Expert: Scotto - 11/5/2009

Question
I am trying to calculate points on a curve.  The length is 90.75" by 6" height.  I wanted the below to look more like an arch.  Here is a illustration:

<--90.75"--->
 __________
         .  
^       .   
6"   .
v  .
 .
.

Answer
This picture would look a lot better if added on in a JPG file.

From what I can see, we can approximate the curve with four cubics.
On goes from -a to -3.  One goes from -3 to 0.  One goes from 0 to 3.  One goes from 3 to a.

Let d be the height on the y axis where the curve goes from concave down to concave up.
It should be the same on both sides.  Let e be the slope of that tangent line at this point.

The four functions will be f1(x), f2(x), f3(x), and f4(x).
If this was a professional paper, I would make 1, 2, 3, and 4 into subscripts,
but ... , oh well.  Let h be the hight of the thing with width 6.

The values for f1 would be f1(-a)=0, f1'(-a)=0, f1(-3)=d, f1'(-3)=e.
The values for f2 would be f2(-3)=d, f2'(-3)=e, f2(0)=h, f2'(0)=0.
The values for f3 would be  f3(0)=h, f3'(0)=0, f2(3)=d, f3'(3)=-e.
The values for f4 would be f4(3)=d, f4'(3)=-e, f4(a)=0, f4'(-a)=0.

Given 4 conditions on a cubic, the cubic in each interval can be solved for.


The other thing to do is to approximate the curve with e^(a+(x/b)^2) where x is the height.
The value of a would be ln(h) where h was the height in the middle.
The value of b would depend upon how much the curve was spread.
One way of doing this is noting the 1st derivative is (2x/b²)•e^(a+(x/b)^2) and that x is the point of inflection, so in that way b could be found.

Is that what you needed?