y=@(T) I*T*sind(A)-(0.5*G*(T^2));
root = fzero(y, 0)
abouve eqation is for a projectile that projected at an angle (gose in a kind of ar half oval shape)
want to find the time it hist ground but fzero only gives me the biginning which is 0

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Walter Roberson
Walter Roberson 2011년 2월 24일
Maybe it just doesn't like buxZED... won't let me submit an answer to this one.
Walter Roberson
Walter Roberson 2011년 2월 24일
root = fzero(y,[eps, realmax])
buxZED
buxZED 2011년 2월 24일
any clue why?
can you post as a comment?
Paulo Silva
Paulo Silva 2011년 2월 24일
Hey Walter, answers stop working for me too :(
We're sorry, but something went wrong.
We've been notified about this issue and we'll take a look at it shortly.
buxZED
buxZED 2011년 2월 24일
??? Error using ==> fzero at 260
Function values at interval endpoints must be finite and real.
Error in ==> p34 at 26
root = fzero(y,[eps, realmax])
>>
buxZED
buxZED 2011년 2월 24일
root = fzero(y,eps, realmax)
gives me a - answer, imposible
Matt Fig
Matt Fig 2011년 2월 24일
Plot the function, look at the plot and pick a closer guess than zero.
buxZED
buxZED 2011년 2월 24일
@matt Fig
that was my inicial thought
but we are not alowed to use neumerical methods or guess work
Matt Fig
Matt Fig 2011년 2월 24일
You do realize that FZERO is a numerical root finder, right?
buxZED
buxZED 2011년 2월 24일
is there a way to force fzero to go to the next positive root?
Walter Roberson
Walter Roberson 2011년 2월 24일
Ummm, make it fzero(y,[eps,sqrt(realmax)/G]
Using fzero(y,eps,realmax) would cause realmax to be interpreted as an Option, and then it would use eps as the starting guess, not constrained from going backwards.
Walter Roberson
Walter Roberson 2011년 2월 24일
No, there is no way to force fzero to go to the next positive root. You can only exclude the area you already searched from the area fzero is to search next.
Matt Fig
Matt Fig 2011년 2월 24일
rt = 0;
d = .1;
cnt = 1;
while ~rt
rt = fzero(y,rt+d*cnt);
cnt = cnt + 1;
end
buxZED
buxZED 2011년 2월 24일
it works
but i want to understand what while ~rt means (what dose ~ say)
dose this stop counting as it finds the first root after zero?
Walter Roberson
Walter Roberson 2011년 2월 24일
~ is logical negation. It is true if and only if the thing being tested is exactly 0.
buxZED
buxZED 2011년 2월 24일
so dose this stops finding roots after it reached the 1st root after 0?
Matt Fig
Matt Fig 2011년 2월 24일
When rt is not zero, then ~rt is false. That would stop the WHILE loop.
buxZED
buxZED 2011년 2월 24일
y=@(T) I*T*sind(A)-(0.5*G*(T^2));
%root = fzero(y, eps, realmax)
rt = 0;
d = .1;
cnt = 1;
while ~rt
rt = fzero(y,rt+d*cnt);
cnt = cnt + 1;
end
fprintf('the root is %g\n' , rt)
gives me the root is 7.11215e-026
but its not posible as the projectile takes around 1.5sec to reach maximum hight, the root should be around 3
can you point me in the right derection?
buxZED
buxZED 2011년 2월 24일
is there a way to systematicaly increase the T value till Y become zero and get that value?
Matt Fig
Matt Fig 2011년 2월 24일
Change from:
while ~rt
to:
while rt<.1
or similar. It might help if you actually gave us some values for I, A and G.
buxZED
buxZED 2011년 2월 24일
it worked Matt
thank you so much
and also ~ would allow me to use matlab in many new ways
thank you again :D
Matt Tearle
Matt Tearle 2011년 2월 25일
Can I suggest using an initial guess of 2*I*sind(A)/G? Just sayin'...
Matt Fig
Matt Fig 2011년 2월 25일
Come on now, Matt! Where is the fun of solving the equation with your brain instead of making MATLAB do it?

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Honglei Chen
Honglei Chen 2012년 9월 17일

0 개 추천

The anwser is in the comments above
From Matt Fig
rt = 0;
d = .1;
cnt = 1;
while rt<.1
rt = fzero(y,rt+d*cnt);
cnt = cnt + 1;
end

댓글 수: 2

Matt Fig
Matt Fig 2012년 9월 17일
Digging deep, Honglei! How did you come across this one from over a year ago?
Honglei Chen
Honglei Chen 2012년 9월 17일
I'm helping cleaning up some old, unanswered questions :)

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