美赛数模论文之公式写作

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由假设得到公式
2 E/ ~: Z) u) n8 Q! K6 W/ n2 R1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式
" G1 a4 y# E* T# f: {! M
Where
& D3 A6 \, @1 o. v! g! D* E
符号解释

According to the assumptions, at every junction we have （由于假设）
5 Z( Y: }1 V; Z+ c* F+ w5 R- q
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）；
; [" ?, b3 o" X* ?6 }3 e; k
公式

; D6 S* z9 ~3 M) D6 bSince the fluid is incompressible(由于液体是不可压缩的), we have
) H$ r& X0 @) Q, g( m" k
公式
- d" s8 {/ i9 L6 [( l
Where
8 k3 s1 p) B# s) W0 S9 e
' w2 ^; S% |% D$ y' c3 p" l) P公式
- O5 p$ {8 o/ P: U/ a4 L; Z7 _
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）
6 V3 W0 ]2 [. {7 h- A0 E/ o
公式

11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
  C: ~) C( q" D+ o+ J' k/ n
! u2 m7 U( Z' I. u# L; V公式

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

公式
6 ?" {' _4 g7 ]5 C$ d; ^0 A+ f
3 y  A% Z5 u* O0 Q. p' ]5 ePutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

公式

Putting these into (1) ,we get（把这些公式代入1中）

. A2 n% ?# o, L6 [7 E公式

' G6 e9 i9 l; l  y5 {6 g+ z5 i' t. LWhich means that the
& Y: h8 w& m  ?& |7 O; S+ a& O
, g9 J( ^0 x  b9 R) K0 NCommonly, h is about

From these equations, （从这个公式中我们知道）we know that ………
, H2 s6 I! E0 m* A+ Z4 J4 e
 

$ Q4 K3 n1 G$ \# c引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,
6 e9 k! ?: z- `  \1 X4 E
We find the attenuation factor （得到衰减因子，常数，系数） to be
+ c: X$ L* i% J/ X
" ~# c* A3 g3 F4 f公式
! ?" ^/ ?' X9 h- Q" J
计算结果
1 D* @  z! M& y4 S6．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）

* M8 S1 S" N/ E  F1 f  r: X/ @. L公式
; F8 I8 x1 v" G, h
Where

7 C. I9 D( V* |() is ;;

7．Solving for VN we obtain (公式的解)
, L* b& C' S* N8 v0 ^
% t/ P  w3 o/ t6 k' l, _公式

; M' }8 Z7 F# SWhere n is the …..

 
+ j2 C2 j- u2 |5 K/ k, d# M
8．We have the following differential equations for speeds in the x- and y- directions:

% l, m/ J; X6 M) O/ T公式

Whose solutions are （解）

公式

4 o; ?7 Y4 F# ~' i% V9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
' l- A2 Q8 m1 n7 v9 q% s6 L! S
  Q: g' e. W4 s公式
% Z( G8 ^9 s$ c/ Q3 {* V8 Z0 C1 r
" C* Z& x% \2 I4 K8 H" u8 ]) T$ W根据原有公式
" g( F, ~/ H1 Q10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is

! U& C9 ]; B. y% e公式
4 c# f0 T6 J) Z" q
The decrease in potential energy is （势能的减少）

公式

! J* ]# b/ t% aThe increase in kinetic energy is (动能的增加)
( ?' I1 y& ?/ X  [9 q' e0 q
! I' z; |' ^' r) z$ B+ }公式

: M) \. X  m* a" v+ H; 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)

+ l3 A7 ^9 z: S. E! _$ H& j2 hWhere a is the acceleration vector and m is mass

 

Using the Newton's Second Law, we have that F/m=a and
- M  U/ S8 ~" O, B+ ]  d
6 D6 G. V8 w$ ~- I0 r/ j公式
3 i& O0 U- `4 _, x9 r: y6 c2 j
So that
, [) [* f) U* i) R- _
9 I+ E) h( l: X9 V0 [3 ?公式
* k6 P  n& F  q0 ]) a4 o
Setting the two expressions for t1/t2 equal and cross-multiplying gives
) H+ J' r% L4 d. ?% z9 g* [7 I
1 K: [8 j5 u+ L& p1 w公式
. N/ _9 b6 @5 \% |
22．We approximate the binomial distribution of contenders with a normal distribution:
+ r- _) e; }2 q3 A' l: {
% r5 Z1 }+ ]6 r, T0 D9 q% \0 L公式
  d8 a# c$ w) K1 e
! G' b& ~" X! X" Z, w8 A/ wWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives
6 y& ?$ I+ n2 B2 L/ m
% d2 f+ G1 X/ Y- K- k" s公式
8 Y# z. }, z7 u5 C
3 R4 ?! i, J8 s6 p+ AAs an analytic approximation to . for k=1, we get B=c
. ]* Z* m8 P0 M' }0 X
 

