标题: 美赛数模论文之公式写作 [打印本页] 作者: zhangtt123 时间: 2020-2-12 17:14 标题: 美赛数模论文之公式写作 由假设得到公式 6 Y, X3 {! z5 f4 j7 X; U G- ~5 {1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)! w: P% y# y8 P
E3 E$ q" u* P2 I& C7 z0 Y
公式( A+ ]) m1 M- t3 X1 E( q$ y
( W% g+ ]1 _' P% K5 t0 ]Where 0 d$ u9 x6 }% U& D* ~" N3 I# |" G2 {9 ]
符号解释 ) W: A% I- e: O4 F" O1 y2 b, g6 |" A. I* K8 g$ g
According to the assumptions, at every junction we have (由于假设)0 M+ A6 h$ u" q6 Z- Z0 l. v
$ v6 ]9 ~3 |0 P! _7 |公式 6 m" P! @+ Q& ~ ^6 E0 B( G/ y1 P' |9 A- W6 @
由原因得到公式% l9 U3 L' w0 x1 s; L# U
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); % `! }. y8 ^/ ~7 O; J% g A: E+ \% J( ^
公式3 a# O6 ^+ e' P# v J
0 K. O: j# q& F4 I" X
Since the fluid is incompressible(由于液体是不可压缩的), we have 6 G% u- f# w' k5 f' E# q3 u/ z$ Q/ G. {
公式 6 j5 d% d5 `+ {/ h3 Z' @; J' P - d) T, V" M% qWhere " f6 x+ X+ w* T) m, e+ i3 H5 j 0 T6 J- n! i, t8 z9 n+ y公式 2 d+ t. K3 u: E/ B 4 r" Y9 v& q3 u/ X j用原来的公式推出公式+ |- L* P9 z( v) C
3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到) & ^+ |. x8 D; e9 \" R, n5 @3 \% c3 A2 S* a$ {: y8 C
公式 ' P# c* A- b. d" B- N3 g5 \4 t2 t
11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields: 8 L: j" y) ~' ^ ?- d. x2 f 9 G9 Y/ A9 c+ ]% |6 S公式 . l x/ n& W) S / s. E! t% O' C; Z0 n12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)+ g4 \' k9 ^8 P1 H
1 i9 ?: Z6 s# T& [7 d- I
公式# l, q6 m! ]# m9 l) x0 L/ Z
' a9 d( Q2 y' u( A; `5 z* I
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have , C# a3 M& v" {; K, w+ }; w $ N- j# _, q% e( W3 N' P3 P+ B Q公式 + i1 j5 B& d) X4 X7 f ! N, r# z; p- @, XPutting these into (1) ,we get(把这些公式代入1中)9 R1 J& M4 H g, a
0 D6 L+ W- M5 g& i+ t
公式 , A2 T; r) Z) a' y8 k* }, I5 ^ ) P/ T8 T4 \: z4 G: k) x$ }Which means that the 4 Q" N8 j* j3 _% G! P* T# } ! ^4 T1 [- x# X" }/ H; _% ]Commonly, h is about7 F6 S u& P4 N; @% u3 ~
! \. A6 N( G) p2 JFrom these equations, (从这个公式中我们知道)we know that ………" _8 B5 l6 }" r# A) m& ~
3 h f# S/ a" T( ?% b
# o% d- E5 U3 F2 R4 Q5 [
* n* d# b/ o, F" H$ `2 ]( m) B引出约束条件! a) K7 d6 F" J+ w/ v
4.Using pressure and discharge data from Rain Bird 结果,( P0 \( u! a; K3 {. U
" v0 A; y' l0 rWe find the attenuation factor (得到衰减因子,常数,系数) to be# Y/ T0 V8 z9 d4 W3 I$ f
0 y& l3 e* x" ]
公式2 m5 d4 u |: D5 V" G
$ K* R5 f, Q' i# G计算结果 p. A7 L A# M- R6.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) 7 @) x% n0 o: t" D; | 8 D/ r( Y: l- |( @* g3 q公式6 z/ H( h/ C! e! t% M) B
; v: E1 b( I- z" vWhere' S! g; v) Z, n
( r: H( Z1 N: a9 Y
() is ;;- C1 T3 R0 D* O$ r) L, z
5 \9 _5 S6 \. c G
7.Solving for VN we obtain (公式的解) # w$ a/ P% Z7 r8 B F: A8 t' h& z ; `3 R& h1 _, o8 ]7 G3 y ?公式 ; d* \" B5 S8 r4 D) s* v+ _7 U2 y6 m$ b2 d1 s
Where n is the …..& p) \5 `9 m- W
. F6 i# P4 C; a3 j4 g" k
