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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 |倒序浏览
    |招呼Ta 关注Ta
    由假设得到公式
    . o. n6 \1 B# s3 \" P: H1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)
    ) |5 w2 i9 n8 R! O% }
    ( Y5 z0 T7 Y% @- l/ f* n公式
    : a2 k8 R8 X, U) q& E5 O1 }! o- {' k3 P* F  E
    Where8 E4 F3 Y+ y) E$ B6 d/ U

    ( r4 G" K; }- i" @$ J5 W6 c9 x符号解释
    9 p: R4 r1 a0 i0 Q, i( S. y: H" s; u0 W+ M
    According to the assumptions, at every junction we have (由于假设): g" v# e" I) Y# k
    6 n8 v  c( b9 [/ W
    公式/ I" }: U9 m" ], R" u' t9 R
    9 Z' F. ~% K9 w9 n$ O8 c
    由原因得到公式6 _: G* i7 f! }
    2.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);
    & ^8 U$ Y' Q) i$ ?+ U, u: |% Q9 v5 y
    公式) d, |5 W, K. [* C* ^
    $ `* B, r9 a4 w9 p7 v8 S3 R
    Since the fluid is incompressible(由于液体是不可压缩的), we have7 W: |6 ^0 L( [  E

    , o) ]! y0 w" @5 ]& d# g公式2 N' {; M4 U: d- G' R

    / I+ Y  V; G9 ?" Y1 e! B! w& dWhere/ J# ^4 y" O9 v9 f& x
    " C, w' z) ^0 T- b: {7 z
    公式1 d5 d. L; ^; n- S

    ) G, P- Q" ]+ H1 ^用原来的公式推出公式
    8 ^" Y* g( E# r" o3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到)
    6 }! ~* M2 T) A, D
    7 T( H% p$ x" ]& {7 [5 x2 L  G( B3 M公式% F. {9 P: H" q
    6 z, b& n( W) \8 e2 W' n
    11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:8 w. ?& L5 s/ C% D/ D5 h. U8 d! [
    # I3 h0 k" Z) j: ]& X: m3 [8 D
    公式  Y# D- Q8 N- |& m; F: y$ ]1 ?& X0 U) E

      k9 b, u: [  h12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)0 a! Z0 O$ z+ G
    * O2 }) ?( x2 t, G7 n
    公式
    ) L* o4 k! k  F/ x8 C/ w" W' M! l, s: X
    Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
    - D0 y3 G! p( p- n# H' _5 c8 ?7 \- o, ?) Z0 k* B9 l
    公式1 `- w# B3 }; |0 [0 q( c( q/ ]: c+ H
    : x; \: R) e. `8 t3 u
    Putting these into (1) ,we get(把这些公式代入1中)6 R( w) `" L3 ?0 R0 }
    9 ]0 d7 T, T# A. O/ r; L
    公式
    9 ]. M: E! t/ E; d1 w# u
    ( K9 T4 O$ |/ L* D9 VWhich means that the! r1 d2 C; D8 x! P! J) Y$ x$ `3 u

    5 |& _2 E, y8 q3 [0 YCommonly, h is about( a; w2 K& \0 }; n

    8 ]) {$ q3 H$ A: `' h$ DFrom these equations, (从这个公式中我们知道)we know that ………9 Y& b! S% Y& k

    # c) J( O) }, E0 } 
    3 B) @( h. E( i, T1 g1 C7 Y0 x" ?4 Y7 M# q4 I7 `; C# @( i
    引出约束条件
    - N3 `) G* ?: {( y9 j) |4.Using pressure and discharge data from Rain Bird 结果,
    4 q" y1 i/ ^) m
    7 j. @5 L3 V4 ^) e# n' Y. x4 ^We find the attenuation factor (得到衰减因子,常数,系数) to be
    ( O. n# _6 R8 a1 {
    ; h+ B2 n4 [; W: ^公式! _  `$ v- ^$ ^" |$ L$ I2 G

    ) r* Y! P- Z6 u: U计算结果
    + U  _6 p  @' W9 s+ \& q6.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)1 R4 ^% h& G& ?3 Z6 W

