美赛数模论文之公式写作

zhangtt123

由假设得到公式
" {8 e5 W6 ~+ j# \, O1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式

# W! h5 x& _% UWhere
% |+ i7 y! d2 {5 w# d) G
符号解释
9 Q4 B6 S. c7 v( a$ S/ T' t
" _# L3 C5 ?$ G8 C- P" P! ~According to the assumptions, at every junction we have （由于假设）

- d) ^2 X8 [, T& Q公式

) S' j* ^2 ~- y/ r% I2 A2 r6 Z由原因得到公式
) ]8 |% l7 n- V: o* S' g2．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）；

公式

; a7 I  Y! W' R6 D" KSince the fluid is incompressible(由于液体是不可压缩的), we have
1 b; R4 ~- _/ C5 f6 g
' s& K2 f" [. P4 i6 J! N  Z公式
( ~" \* _; _, Q9 s/ U4 U
Where

7 F. Z* U$ k% d5 h2 T9 _公式
2 |6 s/ Z/ C0 m. ]  Z2 ^$ v8 x
用原来的公式推出公式
! K6 A% [& D( S+ {. X$ r& j/ T3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）

公式

3 W/ \* f2 U9 D. w7 R11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
  e- A6 q. k9 `$ s! q
' R- K* v, v+ |5 k+ u- X! K公式

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

7 H+ N' Z  B9 W9 i+ B; D% T* R公式
: {0 p8 t7 J' d/ r9 Q6 ?, a; h
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

; b7 o* b6 _. [; F% ]公式

" d7 q/ f. n( aPutting these into (1) ,we get（把这些公式代入1中）
" d! T# S8 F3 g
公式
  h! Z0 |! t& {9 z0 [% h% f- r
, _, J$ g! X1 v3 f1 ]8 I! k( UWhich means that the
8 Q0 K: i9 J4 S% ~) _( d( ?% n
! N7 b( [% I2 a. ~Commonly, h is about
5 Z) C4 ?  z7 g4 b9 M
0 d+ a' X/ j8 z4 BFrom these equations, （从这个公式中我们知道）we know that ………
" w1 [, e! p8 S4 h$ q% l
 

" F1 V7 J  O6 p& f5 |, G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
+ m3 e! O0 Z+ f% I$ x- u
公式
* U. x4 |1 B; R. j/ ^# _
计算结果
% D3 }! e+ 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）

公式

, t% b& S  s! ?' hWhere
5 L7 A- M, i4 e! _0 A2 V
/ F& c" n$ e: E# S( w() is ;;
1 l2 u, K4 W$ J' p/ J6 t. G
7．Solving for VN we obtain (公式的解)

& V* {: t" ~9 k4 s! {  K公式
  Q. x! K. K4 L1 X) s0 n/ w
Where n is the …..

 
* I- z: L$ b* k0 \: j: p7 _
8 o' k% a2 F; v" F, N% N: q' s' I' f3 n8．We have the following differential equations for speeds in the x- and y- directions:
+ {5 O7 L$ f- e1 ~0 }
公式

Whose solutions are （解）
- T$ _1 l; Z6 K. Z( V! z8 I6 J( `
公式
! |- V; s( ?( z+ t$ T' t
9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
" y  b+ N8 L5 d/ @( V: P
公式

根据原有公式
- @, W- D2 [3 y0 K% L% c- @10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is
% \7 Y; _" M; n. F8 F4 q) U7 ~; n, b1 F
公式
+ u' S5 N/ ^; M+ Y4 J! ]
The decrease in potential energy is （势能的减少）
% S9 P3 j* `1 _
) w) K+ o5 j$ n! P公式

1 D. W% J5 a- X. L0 t! m- lThe increase in kinetic energy is (动能的增加)

. t- o, g9 |1 @* G公式
* L: l9 x+ f# P) z3 ?7 x% Z
* l& v, g( ^1 _- j3 ~6 tDrug 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)
1 G" W) ~) d8 V! a
# F( o8 f! y7 O2 V8 JWhere a is the acceleration vector and m is mass
0 H3 M; D' [* r/ f/ B# [
6 J' H0 s' i0 t9 H" n. K 
3 ?  C/ {7 k7 Q" _! j
# U% x5 R6 H/ q/ \% c' j; gUsing the Newton's Second Law, we have that F/m=a and

) p- _- m9 a+ ]# A3 r公式
% w0 b* W2 w" L) P4 o+ u
So that

公式
6 Q% L- \1 r. ?  Z" S
Setting the two expressions for t1/t2 equal and cross-multiplying gives
. d5 A5 C, N& n1 N! }0 E
公式
# L* O+ t3 y: a3 c; |* V3 [/ C
* n% F. C+ U) f2 _$ m- L$ W% [0 H2 ?9 O22．We approximate the binomial distribution of contenders with a normal distribution:
  f  e& q- `  m3 F% Y. I
5 k& _& A  N' w/ @1 |7 [公式
9 t" _  g. j8 D0 ~0 @' s  f  D  T
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
8 A6 w' U6 e" g
公式
8 X3 j1 m. H5 g8 g' K% ~
( N* |& T. p3 q5 S9 l2 w8 A* D7 ?0 yAs an analytic approximation to . for k=1, we get B=c
# J* b/ Z6 X' @: I( {1 l. q
 

