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）

公式

2 H2 \) M5 e9 H# w$ HWhere
7 ]2 z: \1 o+ l3 _9 A5 E$ L
符号解释

According to the assumptions, at every junction we have （由于假设）
% d* ~4 N2 B, }4 D4 K/ [" c$ K
' }: b# w7 {9 U! U/ `) c9 Q公式
+ I" K. u# B6 Y( W
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）；
7 k' Z3 n0 ]% s, H4 N& Y
公式
/ W! u& R0 z' J; g: R
, z7 m+ k9 }. q% W, R; |! P( a5 ESince the fluid is incompressible(由于液体是不可压缩的), we have
0 Z, F) \  I( t8 p' L
公式
5 \3 c& R9 s0 _  z
1 R( Q4 G: k, Z' JWhere

公式

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）

公式
5 s3 ~4 C7 N! o5 |, b' E
11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
" I# S9 l5 S7 S& ~2 T# H
公式

12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）
6 r% F4 t1 W4 }! H# B
公式

Putting (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）

& j# K/ m: I& G) e$ M& _' ^) a公式

Which means that the

Commonly, h is about

# Z7 q" V$ |9 d7 f. B# R) x6 sFrom these equations, （从这个公式中我们知道）we know that ………

 
2 C% b7 R  `$ [- Q
引出约束条件
# b! c! H! M! c+ a6 A4 U5 y4．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
! x( Z- w0 d6 `6 ?# J2 V7 }
3 H2 y- t; B% n计算结果
# t  ^& K, |/ E8 ?/ b9 V" D. ~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）
9 [/ f$ H: e0 @- Q; E( w& b/ t
公式

9 H( ?  [6 Y! lWhere
7 H: R1 L# c4 J
() is ;;

; B" K6 p( e: `7．Solving for VN we obtain (公式的解)
" X1 ^, V  t4 @& ~1 k. k
公式
# u% K9 Q. C9 q' {+ s0 @5 V
! n8 i& R: H* G0 c/ T9 h1 kWhere n is the …..
% \4 t; D* Y4 x
 

- A& J+ ]1 v6 ^) i9 h8．We have the following differential equations for speeds in the x- and y- directions:

5 n- e1 Q) y: @& [公式

* b  S; ]! C) A) ~$ K- WWhose solutions are （解）

6 ]7 r2 U9 k- f9 ^! t. O公式
$ T! R3 Z) @' m$ S: q  r6 \2 ^' z+ {
) ^- Y/ b- C( `3 h# D8 z9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
  t9 Z9 Q5 ^7 X/ S3 Q8 U+ Y( f
公式
# g/ u3 t$ O# f- \0 A3 w; z* S+ H$ l
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
# m* c' [: g7 S& K. [# Y, I0 `* O' d
4 u$ h: j2 d4 a8 F+ e公式
3 Y, f5 C/ J* l6 ~) B9 w+ p* L
8 m, T& ]4 I0 E, rThe decrease in potential energy is （势能的减少）

公式

( C2 U5 s7 ~8 Y9 V6 g+ W2 NThe increase in kinetic energy is (动能的增加)
, X6 J  O$ Q8 Y' {" m' P, y" W6 \) v/ v
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)

1 w0 V4 S+ D5 s5 H; F! [1 OWhere a is the acceleration vector and m is mass

 

# }( G. Q, ]5 m+ Z3 v) a) oUsing the Newton's Second Law, we have that F/m=a and
; g) c! G' k# r/ {+ f+ G  m7 c: d4 v
公式

So that

2 }% [* [( r# v# E公式

Setting the two expressions for t1/t2 equal and cross-multiplying gives
" W, N' T# y! F: X
* ]% s3 n4 r% C" x- z$ ]% t公式

22．We approximate the binomial distribution of contenders with a normal distribution:
: p5 m7 ]) m0 P" J$ L3 g9 i% g
" j9 j0 `1 B* W) w, Y2 P( W公式

5 ^" N. [3 B. [; x9 [+ r0 s, ]3 @Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

* e* }$ W: y% Y2 C公式
# N9 \% i. B# K
As an analytic approximation to . for k=1, we get B=c

 
+ K, ~( {& H& ^' T0 B
26．Integrating, （使结合）we get PVT=constant, where
$ Z, H: ~7 w: k  b5 r) y, q; k9 s
公式
% {6 `  e9 K5 G% Y5 p+ T2 L
- h' t! ~4 f4 u$ m: e( b) R# hThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
/ ]% |' N, b/ C4 }" z
 

