Thread Subject: Duration of different arrays

Hi

I'm not too sure what I would write for the following and
would appreciate some help to perform this.

I have one command which works like this:

a=((1-((max(handles.runoff(handles.start_date:handles.end_date)))/(max(handles.rain(handles.start_date:handles.end_date)))))*100);

however I would like to find the duration between the two.

i.e. If the max runoff occurs at row 52, and the max rain
occurs at row 37, then I could have 52-37, which provides 15.

If each row (value held in array) is equal to one minute,
then in this case, there is a 15 minute delay between runoff
and rain.

Thanks for your help.

"Adam Bright" <abc_services_2000@yahoo.co.uk> wrote in message <g2i0e8
$s3p$1@fred.mathworks.com>...
> Hi
>
> I'm not too sure what I would write for the following and
> would appreciate some help to perform this.
>
> I have one command which works like this:
>
> a=((1-((max(handles.runoff(handles.start_date:handles.end_date)))/(max
(handles.rain(handles.start_date:handles.end_date)))))*100);
>
> however I would like to find the duration between the two.
>
> i.e. If the max runoff occurs at row 52, and the max rain
> occurs at row 37, then I could have 52-37, which provides 15.
>
> If each row (value held in array) is equal to one minute,
> then in this case, there is a 15 minute delay between runoff
> and rain.
>
> Thanks for your help.
-----------
  If you want that information, you should use the two-output form of 'max'

 [mx1,idx1] = max('runoff stuff');
 [mx2,idx2] = max('rain stuff');

for each of your two maximum computations separately. You can then
complete the computation of 'a' and the difference between these two indices,
using mx1 and mx2, and idx1 and idx2.

 Read the documentation for 'max'.

Roger Stafford