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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* }( d5 b# u4 Y  O4 m8 e, V! G
1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)
( T, n. D5 H+ G' b) N5 N5 h2 T$ T0 f3 l- w' ?7 r* ]- q  ?6 [4 p
公式
: L% _* k: K) J( v- G" }
( U. k/ J: G# F7 sWhere4 T5 c  d1 b& v+ l5 j
' b& d) t1 ~6 C6 m  ]+ v6 v  S5 b
符号解释
5 j7 P/ B. k2 O' `, S; `
+ x" J9 ^& a6 Z2 bAccording to the assumptions, at every junction we have (由于假设)9 Y$ J6 Y+ v  Z+ \

, @( k  |6 X5 F3 S: ?6 q5 E公式" ?- w( o6 @) u3 f

8 P8 H/ P1 @  w1 }由原因得到公式
6 ?% B6 Q5 ]( l& w7 w' |0 `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);
* Q4 P4 }) l9 S, k7 }0 E' D4 B/ l: m
公式: s& {+ |" z' H/ q4 c

, C% w% k/ m' M$ ~Since the fluid is incompressible(由于液体是不可压缩的), we have
- {  U& L( f$ \* j4 w# Q
% o0 b+ R+ \0 @: a( f9 w公式
7 R2 b  o8 K7 |" l0 c
1 n( m, U2 K3 W* jWhere: F; n4 R" ~3 |: _8 \; l
6 @, `. p. f, x- e
公式
1 U8 [# q, v/ v/ F2 G* p
% A3 E  @. O) S& O, Y! w用原来的公式推出公式4 ^9 n. {7 B+ e" W
3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到). m; f) Y) m$ K+ h" C

2 F; N& g7 g2 x公式4 {5 m$ R: d( l* @' K
( c+ k& l$ k& k% w
11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields:
' B# z3 T2 w8 P; m0 b3 `+ }1 z$ a3 {6 h
公式0 X9 R+ n: G# P& H; i7 r
' s( f& r' s& f
12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)
  k  l/ [  p0 e0 r/ R, n- R* T4 u1 d
* J# y/ m0 {3 V  ^公式7 ^, t- R: T9 x, s  ^. t& y2 _
( Q5 K% J. K  K  J! k, h! Q
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have4 k; J( {; c+ i( o
/ s6 y; [% U& K6 c! p6 {
公式3 r$ P# o, r! ?4 G# T5 u3 Z

& _8 w& f# ^3 @- O5 A( `& rPutting these into (1) ,we get(把这些公式代入1中)
" F7 K1 T% m( [- B/ I- a# V$ T" r6 G8 w! h+ v* r/ g
公式
6 b. ~& `% B+ ^& c9 c6 v9 ^
2 t2 s. E: Q7 [' _Which means that the/ y! R. N1 s/ o8 J" r& J

$ X* Y# r) w" [+ L3 y8 }/ W7 NCommonly, h is about
% i$ \" N9 E% S" C  d" s6 q  p
+ D" ~/ r. B4 R* I. c+ G, OFrom these equations, (从这个公式中我们知道)we know that ………
- A& K7 N+ U1 Z& R2 p+ a$ \$ n" c9 C% r' v3 A3 O# J
 , N& d+ B* k9 u8 r
- F$ i3 a4 ^" L- }. ^
引出约束条件
0 ~) o& V, a9 n% M4 P# n8 }3 l) R; D4.Using pressure and discharge data from Rain Bird 结果,3 O, d  L( ~) Y- v  E& N& [: K
0 m, w4 p) {6 A, V
We find the attenuation factor (得到衰减因子,常数,系数) to be
1 a3 S' H: h3 J- M, _# J, o& }
( f6 J' D! u. u- V0 A' {公式
* g1 D* X9 u) {' @3 ~/ m' E) G7 Q8 _- ^* ^! S2 V
计算结果% |2 p4 v/ `8 G" T
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), z6 E6 E4 E# \. c$ v. p4 ~) V
7 Z2 ?) B2 `4 J/ C
公式
3 }( V6 E4 l5 Y1 P
. c  O) K. p  L3 UWhere
) n; |# d! `$ E4 \6 T: _9 T6 _2 B- Z& {2 U
() is ;;" u- U" A* G6 L" z" [# V8 {

