Question

does anybody do matlab? i need help with fixing my code.
clear all;
clc;

den_b = 8550;
den_f = 1100;
visc_f = 0.1;
D_b = 0.0064;
g = 9.81;
u = 0.1;
step = 0.01;
n = 100;
tol = 10^-4;

Re = @(u)(D_b*abs(u)*den_f)/visc_f
bouyancy = (den_f-den_b)*(pi/6).*D_b.^3.*g
Drag = @(u)((2.25./Re(u).^0.31)+(0.358.*Re(u).^0.06)).^3.45
CirArea =(pi/4).*D_b^2
f =@(u) bouyancy-(Drag(u).*CirArea*den_f.*u*abs(u)/2)

for ii = 1:n
u(ii+1) = u(ii)-f(u(ii))/finitediff(u(ii)) 11 years ago

Answer 1

@dan815

Answer 2

therse a lota fu in there

Answer 3

what?

Answer 4

okay what is this doing there finitediff(u(ii))<tol

Answer 5

basically i have to use newtons method. function and find its derivative.
but the function cannot be differentiated, since its a function within a function. i'm using finite derivaive function to find its derivative.

Answer 6

try it on matlab. the function handles just prints out the function. instead of giving the answer.

Answer 7

was there some typo here
u(ii+1) = u(ii)-f(u(ii))/finitediff(u(ii))<tol

Answer 8

why do u have a boolean check in the middle of an expression like that

Answer 9

because i've suppose to set tolerance

Answer 10

i'm*

Answer 11

oh that works?

Answer 12

cant u just let it evaluate the expression and then round off your total array

Answer 13

gimme ur code for finitediff

Answer 14

i dont have the fn

Answer 15

its not even plotting f right

Answer 16

i know thats the problem

Answer 17

show me the equation for f and ill go thru it

Answer 18

function [d]= finitediff(f,x)

Answer 19

h = sqrt(eps);

Answer 20

d = (f(x+h) - f(x))/h;

Answer 21

end

Answer 22

please post this kind of thing in the computer science section next time
thanks

Answer 23

okay i made it plot

Answer 24

computer science? but its a maths course

Answer 25

now u gotta make ur newtons method work

Answer 26

how did u plot it? if the function isn't working

Answer 27

clear all;
clc;
den_b = 8550;
den_f = 1100;
visc_f = 0.1;
D_b = 0.0064;
g = 9.81;
u = 0.1;
step = 0.01;
n = 100;
tol = 10^-4;
Re = @(u)(D_b*abs(u)*den_f)/visc_f
bouyancy = (den_f-den_b)*(pi/6).*D_b.^3.*g
Drag = @(u)((2.25./Re(u).^0.31)+(0.358.*Re(u).^0.06)).^3.45
CirArea =(pi/4).*D_b^2
f =@(u) bouyancy-(Drag(u).*CirArea*den_f.*u.*abs(u)./2)
%for ii = 1:n
%u(ii+1) = u(ii)-f(u(ii))/finitediff(u(ii))
%end
uarray = 0.1:0.1:100;
yarray = f(uarray);
plot(uarray,yarray)

Answer 28

the function is not even working. in the command window, it prints out the function of Re, Drag and f.

Answer 29

i added some missing periods in ur fn def

Answer 30

fn deff?

Answer 31

function definition

Answer 32

it shud show u a graph, just look at it and tell me if that is the right graph u are looking for

Answer 33

no actually its the wrong graph, its like a piece wise graph.
discontinuous

Answer 34

oh is it

Answer 35

there is something wrong with these. i'm not sure what.
Re = @(u)(D_b*abs(u)*den_f)/visc_f
bouyancy = (den_f-den_b)*(pi/6).*D_b.^3.*g
Drag = @(u)((2.25./Re(u).^0.31)+(0.358.*Re(u).^0.06)).^3.45
CirArea =(pi/4).*D_b^2
f =@(u) bouyancy-(Drag(u).*CirArea*den_f.*u.*abs(u)./2)

Answer 36

do u have a image or a file with what hte real equation looks like

Answer 37

that is a mess to decode

Answer 38

yup.

