d  o# o
    And thus f depends only on h , the function f is minimized at (求最小值)
    & g3 F' m4 f9 g! @. b% {9 T8 f2 [# v6 B
    公式
    ( Y- e0 W/ T: ?8 V! q( W* y- A3 R  U& M" y8 q
    At this value of h, the constraint reduces to
    6 z7 \" o7 R/ x, q1 `" z9 o2 Y. C, M5 w! E8 s
    公式( l3 T* Q* {- f- {2 L, _- [$ p

    2 O- B3 y3 {4 e$ @2 a结果说明; B" H) j2 e% o$ [# z  r) V4 ~( a
    15.This implies(暗示) that the harmonic mean of l and w should be* R( o' {2 c' j' ]. _) @

    ! N" w3 |8 S, @0 j- G: [, v" K公式
    - j# ]' m+ i& L( V
    / l) r- g( z0 d! s0 ~: F& ?2 {So , in the optimal situation. ………5 s$ N9 u3 I$ C0 O. X5 t7 ]7 {  P; W

    8 |# A5 [; T1 l4 Z/ x; z  x5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is, v6 }6 y, _5 g" _

    ' ~' h2 @2 p5 Z% f) u/ I. L5 a公式" E; _/ K$ H3 r7 E+ \$ c/ r' |5 v
    5 U) y8 k6 C7 y% G8 m/ y
    16. We use a similar process to find the position of the droplet, resulting in* U6 E9 @0 |. F9 a

    : r) h$ F7 e( m6 y5 a公式
    : C5 T0 A) w' S  w
    " E3 k0 e5 u" r' K+ G1 l. A0 pWith t=0.0001 s, error from the approximation is virtually zero.
    4 n7 T1 L! e1 V1 M2 y7 A$ H6 \, c& S/ Y3 O  q
     
    ( b2 P) R; w8 O! [/ A- E$ B% z" d( O$ K( X
    17.We calculated its trajectory(轨道) using  g7 k! b# Z/ p2 p6 |& t( w* e

    4 [; ~) G$ U; Q0 O公式
    * m7 ~3 x! W1 \' I" _1 C; t; f1 v. c) S: ^* S
    18.For that case, using the same expansion for e as above,' J. A" x) n; O& `, H6 o1 M5 U

    1 B' {- c' _. J! b8 X# a公式
    ; w( `2 Q1 h+ ?3 e9 Y* O
    # Y+ p' Y! O4 a( L4 I& u1 A- B19.Solving for t and equating it to the earlier expression for t, we get
    , a% L! J: r7 g& J& O
    - ~5 O( u( ^0 L6 Q5 u% d% M  n' \公式
    ! A6 J: N$ F7 _; H* R$ H( c5 b0 J( K, q: }: ~$ d
    20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is( H$ c4 h7 Q) t! J1 i  }

    . i0 ]4 b) o+ j; h公式
    1 b2 P" F. [( x+ r0 u. M3 i1 I6 E/ \* O0 U+ u1 T, l
    As v=…, this equation becomes singular (单数的).
      c+ S5 v' V! d2 J5 P. p) R3 y: `1 P& @
     5 h8 W1 [  u' M& }
    $ v) Z5 G) A, }! l9 f1 F
    由语句得到公式
    : G: F, ~9 ?' t$ S' t; O2 K0 g21.The revenue generated by the flight is
    ) }' k9 k8 ]5 }1 _3 \% x4 N* _5 J' y( l0 L9 P5 Q  I
    公式
    % U7 q% `' G+ R
    1 _- p& }5 C4 i # J. o$ O+ X& ]: W

    - _( Y# x; [: D; `: s. i2 j7 w2 O! F24.Then we have1 N8 V/ Q: J7 L1 |. ^7 {) O
    * p& \  ]9 v& [# [! X
    公式- i& s( M. S$ o$ _! I2 [) P
    ; y) p6 @3 ?+ j' B  O! s
    We differentiate the ideal-gas state equation3 n  t$ V! N1 g* d& x! g
    9 f5 x: f9 s0 L" U+ P2 u: p7 b
    公式
    ; G' G4 i; Q5 F
    ! x' s, e8 O" m8 b6 Z$ ^Getting# S6 M$ s' ?( ^% ?( }" v8 W# y

    " A. I8 E: M# P$ P) h, x公式  ^) e3 X! `0 [+ E

    : A6 r, T3 O- g0 b25.We eliminate dT from the last two equations to get (排除因素得到)
    ; u! Y. |, S% }0 S/ S3 S$ Q
    ( C9 E  [$ J6 `公式( D: ?- Z0 c1 t' x- ~2 d$ m& S
    $ Y; H* L7 [0 ]' s) Y3 N
     ( q, h5 h" X/ |
    $ E8 h! L9 l' F5 P" e8 v
    22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
    ! [8 ~4 ?. L3 P8 P% g
    5 T! a6 h; U7 B" @公式4 L9 L! l! P3 M/ L' n) ^; ]# ^* D- b, w
    ; N" Z' m+ E4 @/ m4 N* T/ J
    Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值)
    8 |/ b" z1 J1 g' @7 R0 g& v/ D* P9 {5 |6 v& O& R- V
    公式
    # H4 F# X/ Y1 y————————————————" }7 Q$ v5 r' f: g
    版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。5 F+ T9 g# k7 b. V7 K1 [
    原文链接:https://blog.csdn.net/u011692048/article/details/77474386' I2 s# ~( ?9 a7 l1 e  P( F  ~
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