美赛数模论文之公式写作

zhangtt123

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）
( d- g1 [3 ]/ |5 e. h$ ~( A
公式
6 w* c) q# z! v6 q7 x7 \9 ?) j
Where

符号解释
+ _7 R3 r* j; q2 I  _& B
% a! o) \; L9 O" _$ WAccording to the assumptions, at every junction we have （由于假设）

公式
+ z* t# |+ T2 O8 `
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）；
2 L' z) V! m4 Q+ @, @! k0 D0 {$ D
公式

% A4 Z# r7 q8 {; ISince the fluid is incompressible(由于液体是不可压缩的), we have

/ w4 H" r0 x: H公式

Where

公式
! P. l% R7 |/ Y  p; R; R  x, t& k& _
8 t& w% [1 _" ^9 W6 E& L用原来的公式推出公式
- K7 n1 B* c; m, v3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
. [9 ?4 P+ W  l3 O8 d: P+ }
公式
* x, r$ L0 [1 R# L) N' a9 q8 k
11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

3 t& l3 m$ w* F: H6 j. i; Z公式

3 Q' z: V( u. l# r& f12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）
. ?: N  \: t# X8 t: ]. H  l
# C0 ^. j( ?! q: y& Q: c! m公式
. ~6 C1 E5 B( c
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

$ `# U6 L3 O) F9 }3 N& t* K! \5 i公式

Putting these into (1) ,we get（把这些公式代入1中）
$ ?% D1 L: I( r% P# @! A
1 {4 V' }9 V7 J9 T; |7 p公式
3 I4 }( P) ^1 Z
2 a4 B2 h, z: m: Q1 P. vWhich means that the
+ ?3 g+ ^1 f! y$ ~7 ^' Y9 Q) g
Commonly, h is about
+ c8 e7 v5 ?  D& P5 q! E; Q& B/ _
From these equations, （从这个公式中我们知道）we know that ………
) C" z0 F9 e3 k. \0 c# k0 V
 

引出约束条件
+ c1 v2 d( j- P$ N  n% F4．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

公式

计算结果
9 r9 r) Z# r( H6．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）

# l1 {6 ^% ?' x8 @公式

- o4 D" X8 b& Z( Q( H5 Z% {% }+ ^Where
! W3 u& O; {. W( }9 \. q/ L* H1 t
4 ]% i3 u6 R4 p3 }() is ;;

5 E% @/ ?7 t5 v7．Solving for VN we obtain (公式的解)
2 g4 M  f' i0 e9 `
; k" s4 ~. w' ~" ?公式
- A9 J3 s1 X7 D6 f4 O3 y! _& |. o
Where n is the …..
  Y3 L0 J4 K6 B$ P: ^& k1 q8 C# D
2 o2 b* t" n# E1 Y4 q# U$ C  D 

  }/ E% h' k0 X8．We have the following differential equations for speeds in the x- and y- directions:
5 s  M( H! n2 s% M
公式

Whose solutions are （解）
- B" A% n8 P8 d
公式

4 |) l8 s! x# p5 t5 T1 m9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

公式
! p9 Z& k. u2 o4 I) N% ^; j7 y
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

公式
( q$ r* s5 \- G. h5 N6 D
1 U2 P6 H, p1 ~. u3 O' w0 X# ZThe decrease in potential energy is （势能的减少）
! V3 u( E  O& }3 w7 g- V! e. \8 p- O
公式
2 t  y1 i' P; N9 G- m
The increase in kinetic energy is (动能的增加)

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% M" g: m0 f- FWhere a is the acceleration vector and m is mass
/ H6 a+ l/ s  O" p/ K+ S5 \/ A! c
 

Using the Newton's Second Law, we have that F/m=a and
) T. \; J; @& \# \
. A7 H- R+ G; l7 }公式
7 Z% |( W; l# Q9 Y9 b
, b/ x0 x0 L, U7 [& xSo that
6 ^! b" i" Q0 H  Q  P& L
+ m" e* h6 T% {6 R$ K) x( P公式
2 L) y/ b+ f; J" V+ F2 x/ w
Setting the two expressions for t1/t2 equal and cross-multiplying gives
! C( E: p) A, V8 M1 V
公式
6 k' R. Z, f" {) A4 j
22．We approximate the binomial distribution of contenders with a normal distribution:

+ W4 \7 D: U0 |! p9 H% T1 }公式
1 i. i+ g  S  j* m0 e% Z, f5 @; I
/ }; i7 z6 \. b# W) f: r1 pWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
" M/ o+ [: S  K  o8 z- K; y. x7 `
( |( N' y3 x6 v4 N, r( u公式

