My friend is an artist – an awesome one! Above all, he loves comics with his whole heart; he even owned a comic book shop once, and I believe he's going to re-open it again one day.

So, while being an excellent artist, he doesn't like maths, like many of my friends anyway... He can draw amazing pictures and create lots of unbelievable art. So why care about maths when you can make a living by doing art, right?

Well, you'll be surprised...

When a mathematician and an artist met:

Artist: "Maths sucks!"

Mathematician: "Comics? Isn't that just for kids?"

"Maths sucks!"

"Well, those drawings are actually pretty cool..."

"I know, right? Maths sucks!"

"Draw me something, please – I really like it..."

"Yeah, no problem. Maths sucks!"

"Hey, it's really awesome, there's something to it – let's go see the new Marvel movie."

"Yeah, let's go... Maths sucks!"

"That's an amazing statue you're creating. Do you need help with that proportion calculation?"

"Err, if you don't mind... (Maths sucks...)"

"Here are your calculations. It was my pleasure to help you. Ask me anytime."

"Thank you. I might need some more if you don't mind? (But maths still sucks...)"

"Anytime... =)"

A true story. This is a sketch of me by the said artist when we first met (2015).

So what was it I've done so far for my artist friend?

A: Plane Geometry

I don't know what it was for, but there was a circular base with a hole in the middle, made of five identical parts. Given a radius, I was asked to find the length of both the inner and outer arcs of those parts.

Well, that's pretty simple. If you divide a circle into five identical parts, every part will be a section worth 72° (as a whole circle is made of 360°). But in this case, we actually already have the fraction on its own, so we didn't need to use the formula where "a desired angle/360°" appears. We simply used 1/5 as that's the part of the circumference we're looking for.

The circumference of a circle is calculated by the formula c = 2πr (where π could be 3.14 if you're okay with a rounded outcome, and r is the radius). If you have just one-fifth of it, multiply the whole formula by 1/5.

As we had two concentric circles (that means both have the same centre and different radii), I got the radius of the outer one and knew how wide the base should be. By finding the difference, I got the radius of the inner circle and could use the same method to get the answer for the length of the inner arc.

The best part is that the whole calculation depends on the radius only. So if you decide the base needs to be a different size, just change the number for r, but still use c = (1/5) * 2πr to get a fifth of the circle's circumference.

B: 3D Geometry

This one was actually pretty tricky at first sight. Imagine you're printing a sphere on a 3D printer, and you run out of material. So you get only a part of it, but instead of printing a new one, you decide to print just the missing bit. But how do you get the dimensions you need? The printer settings required inputting the height of the missing bit. Imagine it as if you have a part of a sphere and lay it on the table; how high would the top of that partial sphere be?

For this bit, you need a lot of imagination, or you can look at the sketch I made to visualize what I'm looking for:

So, we see that Pythagoras's theorem was used to find a desired unknown. We knew the radius of the original sphere (r) and measured the radius of a flat surface (r₂) where the sphere had been cut. By getting a quadratic equation, we needed to use the quadratic formula to find the roots. When using r = 4.5 and r₂ = 2.6, we got two outcomes: h₁ = 8.17 (the height of the part already printed) and h₂ = 0.83 (the smaller dimension — the height of the missing part — the key number the printer needs to be able to print exactly the missing bit of the sphere). Just by checking, you'll see that h₁ and h₂ together make the diameter of the original sphere, which is also a great way to check that the working is correct.

C: Ratios

This time, my friend actually made his own calculations but wanted me to check if they were correct. It wouldn't be worth it to find out that the statue was disproportional and all the enormous hours of work were for nothing!

So, if you have a character and her weapon printed on a picture, you can measure the height of both of them. Let's say the character was 11 cm and the weapon (some kind of spear or what...) was 7.5 cm.

You want to create the character at a height of 70 cm. But how about the dimensions of the weapon? Ratio, ladies and gentlemen, is the key! There are different ways to do it. My friend used one of them (correctly), and I checked it this way:

The proportions of the weapon and the character must be the same on paper and in reality. The proportion (labelled "k") from the measurement on the paper was: weapon/character = 7.5/11 = 0.681818... For our purpose, rounding to 4 decimal places would be good enough, so k = 0.6818.

The proportion of a real statue and its weapon is then w/c = w/70 = 0.6818. After expressing w from this equation (by multiplying the equation by 70), we'll get w = 0.6818 * 70 = 47.726 cm as the real length of the weapon for a character of 70 cm height. Tadaa!

So now tell me — do you need maths when you're an artist? Certainly, yes! (But yeah, yeah, maths sucks, I know...)