1 N- u+ f2 `1 g3 X! p: j o9 ?2 ]
8.We have the following differential equations for speeds in the x- and y- directions:2 L9 A. I& ^4 t, e; E! d: T
_4 g& m7 T5 L+ d3 e3 h/ Y
公式 3 ~/ [1 X* h8 w! T/ I3 k: i7 Y6 r, A+ L0 I l! f; y
Whose solutions are (解)8 `- c- S& r* f0 c4 l
( W" ~, N9 Q, j
公式9 K/ U/ P8 e4 f% K; b! C4 M0 ]
/ j3 ^: @. t! P, U& i5 _% s9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:4 ?7 f: c6 H: ^" K! s
) ~3 l! M5 r! b
公式8 y- e; m$ q, y% l7 s. D8 l
$ {) R) ^ T% A/ j3 x: x
根据原有公式 8 _. f5 h: V& e7 {' d1 b' l10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is 4 J( \: @2 Y) x9 a0 Y5 U # l5 B6 Z( W8 V) c1 W公式 * f' t( o' H; f- |5 a p$ f+ S/ } : O! _+ T2 e) [. M( MThe decrease in potential energy is (势能的减少)% k" G; f/ v# g2 B/ E v, j# d
% v2 E/ ]6 W U/ e" V8 @( i
公式 3 e+ [# J- M$ ? L ' n/ F! ~* h/ D9 A1 z- iThe increase in kinetic energy is (动能的增加) ; J# c) [ P) w* f1 T' O4 w3 c4 \7 ^) r; B* Q- o. h
公式) d7 ` H' v5 ~! y! C+ \
; S+ Y, j# r7 q+ B
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)9 Z* ~0 I4 T- k4 [$ V7 D
5 x) S. \* V( ]; H- N
Where a is the acceleration vector and m is mass # }5 |0 K2 n. P/ [5 P' D) X+ l: g! `4 U# y g
% n, J% w1 Q, A . l W! ^3 x' g9 t' U6 `4 Z( LUsing the Newton's Second Law, we have that F/m=a and! ?4 s7 H4 i- s/ z+ S) U2 q
9 w" y5 M( F% p1 l( e9 y# M公式. {# g. P$ X- R M v
. ~4 o4 S- H; @; n! @" {
So that9 ~0 x" h3 E, X \
' a" o" \ s- F* O( ^公式 ' v E2 g) x" {7 q1 \( y# ]: \7 \% M3 k& w- W% Z5 x
Setting the two expressions for t1/t2 equal and cross-multiplying gives7 ?3 i3 U4 r& a) n8 H, e. J
) d7 s F/ y' v公式4 x9 o) H+ G* M) p ^
4 g8 o0 F1 c; g: C3 @$ _) n22.We approximate the binomial distribution of contenders with a normal distribution: ) d2 q: v4 Q1 T# B ; a b2 U8 g1 x* C$ \) S公式 8 R0 I% w% ^) V4 B! T7 N* y 8 k% Z8 m0 f7 j& E5 T- z0 SWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives# Y$ v4 d4 K) ?% o
6 b9 C1 Z$ Y& s8 Z; c1 b公式 0 K3 m5 |0 l9 b6 ^0 A7 a5 g/ i J& v9 l, Y5 x5 `" N
As an analytic approximation to . for k=1, we get B=c) _3 c" u3 h& x8 V* e
# m& O' P: Y7 @- Z4 A3 g / C2 a' I% h+ G5 y q- B- g+ ^$ s6 h! t1 ?+ t+ d
26.Integrating, (使结合)we get PVT=constant, where 0 \3 @& p* C* a7 P. C% `- v. I4 h# O* [; }% ]( c
公式 % _; R" q. x9 G+ U * b$ p( ]7 G8 eThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so3 x$ l# ^5 g- G/ k! l( x1 O( ^
, k: F& j! H: k P* ?$ u% z
% R* X0 d1 \+ E
0 m8 p7 q2 ~4 X& o
23.According to First Law of Thermodynamics, we get" u/ ?) Q/ v5 J& x, W; r5 i# v
$ i, C: D+ N1 H( x3 V
公式 , i1 S3 b) ^1 ~! l M5 m6 n& U' A# I1 M8 y$ ~, M8 y
Where ( ) . we also then have 1 L, s, Z6 R c L ; g$ r T% {" J2 h' m" K( I公式 ! _4 Y: _6 _/ b' y- Z" x. w# \" T1 j/ E$ e: {( ?' q
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:+ s* w( _# z- B, ?