    9 I! e8 m. e4 g  y3 n/ W; W公式
    5 L( b: T" p$ T2 e* D! H8 r1 g+ R8 p5 P  J
    Where% _" a; _6 W8 j# x% \) I; s1 d

    ( N* E  t5 p1 N# e7 T$ U() is ;;
    8 r$ g' Y7 Q0 j1 _0 \) V
    , n% b& o7 d9 E9 A" O  U2 U4 ~7.Solving for VN we obtain (公式的解)
    , K5 y# g: J& g1 ?) h" `* c' _$ C/ Q, c, c' p
    公式
      P+ C; ^" u6 m9 }) y* u( a+ V
    & Z5 x, W% V( j2 i. d3 bWhere n is the …..! U' ?, @* H* r- a3 o; Q
    + x" E( [, g+ U5 ?
     ( I. O$ U( y7 v, w3 G
    6 d, W* Q( K1 ?8 `3 X
    8.We have the following differential equations for speeds in the x- and y- directions:. z, m+ O0 b5 U: R6 {/ ~7 I. n
    ! h( j& C  D3 h! s* S! e
    公式* X/ t( V2 t4 F, j

    8 \) I3 ]4 v. h8 Q$ `0 j3 S& CWhose solutions are (解)
    ! C+ }, y# \$ A/ x8 q/ o0 n* E) y" h* \3 U+ X
    公式1 n/ W: \) i' O* a. w

    + A4 ]7 ~' W- P* p  k/ k& U, p+ |0 a9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:% g( l+ y+ s0 n
    9 r0 W* Q9 E6 G; B/ {3 d  b
    公式
    " o4 e; D- p& V4 D" W0 R
    9 z9 t4 X1 p' o, {  e/ J% B% `" f根据原有公式% u; G- C1 `! E1 F+ X6 T: F
    10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is& G+ w+ r( T% E/ V% R. L0 {
    # m9 w) E( d6 y0 I# Q& c
    公式/ T! ^( y! [6 }: }  A2 `

    , i" z- C+ G* ~! s  G) W, H8 MThe decrease in potential energy is (势能的减少)
    4 I9 A8 `$ u. H; [- {9 M  l! I" K( s& M0 A0 A$ Z
    公式
    : e- E3 w2 \  |9 T" Z% l3 ?9 b% b2 y8 @/ b; Q' _
    The increase in kinetic energy is (动能的增加). Z5 N. [; d! p# p  j% c
    3 ~& ~' L, P" n) r$ ~
    公式
    $ i) Z3 z5 _% k2 ~. p: F8 K) K' Q' e) x  p
    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)/ Z9 J& N5 L9 H( u0 m
    4 [6 W0 R* ], o9 y  K0 N1 W4 A- q
    Where a is the acceleration vector and m is mass" s; r; C3 h* |, j9 Z
    * [( d) d% m$ ^2 p
     
    0 k+ M2 N* ^" A: Q
    - F0 w+ @; F% _& g5 E- v& F, ?0 W9 D; r$ ~Using the Newton's Second Law, we have that F/m=a and
    2 q- p0 i1 g3 L* U) \( i$ U$ H; m7 Y- V8 Y- H8 x& U
    公式
    ; e% B- n8 I2 E, d; e: A
    & J6 I" Y4 C  S5 x7 P' BSo that2 T/ i4 L- q/ L; O6 l

    3 g# x' X; M- o. b& w0 O, H公式1 g+ e0 }6 |$ f5 K
    4 p5 _/ o: p: u, L1 ~; B. [
    Setting the two expressions for t1/t2 equal and cross-multiplying gives
      F6 D9 G( a: a3 v$ s' B; O
    ; Q1 L. J; k- d公式
    9 q6 _* y$ }- X) K  S( y; S9 b2 S4 u9 T$ L% O) M) m) u# k& r
    22.We approximate the binomial distribution of contenders with a normal distribution:. e: @! g( A( I5 g# `/ h$ Q6 V
    % g8 G" D1 s# _: E7 h7 I
    公式
    ( M4 g; i$ z6 g1 _1 }  \, B8 {6 n* t% c+ n. d9 b1 K
    Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
    7 }* D- I4 @; A# m
    ( O8 ]' n" \6 g+ X, i公式
    ! n( @' L7 ~+ U
    - C3 P5 @/ T$ g8 v/ VAs an analytic approximation to . for k=1, we get B=c
    ' z# ^8 F1 d+ R: j7 ]2 A* r2 x9 F( t+ I, `! I0 e
     
