由假设得到公式 " z( ^ f b0 n3 y' v! \1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式） 1 W& C' J0 V g* w/ S8 ]/ B+ f . ~* }7 p* G4 N# b% `; u. M1 C公式. n! K, @$ k& ?! p
/ ]1 P3 a a$ M% G% j9 o& E
Where4 a9 Z) i5 J' ~( N
0 |/ G# X6 `. N' ]+ l
符号解释 * ^" g; s' t% H1 S9 ]' A : q* D/ {* X% c* u2 sAccording to the assumptions, at every junction we have （由于假设） 6 y/ H. w9 \$ h" K" `0 ~- Y; K* k- U
公式 ! K w. v* h$ {% g; d' M3 g+ G& o! E" s' o& D) E& ?# `: B1 M
由原因得到公式 ) o1 d3 \" F+ V- Y1 J. _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）；: Q& J: m6 r! }) m5 Y
' A0 h5 Y9 Q+ Q3 f4 x; K2 F公式3 o( O# z! i' T9 R. a, e. C$ _
1 f3 m) J+ G `# o1 fSince the fluid is incompressible(由于液体是不可压缩的), we have& |6 }' E7 X- W: M9 f) W
) u) b* ?; U! g1 P, b
公式$ C5 _# A6 `, g+ |! M
% Q8 Z* N0 j- {% E
Where : b6 {8 a+ r' Z3 q/ J1 V; ^& X% Q9 Z% Y% T, p8 ^3 A/ a
公式 1 k6 r, M6 \" U: A# O q6 Y" o7 w8 I3 l Z* C
用原来的公式推出公式- t# F4 E" l/ s6 j$ I) E
3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到） , n) K& Q- A; t& _0 T % v* e+ N2 ?# ^$ B3 y# T/ `& ~公式 7 J$ t. }. m- e * P9 h+ b; j2 M W. X, y# c8 `11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields： r5 N+ v. v/ r1 O# I! X. c6 O4 r
0 l0 r4 |( x+ E% n# f
公式 6 e: r4 D4 Y5 M1 ` * K6 _% w' `, [12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）+ l' N6 t1 ^- }! a1 [
, _/ G/ {( a% V B5 q8 @公式 % E, I4 \: ~4 H& r # j* s2 ? u. _- ^) v; [8 kPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have7 C1 L5 y) |% S" f+ q' V. M+ A& N
; C6 Y t; H7 \" c: T& e公式 # g7 y; H" \6 Z7 p 3 R6 B6 v: X# L+ n* JPutting these into (1) ,we get（把这些公式代入1中）8 P; v: o2 Y$ Z
+ A3 P6 k1 A& s; S: L; T( \4 d
公式 ) t% Z3 Y: N C4 ^- Y& T0 { $ k! D/ m) x D Z7 x" wWhich means that the 6 g2 w( K q e( g; K) d4 R4 Y. V) p
Commonly, h is about . ~- J( V2 P+ G/ }2 `8 t( I$ t* x* {8 p
From these equations, （从这个公式中我们知道）we know that ……… + C G9 R) z& T: {0 e1 h$ @' X- G, I! f' J+ k/ i& |
% `0 |* _" s& d5 q & c$ N+ S, A a. y" f% \引出约束条件 ' D) X- s# D, u# k7 H3 I2 C3 `; q) m4．Using pressure and discharge data from Rain Bird 结果, " \" O0 b; L2 [! @* F, Z% V$ M! N O$ @# b& Y S
We find the attenuation factor （得到衰减因子，常数，系数） to be; s# S; \: t# H9 U
9 ^% N2 C+ k6 |' H% Y7 E; d公式 $ L7 Q, {, V5 h% v3 Q) A8 v8 c/ _& l4 h, d; s
计算结果 * _ r$ Z7 F5 n6．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) q1 X7 X9 L5 M$ E& g1 H0 T5 i1 e2 }( Z- a9 J
公式1 x/ n$ {9 c5 A& Z' o2 d
' e: v) L( H: u* ~. b! f
Where 0 {; S% e& p" E9 t) A p& \3 d& M M( {# |
() is ;; 7 ]8 a3 r: f' ]7 d! x * L5 y- d. s: y+ F8 i) I/ N2 I! k7．Solving for VN we obtain (公式的解) . U$ x9 o! }/ V0 H; O5 M7 ?/ Y c! c7 B7 Q/ v
公式 4 V) {& G: C. n 9 A0 I: b" [- t7 T/ t" j. Y$ QWhere n is the …..1 m4 ]1 K; \; B2 ?/ V3 i
- @# w2 C# W' F; h( V E H& F8 V7 u. y' z + g, i( _8 `: X- B6 }8．We have the following differential equations for speeds in the x- and y- directions:+ M6 G( l9 z) f+ S
7 i. }; k# G1 I7 t, L/ Q
公式% s3 w3 V- K) W6 Q
5 a" Q& F# i' h$ h1 G) \) B
Whose solutions are （解） H$ \/ L3 J( f4 J! R, l, k9 ?5 g1 n: S6 k3 v# \7 K+ {
公式 * ?" J1 |/ a1 I) C6 ~( I; N' a9 L0 d. U0 N0 b
