美赛数模论文之公式写作

zhangtt123

由假设得到公式
0 o( o, [& i9 V9 }1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式
9 c2 G2 W8 Q# E
Where
$ j3 e6 C9 X" M
符号解释

According to the assumptions, at every junction we have （由于假设）
: A, ~# g5 t/ G6 O4 H1 {
公式

8 u  ~9 C& g, ~1 l9 y4 k, @4 V6 k9 h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）；

: S' P% L$ j! Y* ]' j, O% A* d公式

4 P; |- h8 H: o- c: l% vSince the fluid is incompressible(由于液体是不可压缩的), we have

8 i/ j* }) Q9 p. F" h公式
& P% X) g* H$ E* f- x
Where
8 V8 z/ b) X' J* y8 t
% u/ }3 T. o3 C& N4 v公式
4 R' F. ^6 [9 \3 p
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）

) V0 D' ]2 d, K; y! }, b公式

* O/ \0 J% B" P11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式
; S' D# ]5 O& w+ s) o: {: [
12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

/ G8 t$ ?! G9 w: e; j' c0 p. V公式

Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
: c. {5 w* {1 U- I
4 p. Z5 @4 h# {) o5 n公式

Putting these into (1) ,we get（把这些公式代入1中）
9 ?6 R( l: P" D: K
公式
3 Z9 H0 s- t! z- ~& ]
& n& n) i' J" }! lWhich means that the

Commonly, h is about

' w9 I6 X9 \4 cFrom these equations, （从这个公式中我们知道）we know that ………

; k# M+ Z! g0 V7 {; m0 l/ ~: a  v 
6 I& j0 J9 l! l  Q( m3 P! q
引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,
$ Y; ?. @4 s! M( d# s
* ?+ ?, T) M4 v" Z3 w! b/ V( SWe find the attenuation factor （得到衰减因子，常数，系数） to be

公式
# N4 i/ ~/ s7 }% i8 L/ p; n
1 {$ i4 B. E; \, f3 d% j$ t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）
- _% v2 e+ z1 B. m0 ~; o  J! F" {
5 {; Z  n6 Z) Z9 V5 k$ K* v6 ?公式

Where
' a# |. }+ |! N/ s
() is ;;

! }' ]7 E: W) ~# i- P+ f7．Solving for VN we obtain (公式的解)
1 m! d& ~; \& |! O% f0 o) V1 P+ w
5 b* ]+ K0 X; D公式
- C4 P- P* N1 l( z  u( K. {
( A/ R! O8 b0 I% `+ YWhere n is the …..
3 K* L. x/ A& A& j
 
8 b" ~; k: ?3 b9 b6 k& c1 a
8．We have the following differential equations for speeds in the x- and y- directions:
3 ~9 ?: y8 w3 u4 M% p
0 i8 A6 E4 T2 h( J) c公式

) a6 n- I1 F  l6 r3 Y7 \2 DWhose solutions are （解）

6 y: X% F' S; D. W9 V$ ~公式
5 N9 p% @) e  E* [7 i7 G8 g
9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
( Z7 c/ }( r4 U: f2 s
公式

( n. f% e# q8 c# M, f根据原有公式
. B9 _0 _/ t4 r3 N& p% c5 g$ U10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is
* c6 u3 `  e: G4 B# }: c
公式
/ |  ?& U7 ~5 a- [  h8 v
The decrease in potential energy is （势能的减少）

7 N5 R& U5 q9 `! B. M" A公式

The increase in kinetic energy is (动能的增加)
. E1 ?+ ~! l& B" b* P
公式

; U0 e* h  T, S) VDrug 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)
6 }# S" K# W" D$ O$ w
1 g' D0 I6 ]1 M  ?. n2 vWhere a is the acceleration vector and m is mass
  p" s8 b( W$ x3 Z4 @$ C! v
1 {$ N/ }2 v- x+ u3 d9 R) ~ 
3 @9 E& V* r, @" t6 l$ k( C: L$ @
5 {6 d! ]6 D; M3 p! y% N9 D" h: o3 d* _Using the Newton's Second Law, we have that F/m=a and
: `, W* e3 E* G* x! n6 G
4 `  ]7 \, S" N6 k公式

; Q: f6 U6 V+ XSo that
/ r8 W9 L/ ]" u, ^/ W  W6 \
公式

  @! s+ E8 c7 D# v/ S6 XSetting the two expressions for t1/t2 equal and cross-multiplying gives

; s3 p# ^% z; S% `公式

8 }) V( C/ a. }+ G% s2 C  F  H* W. l22．We approximate the binomial distribution of contenders with a normal distribution:

& ~# e' N2 f6 j- @3 B* @" x. ?公式
$ w7 H% ^4 w$ N% D" y
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
# n! r! g9 w: ^; V/ f- K% v  o
公式
) n" o" ]1 ^7 N" T
As an analytic approximation to . for k=1, we get B=c

