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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1 Y8 ?2 c( a9 D; y) a( v6 g
    1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)- l8 M6 j$ {7 ]' }

    3 S/ a1 @7 E; n8 G, _公式' q# ~4 `: n! ~, X. M
    * F7 x" m. f- }6 o2 L
    Where
    & |' l3 n0 k0 h: {; s! ?' z' m+ }& l
    符号解释
    8 `: R0 @% [/ M& a( ]7 k( z% U
    , r( W& n+ J/ o( eAccording to the assumptions, at every junction we have (由于假设)9 j4 D$ z* q2 l! r

    " {3 n; a7 |0 A% c! j公式* o% D+ X! H( t/ h2 |

    / ~7 i# R6 f( U8 @由原因得到公式- R* C. h4 V" o" T, }
    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);
    ( f. O3 [) H; o/ a1 X/ A- A7 t! h( ~8 h0 I
    公式$ l  e4 \; f! I! F8 s6 d1 x% f

      s! u& q! Y5 E, V3 i$ TSince the fluid is incompressible(由于液体是不可压缩的), we have: D+ S* R" M% Z. p

    3 i6 H% t- O% c' B" b公式
    : L( a) m4 z. H$ N& Z  P
    . }" A4 Y5 `5 s) `/ zWhere
    5 E0 q0 [6 Z7 b0 V/ |0 S2 }. S" I1 ~3 \4 w$ U5 b2 b
    公式" @% q; O; d5 z  q( t
    . i/ J$ ~# |1 z+ W
    用原来的公式推出公式6 T; `* o; k& |& H: f; A# M' b4 d
    3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到)8 b+ B' `7 ~8 W* U/ s* a) x

    6 ~  f* W; @1 o1 \1 p+ a0 m公式0 x) y6 I, \7 c6 k, f6 L4 ^
    3 O) p0 d: t/ L6 r6 r- |: t
    11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:3 L: N$ w, y5 u2 N
    " u' f+ \! M9 }- M% d7 F7 @
    公式
    1 w+ R7 F4 S, v* l6 E
    * J4 R) ?# ?3 D# E2 _  P, ~12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)
    # a: k- m% U7 @( l1 K% P" o$ A5 t( }- f- C; _) [
    公式
    * v  O1 j# J/ S6 P" t# \, O
    4 w2 d2 H. c5 g% p6 E& p- WPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have; |) U6 r$ {, R! p# @
      G3 b: v' ]" |
    公式
    ; q3 O+ R2 d# l2 I* ^. t3 j( f* l
    - i: ~0 `9 K9 O9 \; R9 l  [Putting these into (1) ,we get(把这些公式代入1中)1 R2 L$ b0 j' S& S+ M4 H. D, A

    ! ~7 @7 x8 N! @公式9 D! s" x: e5 ]$ [
    + `. q! t* C* ]& ]
    Which means that the/ ^4 M2 m4 z9 \$ g* w
    : }" ]  D+ O4 s4 Y0 @1 g
    Commonly, h is about
    & C+ H: M4 L, m4 Q, J: S3 |2 a8 j$ I& c3 V7 D
    From these equations, (从这个公式中我们知道)we know that ………
    # Q" c2 `# z% j5 Z5 x
    0 Q, G  {; Z  y; ^/ ^# B. ]$ ~ 
    " x1 U, R7 m8 O5 S5 J2 R" Z& P) n$ u2 F; O
    引出约束条件8 A* t6 p/ A3 t. \
    4.Using pressure and discharge data from Rain Bird 结果,% Z; ]+ r) g. M+ J9 I' a

    9 C* R; [) ?3 w( WWe find the attenuation factor (得到衰减因子,常数,系数) to be- V9 M6 v- [1 U; u6 ]! w+ }

    ' B0 E/ c, T3 }2 G3 a5 j' e- p公式' _) U  y! G: l) V" S! q

    0 `" L7 }" _6 G$ a1 T# s/ A) n计算结果. O$ r! x0 Y' x
    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)0 g, R" Y- m+ ?

