美赛数模论文之公式写作

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由假设得到公式
- c9 l% q0 }  Q1 b! J$ b' W- d1 G: F1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式
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Where
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# P4 {6 e4 L  [! J+ L# _: l' `符号解释

8 T1 e) d, H! AAccording to the assumptions, at every junction we have （由于假设）
/ W9 }2 T* S2 B; U
公式

由原因得到公式
6 [0 N1 I; ?# C3 i: x2．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

3 }# T6 D! T( B& s* d; M5 _! S% aSince the fluid is incompressible(由于液体是不可压缩的), we have
3 S: g- g4 \( h7 L9 G# E
公式

Where
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& b3 b$ J, `: l; r公式

用原来的公式推出公式
& N% l' a1 ^& R( n4 o2 n5 B% @8 v3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
$ x0 f) @8 y; |/ ^, h1 m! @6 s3 {
公式

11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
1 F/ c; o9 j' F: m
公式
+ W+ e2 G# G; ~' a: I2 Q
+ X0 j/ _% m. n6 G' k12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）
0 x' V5 N6 W9 j
公式

Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

公式
5 g1 f; v* R$ Z, \8 R
; d( l8 Q" x7 o. V9 f: QPutting these into (1) ,we get（把这些公式代入1中）
7 s2 l3 [+ B) e" }
公式

Which means that the
* S) Y/ T7 x. [3 p5 A# c& z) n  k4 d
8 v5 E! h5 v3 g7 h! }0 ]Commonly, h is about
$ ~4 K& t( b9 K" M4 d
From these equations, （从这个公式中我们知道）we know that ………

 
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( H& d' [9 p, V6 M8 y4 f# m引出约束条件
" I, v  `$ {) B% z4 f" g4 n4．Using pressure and discharge data from Rain Bird 结果,

- B9 {7 }2 G0 b9 X. |& _We find the attenuation factor （得到衰减因子，常数，系数） to be
7 @: G" c7 a  U6 \2 r
公式
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( K6 J/ l! S, s, `" E& s! ?6 w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）

$ W' D: ~" J$ c8 t8 h/ N; A公式

Where

# R: B$ N1 {/ a) h- M& o* U, V() is ;;
  u; ~/ t1 G  l2 T+ X
7．Solving for VN we obtain (公式的解)

  G  C! Y2 |7 R$ Z; j( e公式

& n6 Q1 Y% H% dWhere n is the …..

7 @) K5 r/ D- \' e) C 

: r+ z  ^# f. M0 o$ m. q8．We have the following differential equations for speeds in the x- and y- directions:

公式

Whose solutions are （解）

公式

! j# A" Q* J2 ^% i, ~0 m9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

公式

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
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! D7 _7 \4 I1 N" j7 U; _% u公式

/ s/ v+ u& x+ Z0 }: h5 B! G4 rThe decrease in potential energy is （势能的减少）
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公式

+ X1 x) `* i3 _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)

Where a is the acceleration vector and m is mass
( K2 w) w/ c- i/ {* l, N+ K0 u& _( B
 

Using the Newton's Second Law, we have that F/m=a and
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/ W& h) |8 z/ L+ O% o# z( e公式

. U; M6 R" p# I9 U* ASo that

公式

Setting the two expressions for t1/t2 equal and cross-multiplying gives
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公式
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22．We approximate the binomial distribution of contenders with a normal distribution:

公式
0 h/ U2 f) k# W7 m( s! j
, t3 Z; w5 Y5 d4 `0 ?. fWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

公式

3 ?1 a9 v: E. l% A  O  SAs an analytic approximation to . for k=1, we get B=c

 
3 p; ^* K+ r- a. k; t
( ~* D% I0 j& q+ {26．Integrating, （使结合）we get PVT=constant, where
' b, o3 |& I) L4 ]' ~4 n' o' b; ]
2 t4 _( w2 O' T1 l- b1 t% [% V* s公式

