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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
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    由假设得到公式8 q' {  u- Q! N3 N! V4 G2 P  ]
    1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)9 H0 {2 h* z8 @* @$ s* S

    , p% [) u$ b- j3 u公式: m9 r, Q/ U9 m8 Q

    * m; u! w) B" G; R4 sWhere
    # [' C% g$ e: v7 s, r+ t  `
    $ r# G( A" S3 J7 g# N/ g符号解释
    - d$ |* m/ `1 U8 n5 F4 O% K8 m/ O' H
    According to the assumptions, at every junction we have (由于假设)
    7 i  I6 X  `6 E8 @$ {
    1 G( ~$ a- W+ W; p公式
    3 h% R  p( Q9 u: J
    ) J, S3 ?6 G' P由原因得到公式
    # }, [) f4 L: E9 m% j2 i) V+ O" p2.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);
    1 M, S! U( O+ H3 k3 N
    ; t& O9 S! [. T9 k7 W$ ?# l- A7 R0 z公式/ Z0 Z+ p* `2 ^& s, h

    4 `( Y2 ]: V6 D( v" y5 S9 s; jSince the fluid is incompressible(由于液体是不可压缩的), we have
    6 x3 Z( E1 m/ B. d  Y+ x' r+ I& B8 F
    : U) W+ z  n* M" }" G4 [( I! I7 |) k" C, b公式
    6 b8 }. U$ L: ?3 c& O/ g" f  f1 B) i/ r! @
    Where
    : K5 S8 F: b! s0 G- w$ ?, k; D+ j8 C! @! ~. f9 ]8 V! u8 e
    公式. j: \" D! f0 k  m7 Y2 ^
    8 s: _0 V: d6 ?/ S& h' ^9 B0 X3 [
    用原来的公式推出公式
    5 L* t' D# z( X% @! b  c+ j, @3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到)6 S' t: `6 Y8 D- W: e
    / Q. {: J0 M$ M( C2 U* u
    公式
    , b9 l7 p; ]% k8 v& z( i
    " c' d% W: h: B+ v+ i& a' P11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:
    " D6 f9 e5 e4 ]5 P( E5 \$ l9 U  I( U3 p
    公式
    ( e4 N, R0 s4 i# |4 [7 i0 ]; ]( k+ ]4 w
    12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)/ N0 P% _- s3 }
    $ }: i, w7 m5 l6 v4 N
    公式
    5 ^" b" w* [- {* S' h( s: k: `+ E* T; I0 E
    Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have  E' p$ ?! v3 B+ [. l
    . G* M! }$ f( a" v/ I# t1 \
    公式- X8 K  b9 }$ y$ }6 T+ ^
    % G% p2 N6 }# z0 a9 O- z
    Putting these into (1) ,we get(把这些公式代入1中)
    " g! `# v0 ?1 O# c4 M/ B- C7 ]) _$ u9 z' U, E
    公式
      N/ o/ U$ g7 r% ]* H6 M! G% v7 f3 h. V
    Which means that the" ]6 a0 N0 R6 T0 V( ^& h% f
    5 F+ @7 E/ T% ^5 d1 k# \3 Z+ i
    Commonly, h is about
    ' }5 w5 c' e: J/ g9 _( S' c
    ! N$ P7 Q, f/ a5 O( LFrom these equations, (从这个公式中我们知道)we know that ………2 b; m2 L0 x/ B# j+ Y' ~
    / `; T: [: X5 R! g7 G: s
     
