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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  J, y1 r$ e6 H1 U" }# g6 \/ R
    1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式). P' L0 c3 C1 }( U7 N% d

    ' ~+ }/ \  q3 @$ T$ _公式1 _# ~/ K* M3 m/ A. L; w7 j

    + \1 Y# n' f5 e& dWhere( }/ k8 R# j, J- V
    1 q/ p+ b7 _; j+ X' `* p
    符号解释
    , o" r$ L& q8 U4 @) l; l" ^# q  h
    According to the assumptions, at every junction we have (由于假设)+ {9 Q1 H+ Q1 S6 i* `$ w: x

    " l: T) K0 A' K" |公式
    " h9 ?$ A3 Y- d8 h" s0 O% p! g" O6 i8 M7 x
    由原因得到公式. a6 E3 g  |0 R& @2 Z
    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);9 \6 a  V# U- K/ x
    ; i; g% [8 m! h* @: N% `/ A" k
    公式
    6 x! @( G. n& |6 P/ F- p) t: R. t1 Y1 w5 o* T& T
    Since the fluid is incompressible(由于液体是不可压缩的), we have: n) K! d" L; D% x/ \. ]) n, O9 ]

    3 E6 z! H9 E& [# P$ _! P1 J2 N公式
    ' R3 `# S$ a" v# V* V$ U9 |# G$ U9 k; b2 k' v4 M
    Where' P8 O  X. Z' t/ G

    ) Z3 |- e8 q* v- C' f: H( B/ W公式
    , T. s) ^7 z9 z2 m) J3 G) S0 ^( |3 q
    + h  e2 G$ E- u1 K: A7 T( g1 h) ]用原来的公式推出公式" G6 U  d$ `( @8 \$ O
    3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到)
    4 H3 q) u* \' M
    3 ?/ r" S. S5 Y, U+ H公式9 E1 E; u( h0 S$ ]! E& s" Z

    / W4 X; \) `! u; q! P11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:% X1 P2 a6 o2 _, k: z& @% U
    7 [+ P# t8 o  G& L+ h
    公式
    / l8 X1 s. v/ a; m% B' C% j0 H, O* k* `8 ~6 ]- x
    12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)
    4 a# S7 X: P' d3 e) f+ h, y# C
    3 k: w$ d( X5 S+ e公式" i7 k# n. J+ Y3 f  z9 Q5 F
    0 a% {- K8 d- {
    Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have) N& X; i" h6 U1 @8 Z; G
    ! g" ~9 K0 ^2 b/ f
    公式5 p* o* F. W4 U, ^2 {$ D: o( n9 R

    # ~* D& |* T' t6 ^; _Putting these into (1) ,we get(把这些公式代入1中)# F; }- V1 H% @& @" I# M% {
      ]/ L% j" y9 {+ u$ C3 X
    公式! a. a6 ?# ?0 b3 b" X
    " S% b) D5 q7 Y
    Which means that the6 L8 [3 p# ?* _5 x+ O% i
    ) ^! q0 w! ]' X& C, h( b
    Commonly, h is about
    : ?9 B) z# U% ^1 C) t8 Q2 }; X. x: \# L9 P* v. w- b' P
    From these equations, (从这个公式中我们知道)we know that ………
    ' y9 i# E3 }2 J3 H" M
    & o! U) K+ V( {! V ' s0 K, n: B; B+ |
    8 R. D3 v5 ^9 _  U
    引出约束条件
    5 c6 |) O9 ^" j: v4.Using pressure and discharge data from Rain Bird 结果,% X. w. {# m4 ?$ c  Y
    ) J" d- D; C$ K/ K( I
    We find the attenuation factor (得到衰减因子,常数,系数) to be
    2 s5 {3 F. I$ U- X$ ^5 V3 v4 o
    1 k0 t/ I% I/ y. n- l+ y0 T& L# q公式
    ) Q- O+ l+ z' i) m% Y
    # w! F% {) u/ k& r/ Z! K6 z# X计算结果
    6 a/ ^: \. p# l9 a6.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)
    . a. o' k1 a9 J+ x1 `( f! x" e
    0 B1 z7 g! d5 d& k, @2 N- i& v7 Z4 x; e公式
    * P+ w1 C" g* G) n1 N( B( b& w  d& g4 x8 \1 W+ T, D7 ~2 P! Z
    Where
    5 k  W* M, g$ ]& t3 Q: R- K, g! a7 ]. O) y
    () is ;;
    1 s- E# `9 p" m3 ?2 C. W
    8 q0 a) o# N- Z7.Solving for VN we obtain (公式的解)( i- d! T' l' T- y$ k# q
    : v2 E6 k4 X' e9 a, n
    公式/ u' h: O5 f1 }# L% R' [2 [