2 r* ^" G( M/ v1 i# D23．According to First Law of Thermodynamics, we get
8 b, A0 T# W( a" M2 U, T
公式
7 n9 z0 Q4 s: G  E* m. e8 L
Where ( ) . we also then have
% @, c: `+ B; T! C' u5 \, m7 P( S
公式

  V1 I1 O9 c$ K. C6 Y, v3 GWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

' Y) j/ c0 \/ ^, K% t& r; K公式
& j. G3 x3 i! b% ]6 H! q1 \0 y
Where
5 Q) r  |1 [- m6 ~6 z+ B8 i6 g0 r8 Z
% h% q3 e9 _8 h4 P2 P$ r 
5 S4 O1 |% s/ q4 N- k+ D8 s
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 p. z, S" F3 z" \
7 f& B# w7 N9 z1 v! d  J公式

# m7 u! b: g) w4 v0 lWe 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
( A, J. P' [% V6 _- J0 b" o# `
公式
, G% i( o# o  i
7 ~0 U7 T* J3 j  e& d; O$ d' s" P. n使服从约束条件
14．Subject to the constraint （使服从约束条件）
$ F! `' v/ B' S2 S* q
公式
) r6 ~1 q" t1 S+ _
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）
! h+ k0 U4 v# y' l8 A# }
公式
2 l3 u; J% `: X. W8 \
And thus f depends only on h , the function f is minimized at （求最小值）

公式
  ~3 Y$ M( L, u8 L
At this value of h, the constraint reduces to
9 g, h9 G9 u9 ?5 m3 J% C, d. x4 J2 n
4 c2 W7 Y; R5 ~公式

- v) `- Z, g/ e1 P结果说明
2 g0 D2 O0 C/ P, r) U, T8 D$ T: |/ i& K15．This implies（暗示） that the harmonic mean of l and w should be
; ]! m5 g# ]" Z0 f0 }
1 ?! f% S9 P6 f3 ]( a) Z' J公式
3 ^% B4 @, L! v5 o* p  Q4 u
So , in the optimal situation. ………
* A7 p9 G0 |: j
- ?0 `& o' d# P3 ]: z1 V" Y5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is

# f+ z, \6 T" o公式

3 R( I* W9 `$ v' r% k16． We use a similar process to find the position of the droplet, resulting in
& _7 |* @0 X- d* h
  K3 l. ?1 y/ V7 n! ~. p8 K$ x( }0 T公式
; H' H6 I- K" N. Z: s0 d9 b
With t=0.0001 s, error from the approximation is virtually zero.
2 n+ M5 G( s5 F4 e$ }
 

, o6 S/ G# i2 ^! O( X17．We calculated its trajectory(轨道) using

公式

18．For that case, using the same expansion for e as above,
3 R/ Y' s* t: d. N. D. K2 v
1 v6 V8 X8 p3 B6 L1 @公式
9 m7 w2 Q2 ]: H8 g
19．Solving for t and equating it to the earlier expression for t, we get

公式
! Q  ]0 y! }! V( [2 q: ]' H
4 `# M9 t# }* v  n& y20．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

4 l& p/ P/ i/ J: lAs v=…, this equation becomes singular (单数的).

 

. L; j5 P9 ]0 B" \+ s3 B由语句得到公式
$ [' e8 a* Y# {' n! m4 Q. i21．The revenue generated by the flight is
4 e& O* D  h: q2 \
公式

 
- _) U, L. P5 P5 B
4 D& o' i: J# ~( o! R( x# e24．Then we have

公式

We differentiate the ideal-gas state equation
3 O3 B$ ~8 f. g+ ]" V
: |3 l+ V' q0 ^" F$ B$ F+ K+ R& w1 I: C公式

4 H+ _7 \! j9 |4 {: e& q, i. VGetting
, e& \$ n2 \3 Q' Y  _
公式

25．We eliminate dT from the last two equations to get （排除因素得到）
( r" z3 }3 c- X: h
公式

 

$ I. y4 m  J+ }, I+ K; R22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
( p9 s: t4 R, N& i# r5 |
0 A+ _# e: M" s, q6 `6 d2 e. n公式

Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
, h: {8 v# ]* X" s6 c) L9 {5 i
公式
————————————————
版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
0 }, J# o8 }( D" Y9 a* `7 w$ m原文链接：https://blog.csdn.net/u011692048/article/details/77474386
/ k0 m3 O, M& w+ ?& G

1369728843

感谢+++++++++++++++
! E9 j; a% ]5 T# t# O2 v4 \

chace

学习学习学习 谢谢