How to solve the loss of values while using the diff function ?

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Sethumadhavan Suresh
评论： Sean de Wolski 2017-5-18
Hey guys , So i was given the altitude and time data for a certain incident and i was required to find velocity and acceleration . This is what a part of my code looks like :
t=0:10:250;
tv=(t(1)+t(2))/2:10:t(end);
vel=diff(c)./diff(t);
acc=diff(vel)./diff(tv);
But what do i define my time for acceleration as? is it ta = (tv(1)+tv(2))/2:10:t(end); because that did not seems to work as it would have 25 elements while i only need 24 as the acceleration variable has only 24 elements . Please help ! thanks in advance!

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回答（2 个）

Stephen23
Stephen23 2017-5-18
编辑：Stephen23 2017-5-18
Use gradient instead.
KSSV
KSSV 2017-5-18
编辑：KSSV 2017-5-18
Why you are worried about size of time? You have to use time step (dt) which remains constant.
t=0:10:250;
dt = min(diff(t)) ;
% tv=(t(1)+t(2))/2:10:t(end);
vel=diff(c)./dt;
acc=diff(vel)./dt;

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