    / P& T4 Y& p$ T+ [; ^( R( I' r4 f- W: s- `- p% ^$ ^! H
    引出约束条件. c5 k; \% J6 M% d9 D) Z
    4.Using pressure and discharge data from Rain Bird 结果,, z9 M6 R% v& J* U+ \
    ; `. h- c! l# ^7 W' O% m
    We find the attenuation factor (得到衰减因子,常数,系数) to be
    & u8 N6 v; [  A! T. C- i; m9 e: n$ [0 C
    公式
    9 q$ t$ T7 I' I% r7 n) |$ `% D$ s
    计算结果" N  N' p! s0 g% G- n. R5 I
    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)" v( s2 c1 G& y
      d, e' p2 n6 E1 R& @9 r
    公式
    ' p0 N& M) ?3 u- ^' z) Y' L7 |6 `1 v* ^# X2 W  u
    Where( q3 S+ X% [+ f' ?: j
    # z( H$ a- {1 Q- _
    () is ;;
    ! e3 c" \  V( x& O# l8 I( Y# D- Z1 h" A" u3 F/ Q! m0 A5 i4 `
    7.Solving for VN we obtain (公式的解)
    ) P: h9 U( j' s. \7 Q, [. F+ W& Q& I. y% r2 p  K: _# r& t: A
    公式
    ( i) g" @( R/ l( [- u; G+ i! Q5 G! h
    Where n is the …..
    / L9 j; M: K) ?- Y* y# @# P, \" p& A- c7 r2 ^
     " X1 P  H& F2 B( ^& A8 ~

    - z: O$ z0 [, R; B7 T2 e! P: l8.We have the following differential equations for speeds in the x- and y- directions:
    " V4 r& l1 D3 N0 C; ?$ _- K1 J
    # `9 k7 H1 a! K8 w. g9 p# k1 H公式2 a2 ~+ n. Z7 z- f4 w
    ; h" R( u  X: Y- I6 J% b, t
    Whose solutions are (解)& L7 H% s: }& a, p6 l, o9 G9 d
    # e( s" X6 x1 p
    公式
    1 Y  {( u5 F  x, \; M6 R# i( x( l% F
    ) P7 I9 V2 c7 x5 \. d7 B- M& w4 o. d9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:
    " |2 j2 G4 y. T5 W4 ?% d# `
    3 J& p6 v, d% E1 B6 h公式
    ) M% M1 \9 j7 f5 W* U, ]# K  l" M5 J/ l2 g8 P) A
    根据原有公式
    : C6 v. R5 M$ l% M10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is9 B' ~- o; J. b# E$ ?( F' j& n5 ^  c

    0 c+ m2 k! N8 {& h2 ^公式
    # c1 ]! _( ]1 u% I  B' c2 Y6 j
    * t8 _0 D7 g/ l. j; K( H4 t$ o3 FThe decrease in potential energy is (势能的减少): T0 G2 v! i$ m4 ]6 w' K; Q3 Z
    3 M+ m& \5 m5 w4 u8 h- g
    公式$ @3 r7 t/ |) ]5 a0 k  [; b
    ; v# l+ A* F0 F' Z& Q. [+ q$ h
    The increase in kinetic energy is (动能的增加)
    2 U1 k' X4 z9 ]; L1 w
    ) m0 Y  h, z  d& p: ]; D* X: Q公式
    ! f$ r( X5 J* ?* y5 q+ Z/ q- _
    ) b5 e/ x$ v/ L8 f: DDrug 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)
    0 j' s: X5 f8 y* H1 n# ]# x! f; \* k" M( l" W
    Where a is the acceleration vector and m is mass
    1 m, T5 t# @( w7 k* t! e; ^) r7 x6 w* f. d$ O2 g& V, l
     ; B8 ?9 t5 g& O+ j3 g' c& Q2 `
    & ]9 w* X' ^8 i4 Z# y% s
    Using the Newton's Second Law, we have that F/m=a and( h8 U9 H- h  O" m3 d, J
    . {6 v, z; b3 V
    公式
    + Z! r/ q% Q2 _! k. M6 a9 `
    : v& h; M% l5 p9 y+ M4 x( P8 hSo that
    * H6 A. S" k+ ?& C8 g( r4 F3 {3 S0 u3 r0 N; t- {
    公式- E" q. y- q1 J8 q7 B! W' X
    5 d# L/ a1 i7 q+ M1 ?: ?
    Setting the two expressions for t1/t2 equal and cross-multiplying gives
    # h0 g, p, O* C$ A! [8 ^+ }' x
    0 F* g, L- c! W# i8 ?公式" \) q2 H2 Z+ p  a0 O
    0 [% Z2 l7 K7 b# s. a% k# w: J
    22.We approximate the binomial distribution of contenders with a normal distribution:
    ( B9 w& l+ ~! Q' _# w
    + l+ Q9 y- a0 d% z( g公式: B# T) E* E% j! ?% p8 U2 ?; v5 n
    1 l# w! |: g" [5 h( ]+ U
    Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives: w" \' c# Q0 w6 J% w. R4 }& S