    $ F% E- j& q) {7 a1 S; Y- _  G
    . O6 X: e  `. p# t0 ?26.Integrating, (使结合)we get PVT=constant, where
    2 C7 r2 o# U# W, U4 J8 C( Q: o! g
    4 X- ?% K, l; x* Z! r% l! m公式
    * `+ }: B( g% \. i$ S
      s6 h8 w$ a/ }! hThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so  t  O2 A  |$ X& ~: Z
    / X; E. K: E& m4 q7 ?) L* c7 p
     * ^  u+ R* w+ z: K$ }0 u8 R
    ! M9 [  x  u* Y1 o7 z
    23.According to First Law of Thermodynamics, we get
    7 v! v3 V' v6 @5 V. V% u3 V* g; @% ]0 C  @: F7 e/ A6 r8 d3 w
    公式
    " Y) A* n/ P4 }! g+ X! u. q  h% p8 Q$ i! \
    Where ( ) . we also then have2 A6 X5 P1 T% \& Z( V

    % z  ?6 i) ^/ D公式1 M* H9 T' _; N4 I/ Z$ O

    3 K( O" P9 n/ h0 i0 `Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
    ( g3 Y( I" S; M# @6 I/ U- U) g* i  Q+ @! @+ J" }- `! e6 k
    公式
    % _4 O% Q3 H+ y$ n) e, \1 ]/ Z" u
    Where
    ; H7 {- S5 o5 H/ h4 R3 G( y9 L0 {% R4 f/ D7 `( G" g% a
     
    ( z  j$ `% F; m, S$ e/ @, _% w
    ' |7 A  O3 N0 X0 b& T对公式变形
    4 C( r+ w( l& r* ]! g& a13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)$ g6 ]4 s; h( h1 U
    1 D: d, j) p1 y) U6 Q
    公式
    - @4 y- Z* T! Q& a- y
    ; K3 r* u) }' q# U  W# QWe 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 minimize1 M+ y! _- d. J
    ; r0 `8 B9 T$ a4 [6 m7 Y, f
    公式$ l; x& z) ?! u& J: I
    5 K# u* M& K2 |
    使服从约束条件4 L. o6 B* h1 M1 K9 K
    14.Subject to the constraint (使服从约束条件)
    ( z, }2 j1 n% A: g+ ^
    / U: i$ v% G! n" }公式6 K+ ]* q. b0 J' }

    2 ~! s  m4 L6 {4 ]) B+ ]Where 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)# {1 |; A3 `! D8 `0 Y" Y

    ' W! z* p7 W3 O公式5 C6 g& M* K- \6 }$ t0 O* g4 {9 F% |

    2 r5 x8 @1 j$ [8 IAnd thus f depends only on h , the function f is minimized at (求最小值). W% h* l  K7 q7 ?) Q& X  X( I4 U' q

    . H0 \" J% d9 A# q公式& X8 J6 _# ^, k( Y0 z; d- q- ]9 [

    0 j% J) T2 h: W2 J9 T; YAt this value of h, the constraint reduces to
    % J4 r0 {/ y% U8 G0 I* Q1 X- l3 n
      U$ u" U9 @9 L; |公式" b! z6 M4 v- G/ v2 Z+ {
    3 c: Q8 ]/ W. `- E
    结果说明# D( s" r% [' w0 z- l
    15.This implies(暗示) that the harmonic mean of l and w should be6 z! a; }! n; D  w
    / z; U5 l( R; h  @! ]' O
    公式4 F0 \4 ?: @8 J2 w* X( b

