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）
- X. [7 h$ S/ S* V- f6 x4 B
公式

Where

符号解释

% y  N# @; ^  G7 S* |5 HAccording to the assumptions, at every junction we have （由于假设）
& S0 G+ l& U$ K2 \0 o
公式
. u- W  \1 m, Z5 {, ^, B3 W
由原因得到公式
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）；

: g; Z" W4 x& @8 w9 ]5 k公式

. Y  Y: X/ m( `' kSince the fluid is incompressible(由于液体是不可压缩的), we have
3 U9 Q: f) Y' z1 T! t' \9 {$ L
公式

Where

公式

用原来的公式推出公式
# t) w( f2 \& l3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
! {4 F  n6 m* ]3 w0 H
  _! {' G% Y6 T2 L' K公式
  ]1 `) z$ `# d* z# ]8 j
7 |8 z; y+ q2 u/ f11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式

12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

公式
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Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
) B7 W0 R* M9 l9 q) q0 q* q
公式
9 f/ j& _& z* T) G% o
Putting these into (1) ,we get（把这些公式代入1中）
- h  B& V, e4 @4 H9 _
公式
! _% @$ V2 `& N: n, n
, {1 _1 _/ l0 ]% x" E, C) p7 yWhich means that the

Commonly, h is about

From these equations, （从这个公式中我们知道）we know that ………

 
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引出约束条件
% |, @# s# l' L; P2 z+ d4．Using pressure and discharge data from Rain Bird 结果,
4 P0 X  v+ _# }6 e8 o- D
We find the attenuation factor （得到衰减因子，常数，系数） to be
# Z( r* j' N& b! O4 _
9 N$ A6 a! U( Q; W+ `公式
  g- x1 E9 H) F/ M) S
$ H* Q# d5 W* H! {, |8 B计算结果
# x: `% a5 z; @2 T$ ?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）
. }% q5 H* [; d
# s  |/ x0 Z( R/ u( k公式
$ V/ M# T5 C* \" U, P5 v) p0 s
Where
9 U  o3 d5 h  R
5 X. E5 [3 d7 P( `& |% e! @/ O() is ;;

7．Solving for VN we obtain (公式的解)
8 g2 E4 T; F0 t  C+ k
公式

Where n is the …..

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2 j' g& H, q8 ~) N: w* i: {  h6 \8．We have the following differential equations for speeds in the x- and y- directions:

公式
% A* v" H( c5 E2 x
Whose solutions are （解）

公式
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9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
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公式
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根据原有公式
' M) U9 f+ Q7 \6 q- ]9 G/ F10．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
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The decrease in potential energy is （势能的减少）
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公式
( L/ W9 b1 r" O' n) m
The increase in kinetic energy is (动能的增加)
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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)

% o  n/ ?. D, v5 i2 v# N' a1 s8 wWhere a is the acceleration vector and m is mass

 

Using the Newton's Second Law, we have that F/m=a and

) D) t* w! i$ [1 }公式

1 V: m4 Q5 o8 R2 |1 Y3 rSo that
1 d" Y1 T6 z1 U' g
) M' {5 U) s! Z" P公式

Setting the two expressions for t1/t2 equal and cross-multiplying gives
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公式

7 w; p  u" k* U2 w/ L7 Z& U22．We approximate the binomial distribution of contenders with a normal distribution:
2 Y; Y3 C1 T9 J  K
公式
. s7 {& u* w3 n/ w  U' f' J$ F
0 `4 G2 H7 m7 z5 `5 ^Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

公式

As an analytic approximation to . for k=1, we get B=c

" `( H6 r; D) s5 V! [4 f 
- Z  D8 K" N, [- t. D& v' F
26．Integrating, （使结合）we get PVT=constant, where
2 f- ~4 C2 C8 M2 ^
" {- j7 p& {" N* [: h/ j& m/ u公式
) |. O+ j+ a6 D  T
" @- g; p# w; z0 J4 wThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
. X& N" `& j- `, C& d$ l- B" l3 h2 e
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23．According to First Law of Thermodynamics, we get
6 |. a5 ]& F/ k4 P, d
公式
& U: u3 J2 T, B
, ^; m2 q- k6 J! w9 x; {Where ( ) . we also then have