    $ D6 t- O4 |6 q8 k. n/ aWhere n is the …..
    ; L$ ~" n0 U( x
    5 u, V) `! h9 Q1 T# w  h+ F+ d# P5 r 
    1 m7 n3 Q! L$ j6 W0 C6 U. z$ g8 W6 f( H/ U9 a# H
    8.We have the following differential equations for speeds in the x- and y- directions:* T6 A4 P7 K8 |( `! ^

    " U. n" V5 L* [) X4 t1 u8 q公式, W4 l. s8 V6 O% r6 n! U
    " m/ n0 O$ a# y( B4 y: V7 s
    Whose solutions are (解)6 m) h, E$ @. P% W$ Z
    & W6 G' b7 p9 G1 W
    公式
    , @/ i! k: K; V4 b3 [* a1 [/ S% b: E$ j- ~0 s) a% s. H
    9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:
    $ C7 G# r& u, X) w; Q- z) u! p% V. q
    公式* `2 K( q( {* |8 h' Y, p

    ! a6 I, C( [8 P6 l1 S( d根据原有公式$ |( Y2 i6 g+ u0 m
    10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is: Y# i1 e' y$ m0 }; n; b, Y
    . m: G  Y8 l7 {! Z9 I
    公式) e7 f. C' E/ `6 k. q1 P9 n3 O
    6 v1 R8 i4 v& t6 z
    The decrease in potential energy is (势能的减少)
    . K, l- H( J% z* B- g% h+ m& Y1 Y* ]& i0 C8 @& @( p' T
    公式
    & t4 Q+ @1 o# u! T1 @; B- o$ r! j$ {# @3 C/ P$ }" S9 S
    The increase in kinetic energy is (动能的增加)
    " {: }; L3 Q$ R1 g6 I* h% }
    + f/ @5 O0 _$ w. i" ?& V公式* A# P+ d6 Q" `+ `
    : ]% d& {: Y6 Q& j' U' D- K' q
    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)5 `1 G, E4 I5 T) l

    9 G' t2 F1 ^* g$ Y: _& eWhere a is the acceleration vector and m is mass3 }, Z5 E7 k5 B2 n1 a6 b4 D
    4 c* S' k, |' f& O* x, u
     
    9 K% V. f5 ^) d  v3 m  |- Q) E) j! k- |  `& r' [/ _
    Using the Newton's Second Law, we have that F/m=a and0 l1 @! r# N3 u% O( ^
    & O- Q9 U2 \! k! Q, b6 T$ f+ R
    公式
    1 M  ]: Z6 ]/ |8 p& J7 z* I; q- @- {; w+ M
    So that# F; o3 O7 j( {2 Q$ d
    ; D* u0 K4 L5 }
    公式! [" C  E3 D9 X' `  v% m$ q

    & ^( x) ?. Q  \" J7 Q) k! [Setting the two expressions for t1/t2 equal and cross-multiplying gives7 }1 V* |3 C, ^7 G$ V* f

    , A% c* ~# S& e5 Y- M" r公式- Z( Q+ T$ v( A. E8 p. @
    $ A! w: y& t  `; p" \# Z
    22.We approximate the binomial distribution of contenders with a normal distribution:
    9 h/ B, V7 }; L2 v# v# X% E$ B) t3 s( X8 {9 |5 y
    公式" m3 T( D/ n5 W8 P

    & @8 F: F* i% o3 r' cWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
    % G# T* i& N) U2 z: w& b
    5 P. U, b! @' Z- ]2 P( k6 m6 p9 [公式: G% ?3 ^$ m- v# A