    8 ?1 \) Q! [2 _: X- Z# O公式- J2 |, A! Z  O0 S

      D  V! g+ H- e& K1 I: bAs an analytic approximation to . for k=1, we get B=c
    ) Z' x, Z, ?' Y) A4 O$ ?
    " Z8 h: w# Z% | + E) N3 A3 e) ~  l

    6 e3 ?  a0 L0 ^6 W1 I3 |1 s26.Integrating, (使结合)we get PVT=constant, where
    , R' K: |0 T- u+ ~/ i; j. t9 K; x8 @0 H; X% A: {
    公式4 @! u6 Z! \+ A$ j7 ^

    / P: g4 O! B1 k# y1 NThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
    ! M1 J: T: w; L( K4 ^6 x0 ?# @' a2 K* Q  Y) _; x! h
     
    ! t, I; u6 ^3 n$ L1 r0 _* H0 I# k
    23.According to First Law of Thermodynamics, we get4 w0 U, B6 C1 B( C) d& R
    - a8 ^. f2 h% x8 {
    公式+ p- g. v- n: q( [5 D
    9 I2 \, p1 J/ i9 X0 f; r  A. U
    Where ( ) . we also then have5 g4 O. L  |" O0 ~; D& n
    0 j4 u; |( x6 o) t4 L1 s6 r
    公式3 ]7 E6 B/ k" {, |' G& ^

    ' ?( O. {1 Y6 }! Z' `" M- xWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:! `/ k' u5 f" S( n; P! {
    $ V7 |# P1 J/ A, E! D  t% v$ J
    公式
    - X+ Y+ C' v3 \, Y1 ^1 v% ]" g5 i
    4 a7 A7 @' y1 L+ B$ oWhere
    0 X# x: I3 n% S7 v
    ) D% i! [. W  n& D5 i$ [ 
    3 `3 R: S/ `% ?- K1 x. F
    8 }7 k# U6 |4 n对公式变形. ]0 f' U) N4 T! k5 M
    13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到): K, t* K. o0 E+ r0 j5 j7 P

    3 I% p8 j' ?7 h& [) o公式
    $ P* X: v) t! {5 A. n: @* q9 l3 U" a4 ?' D/ p% e2 G) w8 ]4 y
    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; ?+ ?! h8 j9 ?, d

    3 f7 t9 v+ h0 x# |2 g7 ]; i* s公式, v0 U" @/ I! O4 c0 L

    0 V- }. i1 r' E( L6 b8 H5 N使服从约束条件
    * P2 \8 K% j% h6 {14.Subject to the constraint (使服从约束条件)
      u% l# l2 o: D, Y5 G9 F* U" ?# H' S6 ~6 m1 b0 B' B
    公式8 h5 g+ n9 f2 b
    ) p2 H+ B" n. l  H
    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)  f" j, v, m% u3 k+ b

    3 T: N6 u: M5 Q公式* A! |' [$ [6 O$ n" e

    & T) U5 |& q+ `0 k. TAnd thus f depends only on h , the function f is minimized at (求最小值)& C! q& s5 {& q( z9 x4 l

    ; g, @, o. `; R6 \) G" v4 n公式: U# x0 j- q, X( g7 z) T3 H
    . l3 W2 Q+ @0 O) N1 D
    At this value of h, the constraint reduces to
    / N" z* Q1 H# b# y3 m2 h, P8 c3 x
    公式
    5 V! S  {  Z2 q: {
    ; O" d' q' \# ?' a5 P- l结果说明
    1 ~1 E7 U5 E; O, W* e15.This implies(暗示) that the harmonic mean of l and w should be  _5 u3 Z9 h9 m* g" q+ r( A

    2 Y# ~3 P+ L( p, {/ X' m% F2 b公式- D! m( d- e& g7 Y

    , \  B/ s# g; L% ^6 g7 USo , in the optimal situation. ………
    * Z# D- y- c9 t# _$ {
    + x5 i+ u4 S& e; J5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is
    5 t) I( q8 t3 m8 a+ |& c' }& s" f- J: V7 @, }+ b
    公式. ~, X1 a4 R, ^/ M, `" |