公式

" K8 g4 v- O$ u& aWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

8 D; b' P1 @& G公式
- ]0 s' F  P+ {2 D; ?+ b
Where
3 b- F1 c9 R/ ?, u
 

2 \4 w) H1 H/ D/ _对公式变形
/ u* f3 G6 r6 k- N2 }. C0 M8 d, m13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

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
  ]  G9 a" l+ I9 ^! f
公式
; u) R- t" |  M& ]1 W
$ I  G7 `# N7 F) S$ R8 I使服从约束条件
14．Subject to the constraint （使服从约束条件）
1 G  R* c" J$ U# q3 o0 c
公式

# M1 ^$ _9 L3 K0 IWhere 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）
/ Q; `3 R1 [; e( N3 g
公式
0 j( _2 k- R1 e. `4 ?/ Z! m7 o8 M
And thus f depends only on h , the function f is minimized at （求最小值）

公式
4 A* L5 q% x4 P- K
4 }4 E# B8 s( @0 o# P# pAt this value of h, the constraint reduces to
7 `2 D4 n3 e  ~( Z
公式

结果说明
15．This implies（暗示） that the harmonic mean of l and w should be
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; t" M2 q9 }4 r$ w, l8 O公式
8 \$ S$ \' C7 H" O4 x) z
So , in the optimal situation. ………
9 M3 ]' Q/ v3 J+ U; ?+ V
: s! Z# P, u3 }1 A( B5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
# v2 ]% w/ Z% i0 @- W
公式
2 G, G& W& v# o/ U
16． We use a similar process to find the position of the droplet, resulting in
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公式

) V/ Q! L/ D6 t, a  ?  q0 J* NWith t=0.0001 s, error from the approximation is virtually zero.
8 d9 `& H" K+ s: p, ]# t
 

17．We calculated its trajectory(轨道) using
; V; @( c9 H8 X: k5 _( z4 z- v
7 [9 s, Y9 c7 F& {. V4 v: P! v公式
0 S; u9 X$ r, t( c
: a2 f) _* b" U  ~1 M; r" |8 _18．For that case, using the same expansion for e as above,

公式
7 e: w6 {: n* p# R  x4 ?( _
19．Solving for t and equating it to the earlier expression for t, we get
9 E6 n$ y" }& q, F- X/ i# s( J
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

As v=…, this equation becomes singular (单数的).
1 J' S# b6 E6 z& i8 a- A* X
7 K6 Y) M6 y; g6 b 
' Y& F4 u0 t+ {/ [; y; Y% F. \: ?
& |7 a* D4 B" S由语句得到公式
21．The revenue generated by the flight is
( U1 I5 p1 x( a5 I2 u) R' ?4 f
公式
0 Q; \0 `: P- I. r) G% O+ e8 _
 
0 l' [/ p* K7 u* J4 U, r
' ]/ S6 B, v* h$ V* j2 E) X% G4 ~24．Then we have

公式

We differentiate the ideal-gas state equation
" g# K- g9 x7 ^$ a" |$ s. ^
公式

Getting

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, C: \' W, r0 q/ c' ~25．We eliminate dT from the last two equations to get （排除因素得到）

3 e9 E2 e1 T  Q" i公式
6 D$ M6 l1 W, R0 h8 c
2 G2 H% n: K. c9 A3 C 
# C" K0 M4 v+ I9 [. `; V
22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
) V+ z$ p/ N: t0 q; v
公式
8 V, {, V- b8 X. m8 g4 P
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
0 a* ]- b; y' v3 P
; E! n- W# O6 F( Q% Y; |7 |, n公式
# {6 }: w  z/ |. U  x3 _————————————————
版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
* N" x1 R4 H; j5 K, y+ Y4 w原文链接：https://blog.csdn.net/u011692048/article/details/77474386

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