3 x0 ?$ S: Y; d/ p/ u; _公式* L' o7 }. Y4 N; ^! n
& r! Z w: l) a# w. @
Where$ t1 `& @1 |+ q4 y
& ~5 b& J1 ~$ g5 H
$ v3 p3 H" P$ }, t2 {. M) W2 Q1 P! }: j3 N/ x5 I; j
对公式变形+ s5 P Y2 ^( Z+ S
13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到) 6 w4 R6 G8 v. k+ p 0 R) C& T9 F8 r公式/ F3 w6 C$ d6 }9 P0 k
" A" p3 b1 {: U* e6 t" j+ Q
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' G; ]1 t0 ?* ?; r9 L5 [
; ]# D) Y8 s: a/ s公式 ; D! p5 d+ Q6 r6 c6 q6 L6 W$ j ( a0 n" ~7 n" D* q- K4 vWhere 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) / ?' M" D) K2 t) ]; \: H2 J \2 R2 H T. @. b4 T4 I: Z# O公式; a; p5 Y' t& h$ C5 g7 ]4 N( B
6 Z8 k& m( O6 w# F
And thus f depends only on h , the function f is minimized at (求最小值); A; J& y+ s, F
/ ~) U- x5 I" H/ e7 V2 \- [. L1 \公式 7 \% x! r/ w" h6 M$ w* N # f: b# F( d6 n f$ `At this value of h, the constraint reduces to ! M. h* E! ~) ]& ~ 3 g d* D+ v. X' N0 C' R" a7 |公式6 D L: `& I# c4 Z2 J0 ^7 F
5 o- e$ {5 M. @结果说明* j9 x* d5 i& \
15.This implies(暗示) that the harmonic mean of l and w should be* y4 K" Z* l3 _1 C. y4 M
6 M8 }! w; K3 p+ R5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is; B. R0 U0 w2 G9 v7 v
6 W- M6 i3 N+ Y
公式 # K' N, j7 `$ \/ Y0 n- |( W. Z( P1 v. C* X
16. We use a similar process to find the position of the droplet, resulting in: j" ?, B1 D) E( [* X( a
8 }) A$ `5 L7 V6 z
公式$ a* j/ i! p }6 V( O: G
5 {3 n6 F, S1 {: A+ H
With t=0.0001 s, error from the approximation is virtually zero. " u* y' }+ h" g9 U j( H3 V n( Z, k7 { j( q" f* V1 f } ( L* q1 [0 Q6 j$ F2 e17.We calculated its trajectory(轨道) using9 Y( q. c, z$ Y( m, d0 T ~8 @% }. W
& C) U( E6 O: }公式% n, Y( @8 I7 K! }/ F0 Q
' L' d9 Z6 I$ L8 `4 w: _18.For that case, using the same expansion for e as above, # O3 [) c# U5 D7 S$ t n2 [+ b, n+ a Z9 S0 o( ]4 H# i
公式 5 Z% r4 i. {+ k ^ % O7 Q# t# n& ?! ^7 o, ~19.Solving for t and equating it to the earlier expression for t, we get9 s& n" G( z$ K9 t9 J
; U3 c1 g/ o: G; b
公式6 l/ Y. g0 a0 h) {5 L3 Z
; d; Y% e- Q* j8 M: `$ b) F20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is 0 ?5 @4 ^( H( q' i, J/ i) E) z" C) g" g2 x- \- R. C, F
公式1 n) P- e k' E5 F* d' l4 p
, f7 D8 x; p" V* r
As v=…, this equation becomes singular (单数的).! U1 J' w8 m5 F0 I$ ?
O/ u% M( P7 N. E. Y * Q% Z: I7 [" @
K) v' {2 M% S% i' t
由语句得到公式# `% [( v2 s0 t2 I$ _$ D
21.The revenue generated by the flight is/ t! t) U2 y3 @6 Z
! A" S: c* a" Q6 b) }* ?3 A3 x
公式2 `% c9 t( k+ j) r6 p
8 N* x9 }6 v0 U5 d- \" l
. _% K" K0 H& g4 F
& y+ a, k4 g) U24.Then we have + j4 J& z3 y$ O- K. g& N& S6 L; I6 Z3 f
公式* g+ L% z6 q5 r T/ ]4 t" q' j
4 j0 J( q0 C. T& A% u* s
We differentiate the ideal-gas state equation8 Y: i I1 J; v; ?* h9 V' q
- v) g. S1 C+ e5 p公式 ( R' F6 w; v+ j# r' y6 a! M 7 z% K9 L j% j: PGetting1 R5 _+ L" p% M6 c5 D
3 h {' C& N% E- _- b+ D公式 N$ s0 ]/ f$ y8 y * M; j5 X3 C- `& Q25.We eliminate dT from the last two equations to get (排除因素得到)( q/ o* Z" O# x# Z
+ E$ c. `8 N! T; _+ A8 s' X) E公式% q( c7 _$ n+ k! K# X
) W! Z$ E( j6 O7 ]+ T
J$ Z6 \, T7 D Z
. X3 ~0 y* Z4 l+ O- Y, _: w
22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations 2 x# `+ P: V9 H! ~1 F9 x! K% c% R% R7 ] W4 z# F/ u2 o
公式, Z) Y! E0 ^* d' E- I
8 O/ N Y5 b( E% xWhere P is the relative pressure. We must first find the speed v1 of water at our source: (找初值) 3 |! v8 Y) {3 z& K/ T* k0 G2 y) B 5 j- m( g0 E8 H8 [& u% R. o& R2 C& f公式 * d# N+ d5 H/ |. \& u2 X) p! J———————————————— * d9 l m5 J4 ?+ W1 d1 b) I版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。; S, z* W. {: j+ O' ]; Y: |5 N
原文链接:https://blog.csdn.net/u011692048/article/details/77474386 ( G: H, x9 r5 b* g- Z作者: 1369728843 时间: 2020-2-12 18:25
感谢+++++++++++++++ ; ^+ R( u* c" |( `. H1 ^作者: chace 时间: 2020-2-17 15:19
学习学习学习 谢谢: N4 B8 y* s" H% v/ R4 t