    / u' |* ^$ y' c1 k4 I7 V: xSo , in the optimal situation. ………
    % F2 `2 N; N0 a# |, e  L8 f
      b1 \  g6 J$ I2 d5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is
    + k: ~. |0 R, m+ j6 d$ D$ h  x: D7 k5 V0 _3 I( v8 O
    公式+ E. Z3 [8 n3 Y+ r. y8 n
    0 r* c2 |: z0 J
    16. We use a similar process to find the position of the droplet, resulting in
    $ \6 {0 N1 B  X, i/ e+ d( T7 y8 e" g' w# D1 G# j5 u% I
    公式
    * Q0 N- c- ^/ V$ d) f: d
    ' W; H0 Z8 y! m9 o) b: sWith t=0.0001 s, error from the approximation is virtually zero.
    " a3 N$ I3 U5 L9 b1 t+ A. Z8 E  h4 L% `" g9 N4 m
     
    : h$ M# p9 T% c' B+ s' _
    * V( O4 J  a% O& j2 Z# E+ _17.We calculated its trajectory(轨道) using: \- _* R# _# q
    1 w( @% D( A5 ?8 X5 t3 h/ }4 m! U
    公式
    $ ?& G. o& N  H* J1 {( J/ s0 D! S6 t5 V! {% C1 @3 D
    18.For that case, using the same expansion for e as above,
    " ?# d! |" B9 }5 }6 @) i0 B8 x2 E& K, B; y
    公式
    * j2 z' d! L2 j' a' j
    6 M4 C! z1 `) {4 E" O% X" u7 c: U19.Solving for t and equating it to the earlier expression for t, we get
    6 _% \8 c( I; B+ z& l- Q! [$ M+ u$ n; Y
    公式; R7 @. ?8 w9 C2 K* j# r, W

    9 {  d' q' s' y5 Q5 @8 }20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
    . M( k! v; }; s" H
    8 E  m' y9 j6 U" `* n, K: r2 J. U公式' g6 ^4 z/ D" x# A. o1 @" U; z! Z
    ' ], ~) Y) y1 b3 [' T9 _
    As v=…, this equation becomes singular (单数的).
    2 t. u- C" q2 V& |
    9 V1 M' v1 t8 x" X# v! `  { + {% ?/ u' N+ }  ?+ L

    * N" a, I* T/ x' ~& m1 g: R由语句得到公式
    7 o* ~' }/ d. w1 N; D( S! y21.The revenue generated by the flight is
    6 M0 K% T& @* E( j3 Q8 o/ @
    , R5 _  b6 f9 }3 ^( u% K/ Z公式  L6 o7 D' ]( A/ T/ [

      A( @8 M* e# ^& Y9 i" j 1 _0 I8 q. j( r, q! W/ i

    $ M: H  L/ `4 K1 `( ]4 v24.Then we have
    % F& E% Z$ s1 }7 p/ a
    ) J, G% `5 B+ B3 O1 G. K公式3 c) P/ [  ^1 i! V
    + l# ~' l4 c0 e
    We differentiate the ideal-gas state equation$ B+ @" j& }6 z( C6 P) R

    ! A# V* s5 L" l, [! C! D9 l公式4 L/ M' k- g: q" L4 a# C
    # o& I* f  y7 y# y- S, d0 |( U- S% f
    Getting4 L" _+ C  K, g
    ) a# A% u% \5 S7 ~
    公式
    1 M9 w* W" C& M7 k4 E3 P; T; H/ {" y9 r% t7 T+ `$ K
    25.We eliminate dT from the last two equations to get (排除因素得到)
    * R3 b; Q  E( ]* q- D. U" M0 ~4 {/ Q
    & l& y" x0 B& C( p" U1 L. h公式
    1 N- B; \: ?& m% ^5 B! R1 L- o1 y; n  u0 T* s/ R+ P1 Y
     
    ; s; H9 f. R' I/ Z8 p. t' f' }: E: b( f; Y( {: Y
    22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
    0 B" D8 W- @% a6 N$ e# n, l( b3 j4 ~- b
    公式$ X: J. x; x; `: O& S0 _
    / q  a( S$ C5 t2 ~) R2 p3 v$ J
    Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值): c* |! N5 D5 n2 o8 k
    " E+ A8 N- H5 `1 B0 r' }
    公式6 F5 j# @9 q7 B, @6 B1 x; `
    ————————————————( C! a! J7 z# r& N: S4 j
    版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。
    , R) t! p. h' B. T$ e+ Q% r原文链接:https://blog.csdn.net/u011692048/article/details/77474386
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