    " C2 V" n7 J5 y7 U! U! j公式
    2 M' d" i1 `7 a2 b& c1 x
    & S$ Y" \7 N% Q0 h0 ZWhere
    % s& @. A. ^5 l3 H3 y
    + \( X) t2 U" c0 j# [8 ]0 Q() is ;;# \4 o% p" l+ b  P& u! |
    & {: }1 C8 O# F* D: K1 o+ V% B
    7.Solving for VN we obtain (公式的解)
    2 p3 X9 G# w  C. g1 f3 L, _7 \' G7 I- a! Q* G6 M8 y0 |
    公式$ \- @( n* t2 ?( `

    " g% M6 A7 l6 x# y# @, E) {& qWhere n is the …..7 Q, H  X$ \2 K/ n
    3 [1 G0 e! `4 _3 h
     + C/ E3 F) h2 k* E* Y
    0 h9 E) U; j! v
    8.We have the following differential equations for speeds in the x- and y- directions:
    ' U) D$ ?; g2 m8 u* u; `9 Z$ @8 x
    6 _  z8 P! N4 I; [5 S公式
    8 i$ |2 r% M" A
    ) R" d& I) A& `; o0 @3 g( B3 W; IWhose solutions are (解)+ x3 l, x  J+ B3 @6 _
    ' @( J: r; {% ^2 U6 q* y. U$ U  ^4 p
    公式
    + l# ?( a# ~& L% F
    3 R3 }2 F$ w. }  Y6 B9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:, K+ b7 e- [0 y
      k$ c5 y# R( f4 o* [6 d; t
    公式5 i4 ?4 h; Y# @2 ^

    6 [2 b9 D$ {7 x$ `根据原有公式, ^1 f3 p2 ?7 k; ~9 @' H: c% g% R9 E
    10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is
    ) G4 M8 j& Q6 _. k+ C( ?, L% w( S7 |; T/ S
    公式
    . z3 M0 M* ]/ R0 y+ Y" \
      w5 p! ~: u4 d5 SThe decrease in potential energy is (势能的减少)
    $ q+ M' f$ J5 W" y: j5 U% p; z/ ]6 J4 W: r/ t
    公式
    ; x7 n! o, K7 ?# N0 K+ N" x* c
    0 j8 G$ _- n9 Q+ IThe increase in kinetic energy is (动能的增加)
    ( Z% Q& Y; ?3 G# q/ u1 q. v# {1 w% s8 z  w* f
    公式
    % N9 _) X  c4 s" y* j* D) o* q; w5 j. m; G7 r1 p3 H' @, {4 T0 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)
    : ~9 B) L# G" u+ j9 F+ Q& \, s; A# l- j& ?& L* i
    Where a is the acceleration vector and m is mass
    2 _1 W: s* f# i! a4 T; \$ x% {; [% j, [1 L& Y7 {
     
    ' o: \/ u  A6 F  w1 B5 y: ?5 P" q/ k
    Using the Newton's Second Law, we have that F/m=a and
    - W" B) n6 r2 h, o" f4 L
    $ B8 M- Z9 i. e公式
    * A$ p2 w( k% O: f! ~. D  L" G0 U( q1 i$ n8 T" w; [* p5 \8 G5 s
    So that: N7 m" l: ?2 h; \- Z  n

    % x; f7 q; ^6 H6 L  \公式
    6 G2 y  Q! |0 M/ G' C) W) j; o/ \( O
    & H+ Q: ]( Y; m& T8 _5 DSetting the two expressions for t1/t2 equal and cross-multiplying gives! \4 i9 g. P6 R) V/ F' w( c& S
    $ l9 \0 O7 f; C( I3 q- N3 t- A
    公式0 z: D! g, Y/ ]3 l3 M5 R: S
    8 R. i  H. X0 X% r& R: h3 A
    22.We approximate the binomial distribution of contenders with a normal distribution:8 X2 g- h, u% f4 m* j9 C
      \3 v% ^- [0 w2 u4 T
    公式
    5 M* _# f, z: P4 D
    * H8 c! T" {7 z$ yWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
    2 X/ B4 X( v" }! [, S. X5 r
    ( @5 b8 B3 Y# A" T" `+ P3 u6 P公式7 D8 F  F% f$ H9 R( ?
    ! V" \3 E- z; y1 ]
    As an analytic approximation to . for k=1, we get B=c5 G. ]: ]2 E" {
    2 a( u7 D% `+ l  N
     3 {8 }  r# r, R6 _% B3 v
    . I( H& c7 \" d+ \& O" _" y' @
    26.Integrating, (使结合)we get PVT=constant, where" _9 b) R- W( ~+ ~& v+ `