9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:/ h/ F8 Y( g* R1 U
- g" [4 H' e. d% M+ |- E公式: }5 N. v; \. U6 B
8 V) C( [; ?7 u5 l
根据原有公式& x" I' ^* i3 O; J
10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is 2 R) _9 l7 H: h& f* m # q7 J; a& Z4 l- t1 ^& _公式 7 x1 q( j! l! W! k/ j4 f2 R6 x' U5 d- L5 `: A9 r
The decrease in potential energy is （势能的减少） $ z9 k0 B, b" l0 q . l- P& P1 W5 G- z公式( w# ~$ L9 [$ m! e h; V
0 b7 y; X4 |0 ?; J3 R6 `) J6 }3 _
The increase in kinetic energy is (动能的增加) ' K, }( b" Y" @7 }- k, O2 n * d* g2 r8 B1 l A* u5 f5 x公式 $ t, D/ h7 n W" G' \, |" X' q' e v# e: {
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) 7 I6 E; j G& a& P$ e1 e( q- A1 T& V+ ^& g% K# `; h
Where a is the acceleration vector and m is mass$ Y! d) E( R8 f% c7 W5 [
/ ~; U6 _& k( [. z# ?% s B
$ c+ B! Q% N5 l p* r) p5 ]+ h
5 Z' z. l9 Z' [% BUsing the Newton's Second Law, we have that F/m=a and 3 D4 U- x" I$ b- r) Z. q6 c A, a
公式 _# B! Q5 W- Z$ D6 ]5 J5 B% H4 [- u# f" c3 l8 G& F. e
So that! v0 w7 s3 v9 c& R( A5 S
: w( g# @/ F/ L, U1 c3 ~, m公式 9 ^! y& X! h# g# G) m$ j& q * J6 N* ^) s7 @% `) _+ ^( zSetting the two expressions for t1/t2 equal and cross-multiplying gives # S3 d! t( T& y' k; ^6 H2 ~6 Q/ ^& F
公式 ( C+ s2 N) `6 N8 J, C5 _3 _( j/ }9 d# f) `/ o) b6 @
22．We approximate the binomial distribution of contenders with a normal distribution:: T- C& z% R; [! z I! w
% Y+ Y. B/ I$ h! x0 Y
公式 7 p( K) P3 t2 ?+ f% S- b8 [8 `* @& {! [: C) v' r; G+ c
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives; v( a" k9 z7 Z6 s
( T E; e4 o5 n0 k7 m- \5 M+ ~0 T公式 " S" R# K* r) ]; N 5 {" M0 c5 ^. S+ x2 J$ QAs an analytic approximation to . for k=1, we get B=c 2 i5 i: M9 ~' y8 c% l $ e$ R' W2 `; W) D! r' y4 ^ 1 e7 h/ g$ V |% Y, e
7 H3 i4 ~' D: m c# J26．Integrating, （使结合）we get PVT=constant, where) Z( h# A [$ h# g0 i2 a+ F
; @ A$ g, J* J( y公式' C3 I, |6 w+ n% t
0 g& _8 O( C7 Q5 A1 S$ H A2 BThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so+ A# L) E* d" N9 n9 c2 X
% }0 ]$ K" Q& ~6 Q* J0 _ 9 o% l" k) u5 _4 z/ O) {+ h4 _; a1 x: e
23．According to First Law of Thermodynamics, we get " n3 Z% G: x4 o! ?2 O2 [6 u/ c2 `2 N& Q# c( w' N& O u
公式. s- B! i+ ^8 {4 ]. E" z
7 z9 r% q$ y: l EWhere ( ) . we also then have2 [: p6 M; d3 d" T0 j( [
* s% Q( g1 b: V/ U! I公式 # \+ c. F2 @: k6 N" n2 n' y2 ~; D( k I
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:% b0 R2 s: [( N5 \* M1 m' t4 b$ g
+ M. C. M1 t# R" ]% H( B, r: V
公式 L7 O9 y5 A$ Y
) N& a4 L7 A, z8 U1 \% Q0 K1 b
Where + X2 h: H# S3 r3 E4 a9 v2 L 3 \! D" V U' W1 k5 Q 2 ^5 f R% T# V) K. X
$ n i" o# C1 T/ _" p! T. J对公式变形7 j2 y8 z) ~' F0 x: C
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）0 ?- u5 [0 H/ z5 \) v$ I/ r
5 q; H5 e$ z# [8 P1 t3 d公式 * y; i3 }$ x O$ C% F0 w' `" {- {% q. _1 C
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 1 ]9 J8 Z9 _7 R: }0 h1 U9 B. i; J% P1 W
公式 ; ~. N6 ?: l$ T- t1 ~ 4 j4 w* C& \5 W* x使服从约束条件/ X4 p8 x. P1 n. B X
14．Subject to the constraint （使服从约束条件） ( L; i9 j4 _2 `5 I4 y7 a, m# W- R+ k' \# h' w1 R
