美赛数模论文之公式写作

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由假设得到公式
+ H3 {& F+ R; B% w2 s; A1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）
/ B# v, X, i/ \9 W2 D5 l6 o
公式

. q- z" n' C) e* c5 AWhere
( C, @$ z/ d$ c. ?- c& `' m
符号解释

# s0 |& h1 x, }% ?# wAccording to the assumptions, at every junction we have （由于假设）

$ F0 y7 W& F; Z公式

由原因得到公式
3 y" O! `/ m" n& T2．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）；
4 S; u4 R5 i% x" D& y
% A! _% _4 R3 j6 Q7 p% T! Z) `公式
5 l/ z% W5 ^$ A* F% }
Since the fluid is incompressible(由于液体是不可压缩的), we have
/ `3 y- x3 `4 ^% i( m; n# f
; d+ N) i' q$ B$ Z公式

Where
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公式
+ |/ D4 e2 C. D, n6 ?+ m( h
用原来的公式推出公式
( S2 n9 M3 w1 f3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）

; ^0 a+ b- @' \0 w公式

9 b6 v* O; b3 D1 E11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式

2 B: K( ]  W4 |- T& ?% A% A, `12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

公式

Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
! H5 R; e: w" [# c8 w( k- G! r
公式

, i6 d2 ^+ E' S; ~! |& a2 P& tPutting these into (1) ,we get（把这些公式代入1中）
. I9 @8 ]' O7 \8 n' u1 f3 |
公式
; N9 Q' ?6 S7 J: U% }
& a8 L& p7 M; ]  j6 D) u: K0 _Which means that the

# T' I) U0 ]8 P- u* e/ GCommonly, h is about

From these equations, （从这个公式中我们知道）we know that ………
6 T' {+ J) j# d/ u& y- k4 M* e; h
 
7 r3 d8 ~5 @& S
* u0 ?- S  d( W" W: F引出约束条件
! T- a" ]: [$ _3 I9 D4．Using pressure and discharge data from Rain Bird 结果,
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# L7 E# p& j2 c$ B: u0 s) vWe find the attenuation factor （得到衰减因子，常数，系数） to be
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公式

计算结果
% k6 G% h! U' R6．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）

公式

" d6 ]8 q6 Y# k% n; SWhere

7 ?( {" Z) x0 C! c, U8 p1 ?() is ;;
8 H5 d+ k8 p) b0 H/ Z7 y
6 P. l* ?+ z5 e7．Solving for VN we obtain (公式的解)
( P6 G8 x3 g: o: x- g# V5 I
2 L" E: a& o+ b& D公式
1 f6 D9 i9 r- r! H  l
9 W& ]$ \+ A# a# e% ^Where n is the …..
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8 ~5 ^! H7 J/ _, f
( ?  k3 ]) N6 ]* K5 J5 A! T! ?" @8．We have the following differential equations for speeds in the x- and y- directions:
/ v; f# q# D  H: c! P' }7 f4 R
公式
8 v) \6 ~( `  E
" s- c1 k/ N( E6 q/ S3 wWhose solutions are （解）
7 U( o. T& c4 Y, t8 M
: z  F, m3 ~6 ^. ~& \公式

- m; ]4 j: p' ?9 }& o, R# j- w3 ]  j9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

9 A" u0 y1 ~& x6 s; l# o公式

/ Y* B3 ?9 ]- T根据原有公式
8 f0 ~4 X6 t( c4 [& I0 m& B, V10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is

/ T, m+ J1 p- Q公式

The decrease in potential energy is （势能的减少）

) Y4 q5 _$ K+ c4 h5 Q% o公式
) H, n; l0 h2 N/ \6 o$ f7 E
The increase in kinetic energy is (动能的增加)

公式

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

 
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  E# N' l$ |- k0 V  pUsing the Newton's Second Law, we have that F/m=a and
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5 ]/ n) L4 D6 N7 ]0 y2 z$ d公式

- L. v2 s# P) |* F! rSo that
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公式
' v3 S. n6 N) n4 ^0 k! S0 Y' L
) u6 ~+ h+ }. `5 M4 m/ b3 LSetting the two expressions for t1/t2 equal and cross-multiplying gives
& L  `/ ?3 C+ t8 f% ~
( A3 _# V9 l# \) W/ C8 L0 O公式
& X- F5 W  O/ n# K/ A. v
22．We approximate the binomial distribution of contenders with a normal distribution:
8 o$ f2 C% J9 f* N4 V
公式

