熱傳模組建立

this a test

黃世榮:經緯儀的定位

前言:早前出差去使用經緯儀，希望能用兩種角度去作定位
問題:如何在只有兩組經緯儀的角度下作空間點群的定位，在建立公式以及程式後，發覺程式當時剛學技巧有需要改進且似乎這樣的計算是有錯的，希望大家幫忙想
程式流程:
clear
format long
matrix=xlsread('diow.xls','sheet1','A2:L3'); %% read the data
[m,n]=size(matrix); %% get row and column
L=input('Input the L you get==> ')
%%request the distance between two measure point

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% to build the alpha's array
for i=1:m
A(i)=matrix(i,1)
%% to build the alpha's array
A2(i)=matrix(i,2)
%% to build the alpha's minute array
A3(i)=matrix(i,3)
%% to build the alpha's second array

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% to build the theta's array
S(i)=matrix(i,4)
%% to build the theta's array
S2(i)=matrix(i,5)
%% to build the theta's minute array
S3(i)=matrix(i,6)
%% to build the theta's second array

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% to build the Beta's array
B(i)=matrix(i,7)
%% to build the Beta's array
B2(i)=matrix(i,8)
%% to build the Beta's minute array
B3(i)=matrix(i,9)
%% to build the Beta's second array

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% to build the PHI's array

F(i)=matrix(i,10)
%% to build the PHI's array
F2(i)=matrix(i,11)
%% to build the PHI's minute array
F3(i)=matrix(i,12)
%% to build the PHI's second array
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% to make their unit coherence
for i=1:m
A(i)=A(i)+(A2(i)/60)+(A3(i)/3600)
S(i)=S(i)+(S2(i)/60)+(S3(i)/3600)
B(i)=B(i)+(B2(i)/60)+(B3(i)/3600)
F(i)=F(i)+(F2(i)/60)+(F3(i)/3600)
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%find that limit it or not is no matter
for i=1:m
% if B(i)<90
C(i,1)=L*sin(B(i)/180*pi)/sin((B(i)-A(i))/180*pi)
% to build X's array (X,Y are not the real coordinate)
C(i,2)=L*sin(A(i)/180*pi)/sin((B(i)-A(i))/180*pi)
% to build Y's array
% end
% if B(i)>90
% C(i,1)=L*sin(B(i)/180*pi)/sin((B(i)+A(i))/180*pi)
% to build X's array
% C(i,2)=L*sin(A(i)/180*pi)/sin((B(i)+A(i))/180*pi)
% to build Y's array
% end
end


%% to build coordinates' matrix
for i=1:m
H(i,1)=C(i,1)*sin(A(i)/180*pi)
H(i,2)=C(i,1)*cos(A(i)/180*pi)
H(i,3)=C(i,1)*tan(S(i)/180*pi)
H(i,4)=C(i,2)*sin(B(i)/180*pi)
H(i,5)=C(i,2)*cos(B(i)/180*pi)+L
H(i,6)=C(i,2)*tan(F(i)/180*pi)
end


%output the data to excel
xlswrite('test',H,'sheet1','A2');

for i=1:m
plot3(H(i,1),H(i,2),H(i,3),'o');
plot3(H(i,4),H(i,5),H(i,6),'o');
hold on
end

PROBLEM #2 黃世榮

%problem#2
程式流程:

%question 1
clear
format short
B=[98 100 102 104 98.4 98.2 98.5 101 102 99.5]
C=(B-32)*5/9 % that's a)'s answer
[u,v]=find(C>38);
D=zeros(1,10);
D(u,v)=1

%question 2
clear
A= [10 8 6 4 -5 20]
B=[2 8 5 10 -6 3]
D1=[A>B]
D2=(A>5)
D3=A+B

%question 3
clear
x=[10 20 30]
y=[1 4 6]
A=3*x+y
B=5*y/x
C=4*x.*y*2
D=sin(x).*cos(y)
E=5*x.*sin(2*y)

%question 4
R=[ 10 30 200 400]
Req1=sum(R)
%series
Req2=prod(R)/sum(R)
%parallel

%question 5
A=[1:1:24]
B=reshape(A,3,8)
C=reshape(A,6,4)
D=reshape(A,2,12)

執行結果:

Q1:

B =

Columns 1 through 8

98.0000 100.0000 102.0000 104.0000 98.4000 98.2000 98.5000 101.0000

Columns 9 through 10

102.0000 99.5000


C =

Columns 1 through 8

36.6667 37.7778 38.8889 40.0000 36.8889 36.7778 36.9444 38.3333

Columns 9 through 10

38.8889 37.5000

D =

0 0 1 1 0 0 0 1 1 0

Q2:

A =

10 8 6 4 -5 20


B =

2 8 5 10 -6 3


D1 =

1 0 1 0 1 1


D2 =

1 1 1 0 0 1


D3 =

12 16 11 14 -11 23

Q3:

x =

10 20 30


y =

1 4 6


A =

31 64 96


B =

0.9643


C =

80 640 1440


D =

-0.2939 -0.5967 -0.9487


E =

45.4649 98.9358 -80.4859

Q4:

R =

10 30 200 400


Req1 =

640


Req2 =

37500

Q5:

A =

Columns 1 through 13

1 2 3 4 5 6 7 8 9 10 11 12 13

Columns 14 through 24

14 15 16 17 18 19 20 21 22 23 24


B =

1 4 7 10 13 16 19 22
2 5 8 11 14 17 20 23
3 6 9 12 15 18 21 24


C =

1 7 13 19
2 8 14 20
3 9 15 21
4 10 16 22
5 11 17 23
6 12 18 24


D =

1 3 5 7 9 11 13 15 17 19 21 23
2 4 6 8 10 12 14 16 18 20 22 24

討論:實際上在比較龐大的矩陣，用到很多元素指令，也許可用指令配合迴圈比較清楚

黃世榮r95631009

老師好 我是黃世榮 我想換此帳號 diowking@gmail.com
這是BETA的帳號

PROBLEM #1:黃世榮

題旨：操作MATLAB基本指令 程式流程和程式內容： %%%% Problem #1 %%%%

format long
clear
% question 1 f=200 ;%force's unit is kg
t=60 ;%time's unit is s
l=20 ;%distance's unit is m
Power=f*9.8*l/(t*1000) %power's unit is kw

% question 2
R=1+20+200 ;%resistor's unit is ohm
V=110 ; %voltage's unit is V
I=V/R %the current's unit is Amp
P=V^2/R %the power

% question 3
A=650; B=428 ;C=282; %triangle's three sides
S=0.5*(A+B+C);
Area=(S*(S-A)*(S-B)*(S-C))^0.5 % question 4
numbers=(52000+84300-25000)/600 % (total heat dissipate per hour)/(heat
form Hog per hour)

% question 5
L=4;H=0.9;
Q=3.33*[L-0.2*H]^(3/2)

執行結果與討論
Power =

0.65333333333333

I =

0.49773755656109

P =

54.75113122171945

Area =

4.523315598098368e+004

numbers =

1.855000000000000e+002

Q =

24.86218409261745

結論：使用FORAMT LONG 可以讓精度變高

r95631009黃世榮

帳號:diowking

r95631009@ntu.edu.tw