 

2 s! r% K% r7 Y* C  S26．Integrating, （使结合）we get PVT=constant, where
  [3 n$ H. D! k4 G" n  l) r
  U- {7 j; ?5 p* w公式
0 @* g& ~4 x8 B7 J& i5 v8 M9 U
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so

 

23．According to First Law of Thermodynamics, we get

* @1 H1 ~3 B2 A. y+ N公式

Where ( ) . we also then have
1 Q$ c6 I7 `2 a1 c
1 \: k4 l( [+ v. T! {5 Y; w公式
9 \3 j, k- ^- b8 v. N, @/ _
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

- `- N: o- x; E7 @4 y公式
2 k0 u5 |6 G: Y$ E: ~
1 P2 B" ^( ^8 {* Q9 \/ h$ D8 NWhere

 

( i9 f. d0 a% V' O" y6 O) u对公式变形
% s, I7 C) M- ~# T3 ?+ j13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
& b8 i# K$ e$ {" l
公式

) l+ s* }/ V1 Q+ `4 @% ]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
% D: ^$ i& a1 d% ?% E* y
公式

使服从约束条件
14．Subject to the constraint （使服从约束条件）

( d: W5 Z4 X  F3 e/ L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）
0 e  m& b. K& K, V$ d8 l
: G! a4 y5 m. b# W( I公式
* g. w: r7 g6 d1 o. L6 g8 A
  O  @' {! d+ ~# n& m( |% M: N4 lAnd thus f depends only on h , the function f is minimized at （求最小值）

; J: z: l0 p1 ]6 p公式

At this value of h, the constraint reduces to

+ t5 \8 z- ?7 n/ B5 i" t: ~公式

) V# e3 D& n" n+ r7 G7 h+ O6 _结果说明
15．This implies（暗示） that the harmonic mean of l and w should be
$ B% I5 P0 E5 R( j
7 f1 F* X* j. L7 f; w; B  s' P公式

So , in the optimal situation. ………

5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
0 k& g) v6 C: C$ ^# ?- U
; v5 Z4 {  |1 q公式

16． We use a similar process to find the position of the droplet, resulting in
$ E( w* I  {! b+ y. U. q7 p0 K
公式

With t=0.0001 s, error from the approximation is virtually zero.
1 \+ J0 H7 m# w$ B! v8 ?- }2 Z( v
 
- E4 E' I' I; y0 ~
17．We calculated its trajectory(轨道) using
- A4 x) N: A3 b  d
公式
8 x- C" N& h% f" f; W7 X2 }% @( w. d! A1 B
18．For that case, using the same expansion for e as above,
6 {# Q/ C$ [4 @( P/ H$ Y8 B8 v
6 R% a& Q3 H' r: U; _公式

2 E" y$ K' I3 y" P, q+ y+ H19．Solving for t and equating it to the earlier expression for t, we get

; m9 W: n) i8 O5 Z& Q) }! M/ P' X公式

3 ^5 T/ m: w  Z  X! V$ Y; @" A" l20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is

- l+ s' _% m, [公式
+ T; Q- z# n  [
6 r& h6 z# F4 L+ WAs v=…, this equation becomes singular (单数的).

 

, j0 t  a2 p/ Z: y( M7 A由语句得到公式
! f5 ^" Z$ D& Q$ ]6 |& v$ Y* G21．The revenue generated by the flight is

公式

 
5 A& _$ [' I# C9 t$ o
/ H8 S8 f& y# K- F: @: F24．Then we have
+ t9 l0 v& i% g
6 N; j% _; X/ X4 u, k) z: d公式

We differentiate the ideal-gas state equation
0 B3 ?2 }; J9 R( I* W
公式

Getting
9 f, i/ b) }+ U) M4 {: c/ J
( C( Q/ N( z6 d. v公式
. c4 ~& j3 i" C8 E% N8 J9 n- [
/ P) w9 p1 N( J, s, Y' d! t8 g5 T25．We eliminate dT from the last two equations to get （排除因素得到）

公式

* f2 f$ d+ I3 s# W$ g, f 

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

公式

( s# Y: k& |9 Q# d/ v  cWhere P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
& P6 n! C: @" O* R- I0 F
公式
————————————————
0 n3 c; _& {$ k; @$ D' ?& a; L版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
! V3 }9 y3 Z2 Z1 X: h原文链接：https://blog.csdn.net/u011692048/article/details/77474386
5 q2 y0 p9 i/ s  _0 S

1369728843

感谢+++++++++++++++

chace

学习学习学习 谢谢
* l  J9 L0 Y5 W7 Y  z2 Z  r/ H4 [5 t1 d9 @