Hi guys ,
I need your help.
I wouls like to divide a circle in n parts (circular segments).
I would like to divide them in a parametric way but keeping the area of eache parts the same as the other.
I’m good enough in GH but this looks like an interative solution and I’m not into Galapagos.
I actually wasn’t thinking kangaroo here.
For the sphere case it works out that simply even vertical spacing of the cutting planes results in equal surface area of the slices, which I found quite surprising
Thank to your help we did it!!
This msg is to show you why we need this script
This is the air duct Francesca (my student) is doing for her thesis.
Thanks to grasshopper, and your help, we made a full parametric duct able to respect the basic aerodynamics principle ( we’re designer).
This may be a bit late to be useful. Or useless even if I’d posted earlier.
If you want calculate the area of a section of a circle, you can use the following formula. In theory you would take a circle of radius r, and set its centre at the origin. Then if you draw a line for y = k (-r < k < r) the area above this line equals the following as shown in the image (easier to do a screen shot than retyping).
So in theory you could take a circle, calculate its area, then divide by n. Then solve for one of these equations. Take 2 times the previous area for the next calculation etc.
Like I said, probably useless as other ways easier. But good for a bit of a maths exercise.
Thought I’d add one more thing. Thought of a big simplification for the above.
If you are wanting < 50% of the circle, the formula for the proportion of the circle is just ½ - 2 * k0 / pi where k0 is the proportion of the y=k line to the circle radius. And naturally for >50% it is 1 minus that formula (equalling ½ + 2 * k0 / pi).
I’ve not fully completed this definition (it could be made clearer and rearranged a bit) as it was getting a bit pedantic. If anyone does actually have any further use for this sort of problem, let me know and I’ll tidy a few things up.
Only wanting to demonstrate there in fact is a simple equation which can give the answer.