 fig 1) Measure inside diameter of ball fragment to get PX.

Sometimes it’s hard to estimate the diameter of a cannonball with only a fragment, so here’s a quick way to calculate it using two rulers and a bit of geometry in just a few seconds.  After a while you’ll be able to just look at the fragment.  This does require a fragment similar to what you see pictured… with a “span” you can put a ruler across.  I deliberately didn’t clean this one so you could see that it works either way. fig 2) Using a stiff straight edge touching the inside on each end measure as shown to get UV.

First, measure the inside diameter  (fig 1) with your ruler resting on the high points of the ball.  Let’s call this ℓ, represented by UV in the diagram.   Then, use a second ruler to measure the height to those high points (fig 2.)  Let’s call this h, represented by PX in the diagram (fig 3.)

We are applying the Intersecting Chords Theorem here.

PX x XQ = XU x XV

Given our ℓ & h values and this theorem, we can assert:

1/2 ℓ × 1/ℓ = h×XQ

Referring to the diagram to the right, the cannonball diameter PQ is PX+XQ (all the way across the inside of the ball)

X= h + (ℓ²/ 4h= (4h²ℓ²) / 4h
So CQ (the radius) is 1/2 of this…
CQ = (4h²+ℓ²) / 4h fig 3 – The Cannonball

So, when I plug in my measurements on the ball on the right:

ℓ = 3.25″  (fig 1)
h= 1.00″  (fig 2)
CQ = (4 (1 in²) + 3.25 in) / 4 (1 in)
CQ = (4 × 1) + 3.25 / 4 x 1
CQ = 7.25 in / 4

CQ = 1.8125″

Multiply this by 2 to get the diameter and you get 3.625” … a 6lb ball*.

(*I used this reference chart.)