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）

2 W$ U8 d2 o0 y! R( _2 }  e& @4 Z0 w公式
1 w# x2 U, m! e; Y3 ]
( Q8 T& [1 _! O4 c9 `: YWhere

( [* q8 A# X* @6 N: a  a% q符号解释

According to the assumptions, at every junction we have （由于假设）
  Q0 E/ }% q& f
5 b* i$ m8 W* g; D公式
$ r$ b4 ^' u( {9 m* y, P
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）；
' f6 G2 C  s2 I- n" b7 i
公式

Since the fluid is incompressible(由于液体是不可压缩的), we have

公式

0 }, I0 R3 Y% ~3 X1 _Where
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公式
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用原来的公式推出公式
9 w) ]) }& E7 y8 |' q5 h& u3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）

' x. |. X; W& L0 q# f  L5 k9 ^公式
; t& `3 _' {3 r: [- v8 J4 K! _
11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式
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$ X, m. V3 q; w12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

; r$ M% {: l) Z" g1 L) ~公式

Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
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& G* _7 W3 @) T3 k& X7 G公式

# y2 f9 F- ~( u+ ]Putting these into (1) ,we get（把这些公式代入1中）

* H1 H% g  A+ J- d; ~公式

' U. F+ e: P2 U& oWhich means that the
2 d* a. U% B% M8 N$ A
4 e8 d' O1 b* A1 ~0 u1 ZCommonly, h is about

From these equations, （从这个公式中我们知道）we know that ………
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! D) z' f, U8 \4 H) S; H 

引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,

# l( ^8 {' e4 {" p( A0 PWe find the attenuation factor （得到衰减因子，常数，系数） to be
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$ k' e9 e1 }! }* t6 ?+ `" Z7 l公式
0 U0 l) {, @- P& W/ t1 x7 q$ x# n
# j3 X: P# l* R1 |& T计算结果
' M4 o8 m# b5 l/ s7 L' d+ }" {/ k6．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
4 N5 o1 O9 L7 k3 Q
Where
, X) p( @! N. D& Z$ ?- y; N: \
() is ;;
( j) s  Y5 Z' ?5 T% L, s; H
7．Solving for VN we obtain (公式的解)

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Where n is the …..
6 y& Z) a" p- W2 \
 

8．We have the following differential equations for speeds in the x- and y- directions:

公式

7 U: o+ T' m, {  r) UWhose solutions are （解）
) @) J, M9 s5 O. U. U+ k
公式
: U. ]! t( _% Q! o) u) Y" N0 S
2 \7 D* K/ J# k8 G: X9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
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公式
; [( P2 Y9 X& p1 a
+ n8 Z1 K+ n# P# o$ c7 e- s5 f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

  V/ ?: U# [! d公式
, m! ?+ c1 n  Y8 D) V7 I! \
The decrease in potential energy is （势能的减少）
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公式

The increase in kinetic energy is (动能的增加)
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公式
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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)
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Where a is the acceleration vector and m is mass

 

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

: z$ F; X% }% \9 u0 D# |& Z' U公式
! ?( b0 b7 F$ \, E7 |5 ^
" }+ \9 S2 y8 C9 w2 L+ a$ WSo that

公式
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3 c' }/ k0 V7 u* O# rSetting the two expressions for t1/t2 equal and cross-multiplying gives

公式
7 v( T) Z9 p/ }8 i7 ]- F' Y/ L
3 S7 O* k$ F3 z0 r; P22．We approximate the binomial distribution of contenders with a normal distribution:
2 F2 U, u; {6 U) F0 k' v
公式

Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
9 ~4 l; Q8 P/ }( B1 c
7 l0 X4 s% [8 J/ N/ ^6 B% B公式
' b. V% G! G7 ~' A; @$ w  B0 J7 j
, t2 j2 n  D5 LAs an analytic approximation to . for k=1, we get B=c

 
' X1 D; z- N' Y" Y
, S$ V) j- s! ?. w0 |8 w9 ~& n26．Integrating, （使结合）we get PVT=constant, where