    4 r5 P6 j! [% T9 f# ^6 OAs an analytic approximation to . for k=1, we get B=c
    8 ]' V0 y1 M5 }7 t5 u6 t4 A
    " j$ e" |; n% Q! k  R & M' G/ e1 Y! x" k6 O9 F9 L1 ~
    ' L9 n6 Z) J* S2 C6 K0 N6 i# ]: N1 y
    26.Integrating, (使结合)we get PVT=constant, where
    ' X. Q* m, |% |6 S
    6 u6 ^* h/ N9 w% Y1 e  R公式
    # d6 S, N5 h0 W4 G( n
    # Y* U2 g& P: u" sThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
    9 I5 S0 z) e% b' A+ Q  Z9 U
    1 N3 ~  f% O0 S! e4 W 
    6 p& R: R! Y* w1 f
    * O1 Z! o6 v% Q1 q" L) D23.According to First Law of Thermodynamics, we get% s7 ^1 U* ~! v0 |

    * t+ i+ Z3 Y5 e5 y# U% _公式7 C; _' f" l0 k3 k  b  u
    + A0 Y( o; b, L
    Where ( ) . we also then have
    9 N/ m6 F; E5 O+ A# R. g: `  z7 o- Q" v  A* I7 _
    公式
    8 ]4 y1 U% Z# P9 ^
    ) m  q* {3 E( V! D8 kWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
    5 X' @  D" _* D8 P! [0 S* m. d5 }+ s- L3 A' a4 T  x) p6 u2 b
    公式
    " b/ T* o- a' z) X' L5 _8 v. A$ N
    6 f! W0 k: p) V" {" |; lWhere
    " y9 n) |7 @- Y& ?9 A3 u$ ?2 a$ K; x+ J, R: m
     2 E$ Q- [% h, {+ U. m# n$ C4 ~: X3 }

    " r2 i4 t$ R: j) D( ?- C- p对公式变形- R" C+ G3 m: o5 g- O7 L
    13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)
    ' D# K3 d% J. i! e! {' @5 I# s6 o2 O1 {/ ~9 U4 ^8 p) e1 V+ u7 v
    公式% @& T$ o' \; [! \

    / F" q* c7 _1 O+ n0 D/ W1 C$ v, ZWe 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 minimize2 r' s7 ~' E& F# h! P0 o# h# v

    5 O0 L- g. q# y: y5 k% u9 `- q公式0 ]" S/ p( J9 T% a) k

    4 K/ I# p9 V# D4 h# a; Z使服从约束条件& ]* `, z9 p9 y' z% w, R
    14.Subject to the constraint (使服从约束条件)- G) B- _! |) J* @

    ' o; j5 P2 I/ Y公式( M+ O7 F5 Z9 @- ?# ?* |) ~
    7 e! h( f% Z. y, i
    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)
    4 u& ^! h7 \  O7 H' ]
    / C: t# {6 ?9 ^" D. R" T公式; J2 D* ?$ i. p0 G8 x

    ! ?2 e; n* N2 ~9 v/ nAnd thus f depends only on h , the function f is minimized at (求最小值)
    5 I7 `8 x) @4 r) [' L$ N5 ?: q3 J1 I! P& q# h5 D. }
    公式
    ) L, b( x* s4 E3 l  q) \& U5 U0 ^1 B" y8 }  w+ i
    At this value of h, the constraint reduces to0 Z2 @, h6 j: @. h' P) z$ ]

    2 w: M8 p2 G9 K# z公式
    : k" r+ [+ f$ C5 p% p# i. K
      n0 n6 j  Q* b- a7 s) k1 K结果说明
    3 p% h& D. p2 B/ H$ |15.This implies(暗示) that the harmonic mean of l and w should be' w8 i* C8 D) J. A3 J4 T
    7 b! v1 a7 _" j9 f5 r5 s
    公式$ N% W/ `( e, p
    - f% x$ l- g( P4 C; ~. F
    So , in the optimal situation. ………
    6 d7 N4 u  R, P6 w, l, q
    7 V, a$ O# H9 n3 S4 N5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is
    / d/ ]4 T# B. I4 V; m: n% G
    / _0 o  r) u6 a: E# w公式$ g5 p6 K, ]: P1 Z4 B/ e  A5 h

