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）
- X4 |/ t' v# N% v' m
公式

Where

- ~/ N0 w& V5 z0 x; }符号解释

- F  Y0 x. @6 _' r+ N+ h$ l; r! u1 ~According to the assumptions, at every junction we have （由于假设）
1 K$ D" D3 ^' v, S+ W- S' ^
公式

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）；

公式
2 T2 \- R) O7 ^0 `' M
Since the fluid is incompressible(由于液体是不可压缩的), we have
* D& S' R2 D0 j# D
公式

Where

公式
' p* w2 I1 o! j+ j. Z
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）
) c  ]) f  k5 _  S
. `% o0 d" B8 ^0 G; V! M1 ?公式
# |; \5 B( O3 W
* ^1 @4 C3 t! H1 W8 \6 _11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
$ d# q  W$ |; }5 z) z
公式
3 Z* @; ~% s# U7 ^2 s
8 R( n1 G, d* A6 @8 M* m12．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
6 i, J, J6 k: w/ M3 }
6 Q: J6 ?- ^  f  H- W公式

' O  o! G& K  x6 x8 H* B- tPutting these into (1) ,we get（把这些公式代入1中）

公式
6 X" F4 K) y( M9 D* o4 H
Which means that the
$ Q9 j5 r" d6 N
" o( B- a8 D8 U2 Q9 QCommonly, h is about
; n% ^4 s0 x5 b( ^* \
From these equations, （从这个公式中我们知道）we know that ………

 
  w; C* w' ]- ~" ~' b7 m) F
引出约束条件
( r+ h) V, U% i* S, s* l4．Using pressure and discharge data from Rain Bird 结果,
. ^" d8 X; c, S% I& m0 ^' R
We find the attenuation factor （得到衰减因子，常数，系数） to be

公式

6 I- [6 g: ~6 U; L/ |& I+ e6 `2 O( q( \计算结果
, Z* k! K9 n/ _. r7 I5 U' [& o+ V6．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
& O3 a  {0 w$ n3 o* v. w0 i
Where

() is ;;
5 i# T' U7 i5 Z& x# G& Q
7 K4 J  e8 u7 Y7．Solving for VN we obtain (公式的解)

# v9 ^1 T$ ?" s: s公式
& c$ q% [& L1 L! Y- ^  D7 p. R4 g
Where n is the …..
8 x0 g- }/ s) b3 F: s5 ~! `/ n! s
 

: B4 N2 f. D: g! {7 [2 M1 Y8．We have the following differential equations for speeds in the x- and y- directions:

+ W  S) H  W* H* L公式

" s' U  |* V$ O( @2 g4 PWhose solutions are （解）

% ?, q# o  S6 t2 k* {" ~公式
9 M! i: b- D7 {* \2 E4 W) N1 {
9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

( [9 ?, x2 a) V公式
  X( e7 f( a7 }" z. e
根据原有公式
* e  J, K# e6 A  U0 ^+ S7 }10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is
/ z! l6 _9 U  _$ y& x0 H- `
公式
9 c9 b5 M' m0 O4 V8 {
The decrease in potential energy is （势能的减少）

$ B6 J: }, V1 ?公式

The increase in kinetic energy is (动能的增加)
6 z  ]& p' T' q0 p9 M
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)

+ ]' t! D+ f: u& M' \Where a is the acceleration vector and m is mass

+ N9 \; t' S. {  H" Z0 [ 
0 h$ @: u9 A" r' G+ E$ D
3 Y. z1 \5 F8 v# {# H# aUsing the Newton's Second Law, we have that F/m=a and
; [' j: J2 \" P8 M" x
公式
% e) y% A3 p" W+ C( x
So that

公式
4 O/ ^& z% D( o; ^" i+ l
Setting the two expressions for t1/t2 equal and cross-multiplying gives
2 s8 J( a# }6 V7 V  v$ J
公式

22．We approximate the binomial distribution of contenders with a normal distribution:

& B) Z3 `* c+ o' k0 [公式
- g( h& c0 V1 B9 w: `! h6 ^+ k! y0 h
9 S+ H" F4 B: y& ~+ @, C. d5 E( BWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
* e( W/ K' _$ j3 m
" ]  q0 b1 l1 E+ N# h. K2 U公式

+ J  m/ s# m  @! ~As an analytic approximation to . for k=1, we get B=c
" A( q* {4 _3 w8 ]- ]% g
& f+ C& k7 ~6 P* I5 S  h 
1 Y1 L+ f3 _7 c' t4 U+ D3 V
26．Integrating, （使结合）we get PVT=constant, where
3 u$ b, {# M1 }* g
公式
  n# s. n' E% s
9 Y  l4 q2 @9 ?6 `The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
- `0 Y, I$ a( c1 l
 
