美赛数模论文之公式写作

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由假设得到公式
1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式
/ z; i* V8 _! \( U$ d
* m1 D% `( x' R8 g' b% l+ S4 y/ vWhere

符号解释

8 z" H! H: O+ |6 k1 p, c3 o( oAccording to the assumptions, at every junction we have （由于假设）
' D/ {: N! L7 {* U1 t4 d: D/ n
公式

1 W  c' {6 N) Z9 u) n, F7 L& T/ |由原因得到公式
7 \2 b  [- ~8 v8 e& f6 @  L) n2．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）；
, b" y) D% @+ r2 }8 ~7 j# q
4 e. h" u6 M! ]9 }" v公式

8 u/ B0 e% M+ Z# ]" N# g! `Since the fluid is incompressible(由于液体是不可压缩的), we have
9 m, A1 G& j' Q+ x- l) u& B, M
. U/ `6 s  ~$ t' u* H) ~/ G: T( r公式

% A# t( \0 m" K5 aWhere

公式

( i* M0 u9 z% w0 B" y5 ~0 U5 \用原来的公式推出公式
3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
, L$ B3 F" }% X' g4 _/ {2 @/ ^. ?) ]
公式

2 u! ^" n7 H' g: w: S11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式
& F# d  ?$ U: f" g2 g
+ B2 W0 A6 ^  H8 V- C12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）
& ?7 f1 d0 D$ p. ~2 e7 s- v, s$ q
公式

6 g, ^3 @  s- J( ^, ^# a3 T; R2 n2 jPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

公式

" G+ d5 t2 f1 u) QPutting these into (1) ,we get（把这些公式代入1中）

/ o& R7 I- \! j# P0 i! V4 s公式
: w# z3 Q" p# N5 S$ g! `
5 {* t, E4 O- Y  n" VWhich means that the
' Z- ]# m' b5 B7 P& l! }
$ O0 h. I5 `- T' C4 p2 BCommonly, h is about
3 ~2 A* k$ R; }# U0 w
% H# z8 J, u+ I2 N! q0 YFrom these equations, （从这个公式中我们知道）we know that ………
+ b( G: _: ?4 w% A5 b9 i6 ?
 
" p; ?& T$ \+ [0 t; @0 a6 C
6 g+ u7 J' B4 D引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,

We find the attenuation factor （得到衰减因子，常数，系数） to be

6 @! _$ b9 M) p; _公式
# o" p$ A- `% l  I0 k0 F
# `( _# A- k9 s+ n% \/ }6 W计算结果
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）
! {+ k( A0 k" N) r9 s; t4 h9 K
公式
9 }! K/ S  C' q* x5 Y# ]* e- D
$ m( p* M: J; m9 |6 ]4 ^- {; O0 p2 |Where
8 H3 \) l5 t; E6 n% [: G
() is ;;
! t" ^: ?9 z2 D) @5 a3 c
7．Solving for VN we obtain (公式的解)

) H1 y/ Y5 X; @, Y' `8 t公式
) o3 J+ d$ `+ M6 q* Y
Where n is the …..

 
+ c" h0 V9 c5 D1 v$ O* p1 P
' n, |3 t7 A) C* [# a5 q8．We have the following differential equations for speeds in the x- and y- directions:
( e) ~+ c, v8 _' q9 `& q: F* P
公式

/ T' L! l2 ?" C; q" AWhose solutions are （解）

( I0 T5 b/ B+ S$ q公式
; P; Q4 l6 Z9 d* O. |* K6 S
( d5 x- e& I/ q9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

公式

根据原有公式
10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is

公式
- e2 i. D6 U- ]0 _* T# P+ r1 a+ S
6 X; i# f4 f9 u& jThe decrease in potential energy is （势能的减少）

0 g% N" k3 [6 S/ ?* \; z* |/ @& P$ D公式
  @' w( J( R' i
! \6 j9 C9 }. q, l7 L* yThe increase in kinetic energy is (动能的增加)

- ]. Q2 ~7 \* C- U' J公式
- |; u1 A" ^6 m/ ]. j- t$ w
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)

Where a is the acceleration vector and m is mass
. h3 m3 [7 V5 h" O, z& u
6 G4 h. A5 P! D& I5 F 

Using the Newton's Second Law, we have that F/m=a and
8 I" J/ ]0 F! s9 D* I0 @) e
公式

So that

/ N# X( J0 F0 J4 m/ D" i公式
% L) K; s: l0 N- V* B
, K+ K% K9 q+ K  C9 E* BSetting the two expressions for t1/t2 equal and cross-multiplying gives
( y( P* K1 N* i, S7 b
公式
3 Z/ g1 ?6 W  K! V& ~' h
22．We approximate the binomial distribution of contenders with a normal distribution:
- x! c8 t( f4 J
公式
9 J, v0 j* \! C1 H% N5 y
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
0 s2 h* C2 P' h6 `# {" J% Q
公式