: f9 l5 O: l4 B( Z( B( R6 ZAs an analytic approximation to . for k=1, we get B=c

+ M/ l" i# o0 Y( ^7 b 
6 @. _. r! N( w' L  M4 h, p8 C
26．Integrating, （使结合）we get PVT=constant, where
/ O* @5 B) D2 K
公式
2 X# d) I3 F2 E" Y# p, H9 ?' a' ?
; p4 C$ K" j6 ^! U0 [# l5 FThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
0 Q& f0 g* F/ o" L# C
 
9 y" f/ y5 M6 _4 w- C% T
4 N: A, M7 t9 m$ q* G% P# Q23．According to First Law of Thermodynamics, we get

公式

Where ( ) . we also then have
- S8 {: H5 `# Q
公式
& f" O' Q% z4 d- M# w  D/ J  D$ d# O" t
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
' M" N6 I/ A- Z3 N7 I' T) Y
公式
$ ]* ], {& C1 \# U! q
$ E+ _, [- ~  y* H  J* }Where
, _) L) e) V% L# t3 ]" x
$ u8 @$ M; p" T0 Z) Z 

对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
1 G% k5 y7 B. F; Z
公式

8 F' ^( t  i' o% \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
! U* R: F1 ]# m" j
* m, K5 D+ F( `; d公式
- _$ Z& p9 V* C+ @
" k0 _& N1 y8 P使服从约束条件
14．Subject to the constraint （使服从约束条件）
# F$ ]3 Q% {* \. \+ L1 b3 D
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）

公式

, A2 K. z8 y! u! rAnd thus f depends only on h , the function f is minimized at （求最小值）
& M& K' S8 j; N& C6 R" j1 g" t* f
公式
0 c+ x: b! a. w2 F4 E, K
At this value of h, the constraint reduces to
" f, J* m8 r) f
+ S. r8 I, n- B' ]- R3 I2 L* N公式

4 J8 Y$ d0 h! B* x, b结果说明
' \  Z6 x& X  W) {+ C! r/ ^9 c15．This implies（暗示） that the harmonic mean of l and w should be
; ?5 V2 f/ l. A
公式
+ C5 ^& X- [6 q: @
! d( P4 R& q  [3 j' xSo , in the optimal situation. ………
7 \- ?( u( Z; b+ g7 _
3 v( y" @; }/ v( M+ M9 i5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is

: o3 M8 C1 }9 P2 R# q公式

16． We use a similar process to find the position of the droplet, resulting in
% {3 t% v( e9 q$ I! y
5 L# |) f! |4 ^, k0 p4 v, T  S公式

With t=0.0001 s, error from the approximation is virtually zero.

1 c- e3 s0 a' U6 p4 C7 Q; W 
  j4 B4 w2 K% G& Z( K
/ l# Q. t5 g1 `  _8 g9 B0 u17．We calculated its trajectory(轨道) using

公式
; O1 W* \5 I% t% \8 g
18．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

公式
8 U8 I1 K( c3 V4 q+ ^
8 M/ F0 Q* G: k: J20．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
2 @9 x' H8 b8 A0 _) L5 }# [
, i: t# \3 Z' F1 [As v=…, this equation becomes singular (单数的).

: l; H% Y. V2 N0 D 

由语句得到公式
21．The revenue generated by the flight is

+ h0 d8 A5 M- `1 l8 H/ c0 q公式

4 O/ C! C" n+ n5 e+ V+ Z 
$ p+ G& P8 h4 V2 W7 h$ i" F5 L
( @2 |+ Z/ {4 r, Z, k4 M24．Then we have
- A# P. ]( _+ V- S. t
公式
$ q, G& e1 a8 P' H1 y# v* {% ^
We differentiate the ideal-gas state equation
) p/ H/ L8 a* j
公式

7 e( [9 H/ M* c" a$ T" oGetting

公式

25．We eliminate dT from the last two equations to get （排除因素得到）

公式
" |, g- Y& F$ `6 @& p
& U/ T0 T8 c' C( Z% v; r4 e/ o 

' k2 h3 P* W1 r/ @22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
8 j/ Q$ \$ k% y7 b" {9 T6 D+ x! a
4 w/ h! E$ s! m4 R公式

, Y2 L: O. P4 Y2 {Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

公式
————————————————
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
( g% ^' w3 F2 _( Z  Y# ?" X