$ W" b- `5 g  G/ m7.Solving for VN we obtain (公式的解)
% n( D( d6 w2 m/ Q6 W! h
/ o( N, P7 ~5 u! g公式- ^7 _) t' B8 k  Y7 k7 x
6 X, q! n8 v& X9 `0 s* [8 C
Where n is the …..9 D9 M! E" A! m, f
* \8 `3 N9 g6 }- g7 W6 ]
 6 `1 M$ N  H. ]6 |; z
2 r  u5 a4 G/ p
8.We have the following differential equations for speeds in the x- and y- directions:; L* n6 h! |$ @' d- y* [
! }0 v' a8 G# y0 V3 g2 w4 ?2 m
公式, J2 t0 T5 M* }4 _) F7 n& l8 N

  D, G+ j0 \& c5 lWhose solutions are (解)
' S. W( j# Q% G3 A" ^& Z
9 y) G3 K- {1 T0 G" U公式
# X" Y! b, m) p- s: S6 J+ f. @8 T( v) m+ V+ s$ x
9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:6 J& l# u5 W. T5 G8 G$ L/ o* a6 w9 a

; C, F, a: L/ X: E' h公式
3 L5 u5 T& d6 W# e) B+ ~9 w- c
$ D6 L8 R+ D3 v$ ?2 E3 t3 Y+ P/ O; k根据原有公式/ \" [) g* b6 Q8 H! Y+ R% B6 w
10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is
, X5 L6 Z2 E& Y% q1 Q# l5 O0 l/ F% [) r/ S" Z9 O# V
公式
) ~/ [2 `; w# @, t1 C2 m7 v
  Y$ D' \' ^+ N9 K/ ?: q, @  t0 XThe decrease in potential energy is (势能的减少)0 Q5 L& b' C6 i# ~0 n
% e" i' p7 F. @8 |# n
公式
3 p; ~3 \) i/ q7 Y( g. b. y, p3 b3 q) |: e- ^  B. {( E
The increase in kinetic energy is (动能的增加)8 n0 i7 o( M% P# [+ F  o! w  _

% f. d- j7 ?. D6 K公式
5 ?" n- o; y+ M" v  m! b0 t* F2 L2 t% s* u+ z
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)
, P6 s" a; [) |. S9 g  C; k- \, [$ B' s/ P
Where a is the acceleration vector and m is mass4 \' I2 D% U% Q4 j! A, h

0 D; N5 T- D' W& U+ d+ \9 [3 h 
3 U- \* _' p6 V4 B* r3 @8 N6 L% F0 Y* A8 y( p
Using the Newton's Second Law, we have that F/m=a and/ _! F( L; M2 o6 a
1 w: p5 t) c; o0 e2 z* h# Q- I
公式
, n+ f, T( j3 Q  t) }) A8 x7 I# M* {9 S; Y! C
So that9 c. T7 G. ^2 h
* m; T! x2 S* n* C8 g8 \
公式
7 ?1 o  a, g* g; Y$ W* Q2 b& D. }  Y3 z" h4 z! c
Setting the two expressions for t1/t2 equal and cross-multiplying gives. E2 u6 _: v0 t9 ?& _* S9 j4 }
, l9 X# r4 F3 H  \) \, t* z- v
公式
) o/ B: t6 I# Z2 s- y6 {
  s' G& L# U6 y: f; x22.We approximate the binomial distribution of contenders with a normal distribution:
. J& Y- B% V1 u% C: B+ j2 K( L- f6 I& z2 u% [# J( |
公式4 i" c( v+ q0 Y4 p# r; n! A

+ [2 N5 A( y9 Q1 ^0 w. ZWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives3 Q) g$ |4 Z1 ?4 P; O- r$ c' P
  s+ [; K* D; i; w
公式0 G+ v" A; C9 @+ A! V3 A4 W

6 A& G5 Y0 k4 @& H# G0 j5 jAs an analytic approximation to . for k=1, we get B=c& {3 z) f5 h- j1 Y# f: s. n
7 @6 {, {% c5 K: ?& z
 3 u1 ^/ n1 }' h2 H; O. i6 J  D8 M1 E