    , w- `8 t6 G; V/ z公式
    - [$ C3 H5 h& q& S1 f8 Y) E
    6 @7 `" S& w. l& ^7 T( m( i7 wThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so" `% j. ]6 ^+ F* ~1 b7 x! ]
    : J6 V7 N0 u" ?/ R1 j
     
    : u: k% q1 h2 P' s8 l' x: T
    . w1 T9 p9 J  ^, f0 O% A- I! ]+ @: M9 ?# U23.According to First Law of Thermodynamics, we get6 ~$ i# H0 M" C; ~9 }5 \, y

    ; V0 E6 q- }) V+ h公式& ?. N$ N, b. X8 x' H5 l& @+ B9 E
    1 {3 `. \$ c& H( a' g; k
    Where ( ) . we also then have
    ; g) X/ c1 x4 F4 w: s$ \5 t' U
    ( a* a# @1 [+ i9 @& B公式
    0 n. t; M. {8 B$ Q) U: ^8 K4 M, B' g
    Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
    # w- }, P1 |7 j9 Y+ ~1 i! S
    " g2 O  _, n5 k% g: j公式
    2 {  b6 g/ R6 l0 P1 I' V
    9 B  @0 c7 g# H% z2 GWhere, K: T1 U+ Q5 Z" X  [8 u3 v
    2 v5 ?+ c7 ~( B$ e* y1 K7 n8 r4 z) [
     $ w5 j$ n) H' A

    6 j# v* h* |% w5 [8 ]对公式变形8 F( w5 `+ a$ o. {4 l7 k
    13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)
    6 M) X* f0 j/ q6 v, z) C1 ~. A! a; ^0 U$ v( K( w* l# k8 y6 k
    公式& h( t6 a& }; m

    $ a  _2 j! I8 mWe 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( G. T6 e; w' f- S9 _

    / P% A' K, p6 I' t0 B2 \0 @1 U% M公式
    3 n! \$ e. x9 a
    4 |* h) p$ e. ]. o% F. V使服从约束条件9 ]4 Z1 r- n% ^8 h3 _  N
    14.Subject to the constraint (使服从约束条件)5 l" I* `  e& r9 M/ U8 V/ X

    9 r% m% F2 t9 l) o公式
      L  g& m* B$ y3 T: ^& N! B- M' b; G" T
    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)8 A6 l7 ^. g- |  x5 R
    % q4 N3 J: Z3 l) s) G
    公式
    : x1 \# r! N5 b, x( G) V- L
    / @+ v+ J8 W% b/ D1 v* q; RAnd thus f depends only on h , the function f is minimized at (求最小值)
    " I; {0 [& }  O& |4 l5 ~5 s( \
    # j0 x( u1 B% Z2 A公式( R; U3 w' S0 z" |2 Z7 T
    + ~/ q& S/ V6 v3 b" R# R; u& x
    At this value of h, the constraint reduces to! ^  E' W+ l5 G/ m( P
    4 k* `; m: r; a+ Z% j* W& j' H: r
    公式
    # ~" P& }1 ?" h( I3 p1 z" Y) B) e, S
    结果说明
    1 O/ z9 u1 M; U) A+ j$ e/ V) T0 N* g15.This implies(暗示) that the harmonic mean of l and w should be) B2 }; x+ f2 m4 v4 ^
    ) X' e2 f& h% X8 y9 F. f
    公式  c2 ?4 w6 m" I3 l9 \4 f# ]

    ( s  d( \! a  y1 F, S  e9 DSo , in the optimal situation. ………* w  m( s1 J; y% g4 f
      I" m; G3 h* \6 o7 J9 I4 {" \2 }
    5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is8 G; w' Y! |/ y; @# J; Y
    - y( {6 \+ l% ?+ n4 Y4 l3 J) g: Z
    公式
    - \0 f2 e! R7 D0 e# c6 V; L0 Z2 j5 k" n; M5 z
    16. We use a similar process to find the position of the droplet, resulting in
    6 _1 Y* P! X4 A& w
    # Q1 D# R% s  _9 T公式1 @. U: V3 k! \& O0 n) X

