When modelling the real world, it's surprising how many things I take for granted. It's worthwhile considering how seemingly obvious "given" information is actually derived — the process often sheds light on much harder problems.
Slice a circle into a great many thin concentric rings. Unroll each ring into an almost-rectangle of width dr and length 2πr, then stack them side by side. As dr shrinks, the stack becomes a right triangle with base r and height 2πr — whose area, ½ × r × 2πr, is exactly πr². The real lesson: split one big problem into many tiny, easy ones.
Lately I was lost in a sea of inequalities and variables. Standing back and writing them out on a whiteboard, I realised I was looking at a problem that could be solved quickly with simple geometry rather than complex algebra — I couldn’t see the forest for the trees. Geometry is one of the most fascinating and underappreciated fields of mathematics, so here’s a simple, very geometric example with far-reaching consequences.
You know that the area of a circle is Area = πr². But why? Start by enclosing the circle in a square whose sides equal the diameter. Four corners are left over, so the circle’s area is less than the square’s:
Divide the circle into quarters instead, treat one as a triangle of base and height r, and you get a lower bound (we’re missing the four curved sections):
So the true area sits somewhere between 2r² and the bounding square. If only there were a way to divide the circle that we could sum exactly.
Cut the circle into thin concentric strips. Each ring, if you don’t look too closely, is almost a rectangle: its length is the circumference at that radius, 2πr, and its height is the tiny thickness we’ll call dr (a hint of where this is heading, for the calculus-minded). So each ring has area:
Put the rings on axes: the x-axis is radius, the y-axis is each ring’s length, 2πr. The unrolled rectangles all just touch the line y = 2πr, and the smaller dr becomes, the smaller the gaps. The area of the circle is the sum of all the rings — which is just the area of the right triangle with base r and height 2πr:
We took one large problem — the area of a circle — and split it into many smaller problems, each easy to solve. That’s the heart of calculus: the area under a graph is the integral of the function that defines it.
Many large problems reduce to calculating areas under graphs — exactly the move behind integration, and exactly the instinct we bring to messy real-world models. It’s the same decomposition that drives linear-programming approaches to capacity planning and MILP production scheduling: break the intractable whole into tractable parts.
That's how we approach complex energy challenges — from geometry to optimisation. If you've one worth solving, let's talk.