      ]2 Z) W: K4 A" i) y6 r4 d16. We use a similar process to find the position of the droplet, resulting in4 i3 V6 Q' h# ?" D/ Q+ t- @
    * b  G5 \; k% [5 s& M  [( T* B# [+ u
    公式6 m+ S: V- R; ]% Q

    9 B/ m1 H4 w8 F! Y# A: DWith t=0.0001 s, error from the approximation is virtually zero.
    7 s! d# c2 s5 A  h+ D- x, i! w' y: E/ g* h
     
    / z8 O7 v% P* T1 @5 c
    2 ^+ O8 r, x2 @; i7 l17.We calculated its trajectory(轨道) using
    0 W9 e( z) g- o3 z- ]5 j3 R6 K6 E; u, F* C% Y# |# b
    公式, U5 i4 O' _5 k+ @( f
    $ Q: f2 s+ |$ B* \# T" s6 a
    18.For that case, using the same expansion for e as above,
    / \0 @% @! [1 M8 X' ]! S4 p7 P/ s" d; ]! d6 h( Y' i
    公式
    ( n+ T# t! h% I& J8 y9 L8 i2 V4 s6 O: N) i/ O3 E
    19.Solving for t and equating it to the earlier expression for t, we get
    ( w$ b1 x+ ^  G# s% Q) I( y
    ' d! g/ B6 }& {" o公式
    / R9 U$ L$ }. r+ m, z8 _; m. r- M7 T. J
    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+ s7 Z0 }: _
    , `/ q8 ~" m% l7 T6 v4 U7 h( K' N
    公式% x; \. l. P1 Y8 _4 F6 m) l( o
    6 d: Y  }2 d, W3 z( }! l; |
    As v=…, this equation becomes singular (单数的).4 w" h8 W# U# W

    / Z" n  P9 e+ O7 Z# D' i0 Q ! [7 n% O  h; y9 r- ^

    4 v: S- N4 [: v9 Y( H; v, M由语句得到公式
    * J: Q9 V0 @; N  x$ z21.The revenue generated by the flight is! S2 ~: |. b) M9 Y5 z

    7 p- q) S9 @0 z6 K# ]) c5 a: ^6 U公式
    % Q% i8 W  l& I, Z9 m" u( c3 i
    / z2 a$ D0 b4 |" W5 `1 r6 [ 
    & G& l( D% {* |  e% S: D9 w4 }. ~8 A( e8 ]. W9 Z
    24.Then we have
    * f5 h* V0 d8 K: B6 s" \/ j1 O9 z! ^4 v" a
    公式
    0 I0 o4 \3 h3 u, _$ t# N5 I( M- |9 u. [, t
    We differentiate the ideal-gas state equation
    3 _& X/ [0 b+ x5 Z8 [0 F" P* s5 V+ c- u
    公式5 ]& D+ b% c$ P. X$ ?( \  o0 A
    ' K; S4 }# `( h$ W  B' Q. X
    Getting
    * d7 `6 b# l3 i1 ?  V; C9 p
    9 K4 p5 S0 [3 n( x  M# F公式
    ! u, ?! p0 _+ L2 l* X, y# ]. R/ \( [
    25.We eliminate dT from the last two equations to get (排除因素得到)& r* A7 H- ~7 ?% h4 x% ~/ Q
    8 O( A3 D8 z! N
    公式
    , y- L, L4 |0 U1 G' C8 v5 a* U8 O7 W7 Q; u$ X5 U
     ( Q# z1 ^# Y' k

    , r- j/ D" X* ^- X" ~+ v) e22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
    ; M; ?5 W7 m3 n! O9 Y$ Q
    $ O4 }' m- E1 W2 J* n4 `5 g公式
    , z  E1 ^: y) l# {
    * ?  w1 |& H7 O$ h; ~) k0 Q0 LWhere P is the relative pressure. We must first find the speed v1 of water at our source: (找初值)) q" c2 R7 G' C% q

    3 B1 J* P8 P$ j$ B公式+ {( s2 h' }+ E; Z4 T0 i# A0 k
    ————————————————
    2 W: ?' u3 K  z8 K版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。
    2 N3 x% [3 T  @" l: W原文链接:https://blog.csdn.net/u011692048/article/details/77474386, b' e" R& R$ u  k& {/ h+ k
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