    7 n& Q; c: l. O" o! ?16. We use a similar process to find the position of the droplet, resulting in
      u5 W' I& a2 g1 Q( c4 s: f8 Q# g
    公式
    8 p9 @) \/ w& C3 ?. ~) j0 j# P" J
    & b. @! c6 C' [& f3 r- MWith t=0.0001 s, error from the approximation is virtually zero.
    . u5 x* `: }8 J* L: ^/ y) B( P, Q: q! J) B* Y
     * E" G) n( r0 Z! [& a: b
    0 C5 B2 W5 y0 Z2 `
    17.We calculated its trajectory(轨道) using  U7 s+ M- R" h0 e4 w) [

    : ]8 e8 R# \8 r& [3 B7 e- W公式
    ; o% J4 i/ d" h/ n- _$ S
    8 {* ^, u3 L0 w5 E1 M18.For that case, using the same expansion for e as above,! K1 }) s2 z/ W! N# S5 A/ ]
    & e. ]* w; m6 O$ s9 t! x
    公式
    # r+ L) j* S$ S4 ?% h) Q% \5 ]/ l1 A2 _8 r5 o- h( Y) T+ p8 N3 B
    19.Solving for t and equating it to the earlier expression for t, we get
    7 T$ Q) f: l7 q
    3 d0 p2 W; q5 w/ h! i- p& k- J0 c公式
    ( Q% i2 C) d& Y/ a6 V; k5 o/ H' ^5 p% V8 K, q% n
    20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
    - @& s' p% H& l. l" t
    1 |9 ^2 y$ \- ?) h公式* a% u0 t6 V, o7 D

    , s, I  m8 W( T) m. @) jAs v=…, this equation becomes singular (单数的).
    9 W: R  c2 b7 \  O1 {& n" x, G6 r& z6 q+ X( U' x
     ( Z, `8 ?) w8 r3 Q

    * @6 m! K5 E6 Z* w% y由语句得到公式
    ; j: |+ }+ C/ \! M7 E21.The revenue generated by the flight is" ]" s. U* y# j) _4 s
    : Q2 v- q5 G& C. [' D' O5 x+ m# k8 s
    公式* r6 e, U' i: Z: k% A* c) m# ]

    - Z/ [0 v5 W% a9 e0 F 
    7 x4 I! ?2 y2 u# x; P
    ( T% \& ?3 {% Q3 D2 T1 A, J24.Then we have
    ( a8 }* X, |, f' F3 Y0 O% M7 u. {% o' z" C! I( J0 P8 ~: n" d
    公式/ r) e5 j# x/ K! v, k  h

      u% s* V' h$ Q8 U+ ]7 KWe differentiate the ideal-gas state equation
    % R  P3 @6 Z# s; z8 a; _* i& U% e
    ! O; k' \( y) @% ]; z/ D2 H公式
    ( k1 z* J; b3 j: W
    . E: V: z  [- S# M: nGetting
    8 U+ L; H1 j* w" h: V
    . {2 E0 P0 p. A; l1 n& n7 G公式
    6 S1 e) V; m( T$ _7 _  R* I% U* q% N4 g% ^
    25.We eliminate dT from the last two equations to get (排除因素得到)
    3 \' K/ _, R# g0 o9 C" T0 T
    . h# {) `0 u$ `8 x* v# F公式. \& |4 C9 N* d- Q/ e% n6 l

    ; O- {3 N+ \" h8 A& ` & E+ {) ~% t5 n' G$ e

    % {% ?: O! k) t  k; m, P+ b22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
    3 g/ L1 v7 x# I; `1 x" J$ T
    ' i- G% _' p+ c3 q5 P( }* J公式3 F- W1 D$ G! Y' _7 A7 F. ~% T
    7 w8 p; }/ t6 _( Z
    Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值)4 h: }6 H) j- d- Y$ E/ |% o
    9 C) A" Y# j) k4 @
    公式# O  R" M: M5 \. S
    ————————————————
    + J' I) w* S  h版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。
    & m( g6 T. t3 L原文链接:https://blog.csdn.net/u011692048/article/details/77474386$ E7 d6 V5 |6 R- a- e
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