& m" x# e' ^9 Q5 b- V) E3 C8 i26．Integrating, （使结合）we get PVT=constant, where
% V6 \: |  \4 h6 E' R6 J- y+ _
公式

" B4 E/ x, ^: d& x; f5 Q) rThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
+ x+ h4 T' E) p
' a8 @( i1 u& a+ E* ]! V 
9 l/ }* S) Q* y1 B
  Y4 J1 E" q, ]23．According to First Law of Thermodynamics, we get

, b1 Z' x' S1 e9 u; H" g& r# @0 n公式
" \9 y$ D) o* W% {2 p$ l+ a
; p0 i  Q+ V, P# P* W( P. w6 ^Where ( ) . we also then have
$ B9 W. e& q/ N0 t
公式

' H% [( L, d; T4 m6 ~' m& a- l& [Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
8 k3 G7 A; W' Y6 E" y1 E
公式
$ G0 b' q8 g5 Y( e. b( c
: @& q9 g6 J- J7 f) z: sWhere
& b# a8 q4 B: T% T7 ~
, F0 C% s" y5 ]) x 
- g0 C+ z- \. z+ J2 e0 Z
对公式变形
8 f- c+ W1 m- X13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
/ K- w  u8 H, W1 t$ P3 t
" f4 m& F, z) f公式
) Y% w- @8 F2 N7 B* z
! p  o: z) q% I/ ?0 V/ ~$ EWe 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

公式

# i4 h! y3 E. K$ |+ L使服从约束条件
4 D* g6 {: s8 W* _* \14．Subject to the constraint （使服从约束条件）

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）
5 h- u6 T" y$ e! h  t% S$ I
公式

7 }2 B+ g2 \) F* I+ v# }And thus f depends only on h , the function f is minimized at （求最小值）
$ w$ d/ R; g  U. F7 z9 a4 Q
6 c1 G; x& Z+ [公式
6 E7 M. m8 U" J: A+ |5 T! I& p
At this value of h, the constraint reduces to
" ?: C2 }: T/ ^, h
公式

  {/ j6 r% A5 C8 e0 q# e结果说明
15．This implies（暗示） that the harmonic mean of l and w should be

公式
  n6 g  n1 L1 H& h
) N6 q. }2 v# G3 G6 tSo , in the optimal situation. ………
+ c" t' a- v, h7 q- s1 [
5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
) u" H8 R& y8 ~4 [- i
) R- h2 a! O9 ^公式

' j7 L/ ]) g3 w! |# @16． We use a similar process to find the position of the droplet, resulting in
  h( W8 Y7 Z  H2 m) J) \7 [" b' T
公式
1 _# v3 e' n+ Y% J. \9 F
With t=0.0001 s, error from the approximation is virtually zero.
4 F" h; ~, y( a
 
: ?1 F! y: @$ ?) f% F" e! u+ V
17．We calculated its trajectory(轨道) using

公式
  O, D0 h( r5 I3 r8 _* e; U
18．For that case, using the same expansion for e as above,

3 x( g8 Y2 @1 T& M公式
7 ?# q6 i) `9 e1 r0 m2 M5 Q/ H0 `9 K
% Q% G- l! R3 a' y* o+ G1 D0 ?) U19．Solving for t and equating it to the earlier expression for t, we get

公式

4 }6 |7 _1 K3 S& J20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is

* n  o8 g$ f# D公式
6 i3 k$ j$ n' e; Q6 p
As v=…, this equation becomes singular (单数的).

0 H! n  X0 b. p0 g+ Q; U# W 
; e) q" l: _% Q+ P
由语句得到公式
21．The revenue generated by the flight is
/ R, g' J: z0 K$ i2 V( U3 K; c, P8 z1 m
/ g3 l3 Q+ x+ d& T5 ~公式
2 j7 C& e# O3 {9 a
 

$ N3 ?- z+ S# x: r24．Then we have
7 h5 M( D& a6 h9 ?+ O9 X4 Z
6 M$ V1 @- W8 s! V6 ]' k4 U' n公式

We differentiate the ideal-gas state equation
4 ?6 I  z" {% H& q5 }# J4 Q/ \
2 t/ W5 v5 ^* R% E/ M. B公式

Getting
/ M8 \6 V& s1 S+ K
公式

, h/ }6 l% M2 ~* H" W25．We eliminate dT from the last two equations to get （排除因素得到）
* M6 [4 W% S) u( c5 u, J5 F: a
) Y, @" J  U; Y4 d% `8 K公式
0 [" b; ^$ D  m* e% `( {6 \# m) k& Y: g
 
4 q+ ^$ m& F! p
22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
3 v- W1 v( s: s- U) l- k5 b! i' l
公式

Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

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。
# d( ?% L6 n* ]原文链接：https://blog.csdn.net/u011692048/article/details/77474386
& C: U- v( {# o# l

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