The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so

2 o/ a( _+ o( c* U# [0 g8 W 
6 ?( P0 Q% c% q# F
) T9 L4 L, t$ E; [) e! d  Z23．According to First Law of Thermodynamics, we get
) G9 y" s5 j1 N9 W! O( R% N  m
公式

' v; j; B' v/ H* w, [# i* n, G1 jWhere ( ) . we also then have

公式
* B8 E1 G' l* D4 V  j% P6 j
$ h& U0 s: w: [) E- s0 j" nWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
3 X- {6 t9 j9 P" i
公式
: y; L# |0 g7 W8 m0 n- s$ `# N, b
Where

; E$ ~+ Q5 n% t6 ]' Q! i8 T 

对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

公式
' ^* h! a, T. i- s/ [
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
# m2 Y7 @$ V5 Q
4 t( D5 S% q$ Q6 R# R8 ~2 Z公式

使服从约束条件
" Q& }; e  `9 q2 K1 W14．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）

公式

And thus f depends only on h , the function f is minimized at （求最小值）

公式

At this value of h, the constraint reduces to
9 J# T: E! I4 k3 O2 |
$ z6 F8 p% G6 \9 K5 L# W0 y) I公式
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( N; H$ ~. E' Y结果说明
$ n, s* I" n# U# i15．This implies（暗示） that the harmonic mean of l and w should be
$ ^% q1 k- n" T- U* j0 p- k
公式
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So , in the optimal situation. ………
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5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
- n( H/ S4 ^: o$ G6 a' E$ ]' m# L
公式
& D0 U, {# [  I" \7 o' r" r' _
16． We use a similar process to find the position of the droplet, resulting in
0 X: h! p, y/ f! l' C6 {& Z. r
公式
4 T. ~: u& y" w& Y
2 {/ _8 l5 @! E9 O- q" AWith t=0.0001 s, error from the approximation is virtually zero.
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17．We calculated its trajectory(轨道) using
5 R: U6 ~/ i( p4 H, O6 T7 b+ d0 }
* I( }6 ]+ I' |( b公式
$ j, O2 |$ H5 L3 K) R0 F% S2 F
18．For that case, using the same expansion for e as above,

* A9 q- e5 m: y公式
: e' `; K# u2 g  o" D- |# w
% _% X0 \8 |5 N! j. X19．Solving for t and equating it to the earlier expression for t, we get
$ b" _* E4 s8 O6 d/ D
公式

20．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
( G) b' r% t. x4 w* ?3 U
; X9 j4 ^; W# A; d) EAs v=…, this equation becomes singular (单数的).

 

0 x! R& b# f* r# n1 c0 d4 u: \7 ]由语句得到公式
5 s% P$ P2 s8 M# S; Q5 f5 |: F+ H21．The revenue generated by the flight is

公式
0 V; N0 U5 S! S" C
 
. e) |% p0 B' T' z! O8 U7 {0 G3 r9 |
: Y) \* r& X3 K% R* }' |24．Then we have

公式
2 v- u' [/ X0 |8 z1 R1 I
6 ]( D- W+ p/ D' m& T0 s2 eWe differentiate the ideal-gas state equation
% F1 U  ~( N) u+ K/ o: {2 C
公式
6 _8 _) ~% b+ b' t( p8 A
. \" b# w3 [/ w3 g' NGetting

" {) X! b6 a/ t+ U- B1 _公式

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

公式
6 E7 A  J* ~/ f2 \$ n1 e
 

) q) C. U+ ^0 L) |22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations

公式

, Y+ f9 \7 ^. d) c* O% CWhere P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
5 K; e9 z; i0 @. X
公式
. P% O1 r9 C) I2 n7 B" W————————————————
$ @1 n/ X8 y' p( K/ b2 ]& B版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
/ i1 I. P7 w$ g8 b, s3 O原文链接：https://blog.csdn.net/u011692048/article/details/77474386
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