; r4 x# y# h: M; b8 A$ P$ e26.Integrating, (使结合)we get PVT=constant, where  s" N. n! _: A9 J

. U5 t- F2 Y: c7 }" o- e# n% l公式
( _# l5 Z% C! P5 H1 Y) h; U# o  j6 H& |7 h7 I
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so6 c( a  d$ K4 B. ]: P  J

3 [5 w( U; y* S& {) c& S0 k 
3 v/ _) t9 ~) ^0 v4 ~7 s$ _) J& D. C; |7 x% J
23.According to First Law of Thermodynamics, we get+ H- r* v5 u* ^
$ h+ r! g  |3 J$ m
公式
; e7 B- o# `/ K! Y2 a1 o& Z% s& G: N0 D5 D+ {' z6 h; H8 G
Where ( ) . we also then have# U  R, ^) D3 _

1 L" v# e$ A; C! `; L5 s公式; t7 u) S1 k7 T1 }7 I! {

3 h" P0 h# J; k! CWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:1 H# y" D( a! t! i3 \. C

! s0 Y3 Q2 x0 z% k7 n, p; x( g公式, a- V+ P' v, l1 _' F

. N$ n" |  l( mWhere3 ]9 l7 E% K1 }4 z8 ~5 N

# z$ Z+ f0 q& B& V* k/ s' ]3 I ' F) [" j% Z4 }4 W$ R
% n6 n9 P; C7 B! r% j% g
对公式变形+ ]5 b4 X( t& n1 N$ V
13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)
& _$ D* q$ }- C. @, j/ b) T( l6 o0 ?3 n
公式
' n: O; K/ q3 I8 n7 `: z' A2 V
) p4 x( H/ o3 D% n6 _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- S% O  Z& X5 H. k  o3 x! Y) N% g
) S. c& i3 n, i" |# `2 s4 R
公式
" v% `3 e% K) |2 P. r  S0 w/ n- b5 W1 n6 \4 z9 a. W
使服从约束条件
$ ]! r6 w! H; P! @( J14.Subject to the constraint (使服从约束条件)
# z9 F3 \, R5 b/ N2 ]. R
. l* d  P3 U0 B/ q) h( ~2 F- ^公式
- a( V& X- t4 n* P# e" {/ i& n% L, `3 h/ ]5 J3 [4 p. L  t: B! J& I& ^+ [
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)  x5 R9 p2 _% z

: v) G. \4 i/ P2 P( \2 d5 E公式5 ^% }' U6 |2 [: D0 G
. L( k9 f4 N% d
And thus f depends only on h , the function f is minimized at (求最小值)1 E0 b; @5 F2 G# P3 V# q% a7 S' W
. ]8 p9 i9 f' J! t
公式) T+ E3 m4 ]( S% d  m9 b

& A) U: X0 d" S* ?$ b, WAt this value of h, the constraint reduces to
( s8 E, M$ E; c4 u  F5 k) Y& [( w* G* Z( x
公式
# g6 i! e; k8 }+ I* c' e
+ h! z1 O) ]7 T6 G% S% W* ^结果说明6 n; @1 I5 I& {; s& b( O
15.This implies(暗示) that the harmonic mean of l and w should be
3 p2 Z; w5 ^! o# T! b
# J7 a* B3 H- I' Y# V% S# l) N公式
  U1 j" V9 n0 k4 Z5 a+ d1 F
% C: }& L  ]$ m2 D, kSo , in the optimal situation. ………- V- O0 i& I2 R0 y+ Z0 i: {
3 m2 ^) }0 N7 W, v4 A% j
5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is
7 J# J7 O* Z! R) N
7 e( P$ c0 K- i9 \公式
: z# u; Q! a8 o% B: e8 P+ Q- _  {4 m4 w; ?4 f! t
16. We use a similar process to find the position of the droplet, resulting in
8 ^7 D2 [/ C! u9 m* T; U6 y, {( _
& @# r$ O" S0 v公式; N$ J" ?9 B. p! I