公式 ( i$ K+ M$ ]" _. b + P B* D7 W) W4 N4 bWhere 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）8 ~: o5 E4 W4 B8 j3 _- R! z
* H% R, Y3 w' {( K8 h- e/ l* q公式; Y1 c7 U; _* w5 H, i: P2 m. ^
6 O1 ^* ]8 J8 b- @0 X
And thus f depends only on h , the function f is minimized at （求最小值） : C( u; ]8 \; d 9 I$ E( t& a; C# w; a l; ~公式 4 i+ h, h# c# b4 Z) } 4 t3 {; c! P# [' {At this value of h, the constraint reduces to1 l) k2 A6 y- l( h* o3 m
! P7 D9 p: c# E+ _2 s6 g& t; i公式 : R% V2 Z6 M0 G. @( f y7 d 7 \3 c" l, v/ L& ~* N7 @7 g结果说明 : S% v- v4 T8 n& u/ c15．This implies（暗示） that the harmonic mean of l and w should be 1 s" j1 q( g/ v) D& E+ Q 7 a( h' i! b% p4 h公式 " n0 G' Y W/ c4 F8 v1 q& Y' B# }$ y( t$ z2 }
So , in the optimal situation. ……… 0 Q0 _. V" ~5 B+ z1 f. Z \) o+ X" ^$ x! j; V) G: z
5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is ; c4 S, {5 B, D& S6 q0 G( R5 `6 ]9 t* p1 J
公式 1 q- | M# f& Z+ U2 A0 W( Y, w! f5 K @& O# g
16． We use a similar process to find the position of the droplet, resulting in' Z; e' i5 `. |$ L
! \5 N. @8 c" }$ j8 h% ^3 r
公式 9 ~; S. e( ]7 ~% D $ t6 ] k5 A" u. T( FWith t=0.0001 s, error from the approximation is virtually zero.) f( g; ~$ G; M
8 B* e) I4 n& o: Z3 r: K6 S " r& L9 w+ e5 N W: v `7 s* j
/ a& } K7 Y6 _- J$ P: ^17．We calculated its trajectory(轨道) using& s4 c n$ Q1 ?: o
) w' P; c* E5 K9 t( s g公式' J3 X3 ] v. ?3 p0 s6 _# ^
, x" I3 E6 R5 _; M7 m0 R8 u
18．For that case, using the same expansion for e as above,' V- L: n/ n2 R+ t% e7 m
U4 c' p8 E F* u2 C公式 / N& Z! a0 R% n7 s l. U, b: c3 y( z' C" ? ' c- ?0 r7 u' i19．Solving for t and equating it to the earlier expression for t, we get $ t- b/ H1 P; S' y. d1 h9 \3 R3 f3 M: ^8 \! b
公式 K2 [+ k* c& `) I. ^! P: _. N8 U
; _+ j/ {: F; q# [% B2 _# N20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is / D' e0 _# [! q! z 6 i1 S6 N6 ^' \* Q) `公式 9 _, |* }! F5 m) Y! G- m! S0 [9 T - g+ m- X! f0 M8 _As v=…, this equation becomes singular (单数的). ; R( ^% l9 `% W4 x2 V8 ~" { 6 l! A+ p) e H 3 }+ z l9 W+ ]' d+ L) y
0 S0 q5 N+ A: j% j' ?5 \
由语句得到公式 / o1 k# ^9 O5 g% | D21．The revenue generated by the flight is m3 r' r! D3 g/ V1 u# A2 K8 g& [; t8 l4 k0 B3 d* e6 Z5 w8 m# {
公式 - \! P7 c6 ?' c7 @3 { 5 Q2 K' N9 ~& E' p8 a8 r ; W5 c% L6 b" z1 P }, c/ z/ K/ _5 V( D5 [/ f/ J; R {
24．Then we have0 O" [/ E3 ?& }- A9 j/ R
% v$ \3 a% `) o2 I, X: b. G/ q- v公式. ] e- |# H3 G. b4 ^9 [& v+ |- ^
0 k" J( J3 S/ h% r9 n. b( xWe differentiate the ideal-gas state equation5 }+ x) C" h. I; H, A, T
% M& t0 r6 G5 Q公式 * p) R- g8 ?. `6 [, p6 v! e* F- B c/ T
Getting ' j! @& `* m. m7 k0 a& l4 u& Y. c4 z/ V* X \/ c& ~
公式 ! N8 V) w; b6 O. V; b |6 w9 R9 s+ ?6 j25．We eliminate dT from the last two equations to get （排除因素得到） 0 ]5 d' _) n( u' ~9 x2 v8 O# m% b/ r0 }( y
公式 / ~8 z8 u& k/ H! r8 T, {- ]5 @# _* }
3 `; k9 r& v8 P! n- T0 z * v, C* ^( `$ j& p# r22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations 7 d( H% | |0 S# E( ^ 5 I( g" y. ~8 Y B9 K! l0 N6 M公式5 f4 l$ `! d" B( o
- [3 {! J% G0 o: @4 u
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）7 ]( P! f; u( E: {7 R
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