7 V0 @+ p" p; ~+ h2 _) {
- l0 s5 z. E2 O7 @23．According to First Law of Thermodynamics, we get
) C7 }- U6 T3 O! b
4 U, D* K3 }7 Y. c& ^2 ]公式
" O8 V- K( C, O* v, ]
; N5 n4 R3 e1 Q: QWhere ( ) . we also then have

公式

- H9 b9 P0 d& ^# v- z% YWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

公式

Where

* X* _0 o, I  T5 ?( T  l8 q3 m: ~0 _ 
$ h: j% _! r8 R- Z' d6 p/ X
3 ]& k7 S! ~3 M; y8 C  L对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
. J+ F% _& ?# r6 O
公式
! o  E4 _0 z3 P. {2 A, a
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
' M7 \+ g) U  ~  _( g
% x6 c6 ?% G& J7 K% g6 j: ?- C2 I, U! w公式
! t6 L: u3 I# F, w  k
) S( B) _( |2 v- j( {使服从约束条件
! G( ?& d6 I0 }14．Subject to the constraint （使服从约束条件）

公式
- Q- T* ^  H( m  M5 P
' @0 Y2 G$ O5 `# TWhere 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 _" t5 W- I  m3 v* p$ c公式
- \9 ^) E& k6 u1 @/ o
And thus f depends only on h , the function f is minimized at （求最小值）
1 o7 m! x6 m9 j+ h
# H1 `4 Z7 T  z( {公式
+ N# I5 i7 @' W3 |4 a
0 V: t8 W- i/ e! p5 Q$ A# fAt this value of h, the constraint reduces to

公式
( |- ?$ R8 f+ j" P* L0 R
( n6 \1 e- [& h! p9 b结果说明
15．This implies（暗示） that the harmonic mean of l and w should be

公式
- \4 S0 Z0 a. E# b' ]
. m/ R9 v2 Y5 a0 a& _So , in the optimal situation. ………

- ?5 r/ H3 f7 l; ~- T+ R. h5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
5 O- A- }9 L3 ^! _5 Y5 x- v5 U
公式
* [: [! J' @# {4 q. A  v
9 P+ j8 t8 ?: |2 K5 u. x( @4 T16． We use a similar process to find the position of the droplet, resulting in

公式

2 _4 \9 U$ n! W3 D) r5 FWith t=0.0001 s, error from the approximation is virtually zero.
2 L$ @/ h2 v3 q+ u4 {- P
 

17．We calculated its trajectory(轨道) using

公式
- w) b0 j7 w5 x2 ?8 h) y( O4 p
18．For that case, using the same expansion for e as above,

% ]1 U5 j6 `' T" h# J/ D& ~4 h公式
4 |5 E! G6 \# g& t
19．Solving for t and equating it to the earlier expression for t, we get
3 t8 G; I4 I1 V9 P# w0 j' W
公式

# J) y4 Z* Z6 V1 z20．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
8 O" H) x1 ~+ k7 K
As v=…, this equation becomes singular (单数的).
" J. i1 F4 Z+ V% f& K
* q4 E4 z8 t9 m) w: ~: J 

由语句得到公式
! D0 {- A/ t+ Y! _( p' o21．The revenue generated by the flight is
+ b- g/ G* O. y' L5 ~8 l$ G
公式

  s7 S9 ^  m3 u7 _3 d( i5 d* b 
# r  j$ z" F' c
8 c% {1 i) q/ ^" I( |24．Then we have
3 v4 a8 d3 d8 V
公式
- w( r# E& P0 A3 O' |
We differentiate the ideal-gas state equation
+ q! w# r3 n5 E8 \# }
3 y9 m2 M, O% H, s公式
) a# l0 c5 N- u1 I; q
Getting
2 h9 P6 a5 ]# b4 G7 x
公式

2 y, K, m% B7 r6 y$ b- s  d25．We eliminate dT from the last two equations to get （排除因素得到）

, B) u- T6 u/ ]; M. ^' W/ x6 o6 W/ a公式

7 d- r% w/ _3 i) P- T" O9 s( `( ? 
8 d7 R- i- C4 p8 B* |% g3 z: Q
. M4 s8 I6 U1 N; |1 u, ^22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations

公式

3 I) A1 ]) X) m9 R5 c  r' T! }Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
& ?  G, p' m1 A+ ~0 H+ Y% k& B
8 A" j! `! @/ f# H; x公式
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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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