    - K$ G) F* @8 s: h: F# vWith t=0.0001 s, error from the approximation is virtually zero.
    4 R# |! e) k4 o- Q! d* [
    9 J3 e; M) \# N5 X) c 8 H2 D; a, h7 [9 A
    , T% K% l9 s& c3 w( ]* D
    17.We calculated its trajectory(轨道) using
    8 S" c2 `; L2 h. |$ y" O8 d, ?% U  A
    公式
    2 z0 i) i; {9 m" J6 Z* _# A" u6 N
    18.For that case, using the same expansion for e as above,
    - o/ z! h0 h0 Z) J5 ]) V/ `( ]/ u# m! b, @
    公式
    7 W' o; g0 t& K2 k9 |/ x4 @; b% F" _1 L' q: x
    19.Solving for t and equating it to the earlier expression for t, we get/ P& m- @& V; L9 o# U  N

    ( X& t; W% w" C. f+ B公式! [) a- p4 |8 G, M9 i

    ( w) ^; Y% z: v20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
    : W3 `% ~8 `# S6 O8 o( I2 F$ s! t5 Z
    3 S( Y/ j9 e4 h* U) y公式7 m# m7 m$ y3 p- k. c+ H9 |9 C2 N
    1 g6 R. S6 i" S9 L7 x/ O
    As v=…, this equation becomes singular (单数的).  q& a9 Q4 [0 V" j& Y& q
    . _7 ^/ ?$ Y- T+ Q
     ; r0 h2 h9 J6 k. }

    % W, w3 ?; p3 t! Z" r' I8 F8 r由语句得到公式
    % Z: @; a  S$ z21.The revenue generated by the flight is  u  b$ v2 N0 J2 X  I3 ~1 H- M; A2 U
    : T. |# S5 Y2 h( u
    公式: `% s2 t. E' t5 q# \9 l' x
    $ U& [5 S' F! s& r" a( M
     5 j0 m& S* w( B6 w
    2 n1 e! u; w: w8 h- G, A( R; M
    24.Then we have
    ( I( L+ s' l, L: _
    # F/ E& x& s, _) t4 z2 e公式
    " I- V: r5 s, U+ ^
    * p; O6 [* J% IWe differentiate the ideal-gas state equation
    $ n% O( A# f8 b  D3 K% h
    & |" w6 a& Y0 R% b2 A公式
    1 Z8 U& d. t1 w7 l) V4 P' L# m/ t) N8 y) C
    Getting+ F1 l! m% a3 b$ k( Q) w3 c

    ' f6 Q9 B- p+ p; G- k4 N公式
    ' H% S" D' Q4 y
    " d! L+ ?0 j* G, P8 p$ a7 x25.We eliminate dT from the last two equations to get (排除因素得到)
    8 L' J' C0 C, {3 b/ f+ X, h5 C) j9 X- Y/ E6 I/ j9 E
    公式
    4 S; I2 A  m7 J, N% G$ Z- p" I
    . C5 [$ h0 ^: c ; \" @7 [) F0 A2 b- a: s- K) }* n  Z

    / Y* N. y  [0 }7 _% K3 }22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
    8 ]9 @0 u8 F& V/ F2 j5 B3 ~  l- e" G9 h  {- ]3 y6 I  t1 B: j6 w
    公式/ h7 `: i; E# i9 t, T$ ~' S
      p6 X; m, D+ }
    Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值)
    - b9 |# D  \. F3 J& k8 g8 @  E1 r
    0 \; Z% t" ?# [+ v5 X* Z) X公式
      K4 Q% J9 u$ g1 g$ L1 x————————————————; J; z3 z% i! f  U; g) d) I
    版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。
    ' y% v2 F5 F0 q原文链接:https://blog.csdn.net/u011692048/article/details/774743868 N9 v1 [9 f2 Y1 J! @
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