As an analytic approximation to . for k=1, we get B=c
& A" w1 F3 Z. _, `% U+ a
% v' a3 D( A8 U; [- W  i  G6 H+ t  u 
( N/ c; t2 M0 ~% @1 M9 o
: |* d+ Q; N1 r8 }26．Integrating, （使结合）we get PVT=constant, where

公式

The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so

 
9 b, k, F3 f4 d( E; r2 W  v: X
) z/ @- E$ z( k5 @& A; ?; L) V' L23．According to First Law of Thermodynamics, we get
9 E# O8 c1 i) ~5 t3 O9 s2 ~
公式

Where ( ) . we also then have
4 m- {' X" A  ]9 o8 `
公式
) f+ r  z7 s0 q$ E- |6 F  {5 W
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
8 q9 R' ~2 k2 i( I: L2 w! t' x  N
) b& ?" V  v5 j2 P* ^" F# A公式
& O* f. E$ K4 A1 @. E' ?' L8 B" Z
: z- D, z' f3 u6 z7 h; J; M) AWhere
. m1 q) }+ E! r4 [6 C1 c# V' U
) C) ]: v- j/ X- D( m* \) H 

+ B; f3 f& T. S0 i7 [对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
% Z- ^5 ^6 a# c/ D/ x) }
5 e4 E( `* T5 G5 t公式

# K1 b1 o# T$ c& IWe 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

1 ]! b/ _% y$ j& g, }2 {公式
3 h) g; ^1 F: r: [* g
使服从约束条件
14．Subject to the constraint （使服从约束条件）

公式

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）

% v4 \# w2 d: a$ i& R! ?2 \/ P7 x公式
$ J+ \' ~; E3 e" n7 f. Q% f
( O# C' p# m9 y: u1 _# |And thus f depends only on h , the function f is minimized at （求最小值）
  H; S' y1 E) d6 T8 O
  w2 k) i5 a6 g6 ~6 }7 m+ A4 D8 ?公式

At this value of h, the constraint reduces to

. ^6 |+ e' r# ~' V% k1 _公式
& X8 h' i5 }: \! v) L
结果说明
! n% L. Z) ~# P5 E" z1 t15．This implies（暗示） that the harmonic mean of l and w should be
) w! ^+ {3 F. V
公式

So , in the optimal situation. ………
$ Q$ a, q* b9 v. u+ Q8 Y
* A* i. @. f( C" A, p) p- T5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
. K+ K, A6 v) E  H7 B" k, P# G/ ^) q
公式

6 r, B, ?; q+ D# |# G% v* X16． We use a similar process to find the position of the droplet, resulting in
& _8 Y$ Z9 G0 a" i2 l  v9 C
/ W8 ~2 o/ y( Q3 {- ~5 k公式

With t=0.0001 s, error from the approximation is virtually zero.

 
9 C  _' q% z3 M4 P1 L, Q2 X
* K9 U! x- [; W, T1 m) n/ F17．We calculated its trajectory(轨道) using
) e! R8 J& p, z, U4 \4 e
公式
( C* n/ Q8 h7 n6 Y1 o1 t
3 E) m  f, c! C3 m! v% u18．For that case, using the same expansion for e as above,

. K2 l& i6 o& u1 U0 P! X2 ]7 z公式

19．Solving for t and equating it to the earlier expression for t, we get
3 O% O, f" N* F3 |
公式

20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is

公式
! K, ]! _+ D0 W0 v' ^
5 h. C* Y1 B1 x- ~" }As v=…, this equation becomes singular (单数的).
9 ~4 P2 ~6 {7 _
 
" I5 q1 ~/ b! [: G, R1 R" i6 d# N; _
由语句得到公式
21．The revenue generated by the flight is
# r7 Y0 d8 V* N6 E- ?
公式

! F' T# X& z% D. ?% C! S 
, z' ^* ]5 v6 o7 T' L/ X# j. M
24．Then we have
2 i- R( k9 z" S5 O" @  B1 S
, o9 n7 c9 L' e. X2 D/ e0 h% Z6 F公式
% o" ^2 |* W; c* \4 o  f7 f6 L
& ]7 Z" q8 R& Y8 hWe differentiate the ideal-gas state equation

公式

3 Z0 ?( }" u  F" CGetting
& J+ F* Q: Z, O. e/ W$ }
公式
2 i: t. k& P4 r/ f
3 e9 D2 S5 l* n* d! y# r; D) ]. ^25．We eliminate dT from the last two equations to get （排除因素得到）

公式

 

22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
4 b% s' [5 v- R: ~6 ]3 A1 `
0 Q7 g9 W! e9 b: k公式

6 |7 ~; v, W) H2 \Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
5 {2 e% ^; t, P% g2 Y7 Y7 d
: T. n& h6 Y% ]9 y; @* n公式
" n8 e3 D0 b& D- f5 o4 S7 A( I————————————————
& ~2 E, [0 x# u: X# M版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
原文链接：https://blog.csdn.net/u011692048/article/details/77474386
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