26．Integrating, （使结合）we get PVT=constant, where
; n; _- |3 d8 b9 y
: P! J4 X; m" k8 o: j1 D( d公式

* q% ?$ P/ q/ [1 e. X8 |( `The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
- m- J0 O& D: j$ p& |+ t, O" |2 a
6 s2 u/ \$ `% k  B 
. M# {! m  {9 ]3 a( z7 N
23．According to First Law of Thermodynamics, we get

2 N$ F$ k/ Y4 Q5 i4 x- @7 g& ~0 b公式

- t3 Z. u; ?9 v/ K' fWhere ( ) . we also then have
/ r1 ~( G! l3 Q5 p  A
公式

& t4 m: G, s) nWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

公式
7 M& w9 U+ r/ `$ n/ \9 f
8 g3 U$ z0 v: b" s+ `Where

 
4 f/ o8 u0 @6 W1 A/ E3 m6 l3 I5 v1 |
对公式变形
9 I# i2 C* y4 l# Q1 P/ [! d& g13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
4 D: }+ A9 l; y$ T- g2 @
6 ~9 ?, V# i9 k公式
8 [% a) H& i6 B. t
. N( x  o' b3 BWe 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

# X# A' |- U6 u" s5 H) I" [公式

( a( v! l: Y- g2 d, K使服从约束条件
8 A. k1 e! k% z14．Subject to the constraint （使服从约束条件）

3 A0 ?/ x' U' G& D0 y! `0 [7 i" v2 s公式

: P  ?( T" f: Z& ^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）

公式

6 l: r' ~1 @5 |) Y8 |" Z! hAnd thus f depends only on h , the function f is minimized at （求最小值）
' M( C2 ?5 W% r, e8 Z4 L
! q1 E* ?& M$ l6 M- m  W3 c公式

8 A7 O8 N. z' k5 kAt this value of h, the constraint reduces to
, I$ A' ?- o/ h8 m/ {( ?" R
公式

  t. H1 P% X/ ^9 w+ r* `5 Z, K结果说明
' L+ Q9 c" R- e- C15．This implies（暗示） that the harmonic mean of l and w should be
: @( M4 y2 `, D2 f9 c8 Z* B1 V
公式

So , in the optimal situation. ………

; ~$ [: y4 O/ J, ?8 ~+ }5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
. a* i( ~% B: B& ?
公式

- x" m5 N% B  q16． We use a similar process to find the position of the droplet, resulting in

公式
) B- F7 {7 J/ j& o
With t=0.0001 s, error from the approximation is virtually zero.

% M1 d3 O1 U& }5 G 

17．We calculated its trajectory(轨道) using
/ s8 p: n- u: ^/ c2 O
' C. H, G% w) H公式

8 q# q/ |5 b' E- h1 i5 j: A18．For that case, using the same expansion for e as above,

公式

19．Solving for t and equating it to the earlier expression for t, we get
/ k1 o( {% L9 _% _5 w# {
公式

* ]6 [, Q$ j5 B* N3 B1 d20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is

. U4 k: ~- s9 {公式
3 q4 c* M5 ^" J  O1 g( F
As v=…, this equation becomes singular (单数的).
& B( i9 O2 b( _$ X# N+ Q0 g( W" D
 
$ w( m3 J% M$ B0 x  ]& |+ V
9 @) b# F& }( @# u* n! I由语句得到公式
21．The revenue generated by the flight is
+ h. V4 x" ]/ g" |5 j
公式
4 e2 g4 t0 p8 a0 M
 
) `: W: c6 n, m% N$ \' j# `
8 o) \  ^. O' A5 ~! \5 @, x24．Then we have
& @2 u3 |& c6 v% o
/ X1 n* m) y: x公式

+ f6 b, e; i1 zWe differentiate the ideal-gas state equation

公式
" e2 Y& t; |; c
" ?  _- _# z6 u7 e0 FGetting
8 u0 R" j- U9 K1 X+ F% ?& j3 {
公式
$ h6 r" U7 V( \# q9 S8 }
25．We eliminate dT from the last two equations to get （排除因素得到）

公式
4 ^! q0 V+ b2 Q& c7 n7 k9 Q1 F
3 l+ [$ D$ b, ~7 z% B1 Z% R 

22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations

公式
7 Z5 E3 b- a4 g* {* C5 i
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

2 B) z$ W0 g6 ~" E公式
7 i, U+ J9 m% o+ N# K————————————————
( i) i" A/ O% i, Z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

1369728843

感谢+++++++++++++++

chace

学习学习学习 谢谢