I wonder what was not right about the way I did it.

 

Hi there,


I like the way you approached this problem in looking at the areas.
I think however that you just need to modify that technique a little,
where instead of looking at the area under f(x) you need to look
at the area under (f(x))^2.
In other words, instead of finding the area of those little triangles
shown in the pic of the wave (which have three straight line sides)
you would need to first square them for all time and this would yield little
triangles with one curved side and two straight sides.

I'd like to see you do this again making that one little modification
and am betting your idea will work after that and you'll get the very
same answer. I really do hope you try this again i'd like to see the
result too.

 

I did do that. I squared the Y-axis values (or height) before multiplying them by 1/2*base. I just stopped saying the "square of the function" because I've said it like about 20 times in my previous posts and it got very confusing to follow the post with such a frequent wordy phrase.

 

Hi again,

Well, what you have done is made bigger triangles by squaring the height of the
triangle alone, when you really have to square the entire triangle over time t.
When you use a formula like 0.5*base*height, that means you are sill using
a triangle that has three *straight line* sides (ie a normal looking triangle)
when the triangles that really represent the area have only two sides that are
actually straight and the part of the triangle that is normally slanted (and straight)
becomes a curve instead of a straight line.
You'll have to figure out some way to calculate the area of that new pseudo
triangle (it isnt really a triangle anymore).

Here is an illustration showing the difference this really makes:

Attached Thumbnails

 

 

Ahhh, i see now. A sloped line that is squared...obviously becomes a curve! I see now. It does not scale linearily.

 

Hi again,


Yes :-) and i posted a pic to illustrate this but i think you replied while
i was drawing it.

You might still find a way to work with that new type of triangle
though.

Attached Thumbnails

 

 

Hi thank you for the solution, and also for the explanation here:
here is another approach of solving this problem.

The Delta signal between 0 & T/2 has already the RMS of i /sqrt (3). The constant value between T/2 and T has the RMS of i
Both value will be squared, with the time T/2 multiplied, and add
The sum must be at the root, therefore the result is: i * sqrt (2/3)

thanks again!