, ^0 \" K( I. v" e/ c$ H公式

$ o" i1 N5 j3 P5 iThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
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1 U& S0 `, m% t) U
23．According to First Law of Thermodynamics, we get
, D. h  r- d5 T6 t8 l
公式
; O* e- Y" D7 Y3 {8 A1 m  l3 B
9 S4 e& A- R- V8 v- BWhere ( ) . we also then have
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公式

Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

公式
( ^% i% G, l2 A% Z* b/ _3 T( m, l% n
Where
" y. Y: j/ p' K& p
 
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对公式变形
4 c$ M( }3 F) ]9 B13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

公式

+ H9 s# p1 }  h3 z# ?" T8 l7 ?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
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公式
7 N& B! h5 ]7 A! V9 l
4 q( r7 r8 d6 k) \" O! [使服从约束条件
14．Subject to the constraint （使服从约束条件）
( X7 F+ o& @% x3 l) L' _$ u
. m# f4 T/ }3 c9 V, C7 M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）
* Z' g% Z4 z+ {! P$ o) P
8 d# f9 T- n7 @$ P( e  P1 H. F; Z( x公式
) o3 \: d/ L8 ?4 @: B0 N2 w
. S4 ]8 h+ u% v7 m, mAnd thus f depends only on h , the function f is minimized at （求最小值）
+ H# `  d% b4 H: Y, f( o% l8 X/ d
1 g8 F! I3 C: C/ |7 ~$ [公式

+ n- _2 |9 E! k1 V& \) M4 IAt this value of h, the constraint reduces to

3 R& X1 T; K+ a& @! [; }2 L4 w" w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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So , in the optimal situation. ………

9 j9 }, y' W5 U2 ^( P2 I5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is

7 b: D+ p, l$ R1 ?8 @公式

16． We use a similar process to find the position of the droplet, resulting in

公式
  |- f# o$ e/ o+ t  M  B+ O1 g9 b
With t=0.0001 s, error from the approximation is virtually zero.
# x# W" A* D8 l
 

" C$ A$ O. r0 `% _% r) z17．We calculated its trajectory(轨道) using
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7 e8 D& e3 k) [公式
) R) F7 @  I2 a9 ?& e7 b6 H/ h' I
18．For that case, using the same expansion for e as above,
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9 @( \' W; s# b0 f8 k8 E2 q5 z公式

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

公式
. T' ~  J" [! Z+ H( a
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$ r' Z$ j+ ?6 f6 q* d; R
公式

$ L( {: E) f! ]/ XAs v=…, this equation becomes singular (单数的).
- {6 \% B: t5 M. T0 S
# K7 i+ B5 m6 f6 L6 {7 d9 ^ 
/ k0 E/ L/ Q* ]2 r7 S7 f: `. R
由语句得到公式
, x4 ?1 d# \) Y6 V7 Y21．The revenue generated by the flight is

7 P( D/ l- {* C8 ]/ G9 i公式
9 a# u- H  i  e# r
4 p* Q) c& j5 ^# d 
1 G/ Z% L+ Y1 w1 x- D
24．Then we have

: j' P  A4 i" c' Z* {7 H& q* K8 s3 C8 D公式
1 C6 j' k+ `: R
We differentiate the ideal-gas state equation

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. }3 T' W% Y' ?% H6 _, N
Getting
4 Y+ [! [/ |/ J4 }" J
公式

25．We eliminate dT from the last two equations to get （排除因素得到）
) |# F& X4 D) V" g
公式
7 S" _9 O4 l8 |. `  D
 
2 W5 l* C9 f  Z8 a7 M
$ ~( Z6 k' Y; V% {6 z2 D1 ?- `- M1 x22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
! c% K* e' \/ s: |4 G: M
$ |1 s, Z2 d0 R1 v6 s7 a% Y公式
$ n( ^# q& y. e( X1 M* J
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
+ A. y# @+ k1 G! g
公式
: @8 J+ @# w0 r# ?————————————————
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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