Answer 39

the long one(f(u)) consist of all equations.

Answer 40

okay

Answer 41

is D=0.0064

Answer 42

yeah

Answer 43

for this code i should get the answer:u= -.0584 though my coding is a fail.

Answer 44

oh i see

Answer 45

so u are gonna use newtowns method to find the roots for f(u) right

Answer 46

yeah, the new velocity.

Answer 47

ok hmm and u are also gonna diffentiate numerically?

Answer 48

no not numerically, but using the function finitediff that i created

Answer 49

hmm

Answer 50

lets get this fn to work for now

Answer 51

it seems like a pretty easy question i think numerical derivative is okay too tbh

Answer 52

what is this fn?

Answer 53

yeah its easy to do the nwtons method, but the @ function don't work.

Answer 54

http://prntscr.com/5qccgd

Answer 55

rewrite the fn

Answer 56

what do mean?

Answer 57

p_f=1100
p_p=8550
u_f=0.1
D=0.0064
g=9.81

%u=-10:0.01:10

i=0
for u=-1:0.01:1
i=i+1;
Re=D*abs(u)*p_f/u_f;
y(i)=(p_f-p_p)*pi/6*D^3*g*(-2.25/(Re)^0.31+0.358*Re^0.06)^3.45*pi/4*D^2*p_f*u*abs(u)/2;
end

plot(y)

Answer 58

here u go u can mess with it

Answer 59

got rid of the function way, just to see what this thing is supposed to look like

Answer 60

looks messed up.. check of the function but use that method forget the function definitions for now, just see what the value at each u is and graph it

Answer 61

i don't understand this

Answer 62

i wrote your function as it is instead of using @(u) s

Answer 63

how am i going to use newtons method on this?

Answer 64

so it only accepts 1 input of u at a time

Answer 65

and u has to be defined earlier

Answer 66

u are seeing what the value at each u is and adding it to a matrix thats all

Answer 67

just look over the function, it the graph looks crazy..

Answer 68

let me understand this. i'm lost

Answer 69

i'm just confused. how i'm going to derive it and apply newtions method on this

Answer 70

okay

Answer 71

i'm just lost now. is it possible you can just fix my code and see where i'm going wrong? the function handle is just printing out the function and not working.

Answer 72

there are some annoying matrix dimension problems that are coming up because u have functions embedded in functions

Answer 73

im not sure how to make it evalualte one variable at a time for each function

Answer 74

i greatly appreciate your help.

Answer 75

clear all;
clc;
den_b = 8550;
den_f = 1100;
visc_f = 0.1;
D_b = 0.0064;
g = 9.81;
u = 0.1;
step = 0.01;
n = 100;
tol = 10^-4;
Re = @(u)(D_b*abs(u)*den_f)/visc_f
bouyancy = (den_f-den_b)*(pi/6).*D_b.^3.*g
Drag = @(u)((2.25./Re(u).^0.31)+(0.358.*Re(u).^0.06)).^3.45
CirArea =(pi/4).*D_b^2
f =@(u) bouyancy-(Drag(u).*CirArea*den_f.*u.*abs(u)./2)

%u=-10:0.01:10
%Re=D*abs(u.)*p_f/u_f
%y=(p_f-p_p)*pi/6*D^3*g(-2.25/(Re)^0.31+0.358*Re^0.06)^3.45*pi/4*D^2*p_f*u.*abs(u.)/2

%for ii = 1:n
%u(ii+1) = u(ii)-f(u(ii))/finitediff(u(ii))
%end
uarray = -100:0.1:100;
yarray = f(uarray);
plot(uarray,yarray)

Answer 76

here u go its fixed

Answer 77

there is a zero at -0.something

Answer 78

http://prntscr.com/5qd21t

Answer 79

functions embedded in functions, thats the thn. i was told if i use function handle for each and ttry to put then together like i have it'll work. and i got it to work before. but matlab crashed, so i lost everything.

Answer 80

now to do newtons method

Answer 81

its not wotking.