- M$ X7 o5 x) b. W1 ^3 B$ `% sWith t=0.0001 s, error from the approximation is virtually zero.8 U, Z2 S; H8 H2 E' ^8 l5 t
8 a( a, u# J! l) a# V
 
% d4 i9 ?3 \9 q0 N7 E* a5 O6 i5 e$ e
17.We calculated its trajectory(轨道) using
; Z( t- H( C, I% x- E1 h" ^6 j$ q3 G. A1 {5 i, j, F
公式
8 p, e$ n% F' {+ G% w- L- ?- ?5 U5 G. c) X
18.For that case, using the same expansion for e as above,
" Q. H) w, U; ]7 l1 S5 ~# L4 U
( O; Z  {5 J! c6 ]6 k- U公式
* b1 O  H% g9 L+ y" [0 G  T2 k1 P/ ?  p) W$ |! P$ L1 u
19.Solving for t and equating it to the earlier expression for t, we get  }; h1 ^1 p$ K. b8 G
4 b. f& J, N- u  T
公式- T: a# Y  e$ Y0 ?  ^' ?7 U. j) Q

4 n! u5 W; l! C; s7 U2 m20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
( a& k8 z0 f7 J" ^0 p' J' [4 d
# y0 _1 {9 p# J8 [% Y7 a公式
6 K$ }' @) S  |4 X8 m! B; _: o3 Y0 w. i4 F: d* }' K% r7 a
As v=…, this equation becomes singular (单数的).
9 J/ K; S  y/ w0 Q  v0 U/ y' E- }5 ?( X6 G& o0 h- \& T' ]
 : y( |7 Z6 J  q) T( e$ `/ z( c
% k) ~# P0 V; }
由语句得到公式
0 X. M3 _9 G( ?4 j) K5 H6 D  v2 i& Y& Z21.The revenue generated by the flight is+ |9 E  j( _: P: k7 V

% P3 j) Z1 q. N! I公式) |; {# m  K. q  U

2 [1 `% K/ r% C( l' Q- e, e 1 q3 n! d" ]8 o! g" H3 C

9 ]  I) A; u; X1 r24.Then we have
6 U2 h2 E- b5 O& K. U8 L4 F: \7 |% J- {5 y& ?- i0 l
公式% ?; |8 G1 D( ?" e" G$ m

4 ?0 J( e6 b2 N& Y/ ~0 J  LWe differentiate the ideal-gas state equation
' Z2 V+ c# a  r, m
7 Y) f: L1 c9 A公式
2 C& Z) Y6 K0 Y
; D# B6 i9 [  j8 K/ D) ?1 J0 a5 w$ JGetting
7 Y" a, h$ ?2 {) B3 F! E" r# B8 l5 n, |/ O
公式4 U1 e6 K- [; y: F0 `

1 F, n: ]) r5 {9 i9 s25.We eliminate dT from the last two equations to get (排除因素得到)$ ~6 k/ j0 k' r& M! W$ a
" m. w; d5 ?  O, q- F. {2 }
公式# P- V+ a- ~# d" [
0 S  C' P  F  F
 ) N& m; r# [+ S2 c  X, q, p: K7 a
0 t6 I1 q+ C+ j' a3 |( ]3 I
22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations$ u# J6 p2 ?' L1 A( u
( W2 V" F) ~0 A3 c  I) A: R5 v  V
公式
- J' `/ [) O1 P& g; k4 q, q  n+ L, |8 i$ I: i: [7 q) ?
Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值)' b' l% B7 t' n4 R
0 W' b! K/ x' G/ A" Y9 D
公式8 n! J. ?# L1 a/ `7 Z
————————————————  v7 }" s7 k  F$ O* Z
版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。
/ u4 y# }! t' Q: V2 n: O9 R原文链接:https://blog.csdn.net/u011692048/article/details/77474386: u* D0 N1 ^8 X4 A

作者: 1369728843    时间: 2020-2-12 18:25
感谢+++++++++++++++' U8 ]" U7 ^1 A; I$ w' R

作者: chace    时间: 2020-2-17 15:19
学习学习学习 谢谢4 I- Q# H  z  }' p, _; g4 O3 b





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