# L6 y1 a& e) q# GWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

公式
1 [, g$ s( L/ b# B* ~% @: e5 O+ c2 W( {
9 [. B. k. `8 d( l+ xAs an analytic approximation to . for k=1, we get B=c
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: t! v) X7 U  B- C
# H4 h' P$ K8 o6 {! O26．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

8 a+ `% j7 e4 r8 } 

23．According to First Law of Thermodynamics, we get
5 G9 s3 q+ C0 j- R
- T; b+ @8 a  x# D, c公式
$ `1 k' m! V, w, J* Q$ M0 e
Where ( ) . we also then have
; `9 L' M  }2 w# V! H
公式

- f( T, Q/ m3 h% XWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

) I; W% B) o. L8 A公式
' d" G) l7 g; ]* x3 v2 n
: `7 `) b% Y( C( pWhere
% S# y3 `: E6 ~1 ?
 

5 m  D5 n3 c4 R$ E' `& E对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

3 }. a0 A, j) M0 L+ h公式

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

公式
* B$ N  D( d% m7 |. D* `9 }
使服从约束条件
6 P  y: a, I% v( Y6 t6 M+ M14．Subject to the constraint （使服从约束条件）

0 u3 q+ u8 |7 z3 `1 F公式
# J6 H! f) |3 ~1 _" \
! F9 ]( {0 {  c; q0 zWhere 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）

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And thus f depends only on h , the function f is minimized at （求最小值）

  }' n8 D+ {# X公式
" g6 x/ g! @+ i" w
At this value of h, the constraint reduces to
0 `1 Y4 C7 x3 k) F  n5 {$ M
9 v$ U  o& L6 _公式

结果说明
& ?3 ^+ X$ M+ a$ @6 Z3 w3 h15．This implies（暗示） that the harmonic mean of l and w should be

# q6 P7 c- Y, Y: M公式

So , in the optimal situation. ………
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: K9 ^( W) A% v. l9 G  `7 V- h1 @5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
; W$ s  d5 B; X9 G: q! @. [0 K9 r
, T' Y7 y7 @% C' I2 ^, T公式
+ S) _2 d, s+ a: K4 w
16． We use a similar process to find the position of the droplet, resulting in
# `% ~6 q* l! s: |0 K$ Q
1 G, v4 Q3 K' K& `公式

5 x% y% a+ c5 mWith t=0.0001 s, error from the approximation is virtually zero.
3 K4 L8 \0 a! o" ^
 

17．We calculated its trajectory(轨道) using

公式
: f* _; S  G% g3 E: R
18．For that case, using the same expansion for e as above,
1 g4 R4 c, \# L* w. n
公式

19．Solving for t and equating it to the earlier expression for t, we get

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
$ Q3 S7 f6 |& u' E6 ^: }, G
公式
4 r5 ~5 {' n) K9 ]- R9 a6 `; @* t
( t, l1 [% A  E8 u! S. ~As v=…, this equation becomes singular (单数的).

) Z6 x8 \! m) j, B+ d. w. u+ q7 Q 

由语句得到公式
/ U- m0 ~; [4 W% h21．The revenue generated by the flight is

! n5 B1 w4 Q) m- Y5 n  }  L5 Y$ t- N# W公式
+ ~9 J2 E+ o1 V( B$ ^3 j
 

24．Then we have
; i3 c6 h) h% j) ^2 E# y7 {! A
$ {3 @+ T5 `* u6 P& c公式
0 ~. h) q4 l5 g' b. Z2 p. [
We differentiate the ideal-gas state equation

公式

Getting

2 T& M- W0 y0 b: t. b公式

25．We eliminate dT from the last two equations to get （排除因素得到）
7 i9 s6 |- `5 B& d  C! `
公式
, z! x, x' i) L' z# `1 F
- o' G2 u8 {( B 

22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
* a, x; `2 c0 ?& W0 c6 W2 k
& q! ~$ w  U4 Q/ s- o公式

Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
  T* {% X" `  u- A+ }" u5 r
公式
( G% i+ g! L4 a  n————————————————
版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
  x( i& m+ h5 |4 E原文链接：https://blog.csdn.net/u011692048/article/details/77474386

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