A drive is not the only way to take a pump down a size. On a fixed-speed machine the works turns the impeller down on a lathe, and the same ladder of exponents applies to the diameter instead of the speed.

$\dfrac{Q_2}{Q_1} = \dfrac{D_2}{D_1}$ and $\dfrac{H_2}{H_1} = \left(\dfrac{D_2}{D_1}\right)^{2}$ — Q-two over Q-one equals D-two over D-one, and H-two over H-one equals D-two over D-one, squared. $D$ is the impeller diameter, in millimetres or any unit you like, because only the RATIO ever leaves the line. The subscripts keep the same meaning as before: 1 is the impeller as supplied, 2 is the impeller after the cut. Flow linear, head squared, and power once again the cube of the ratio.

Solving for the trim itself is where the square earns its keep. Wanting a particular head, $D_2 = D_1\sqrt{\dfrac{H_2}{H_1}}$ — the square root, because the rearrangement UNDOES the exponent. Forget the root and you will cut roughly twice as much metal as the duty needed, and unlike a drive setting, a trim cannot be undone at the keypad. It cannot be undone at all.

Two practical fences. Keep trims inside about 10–20 % of the full diameter; past that the vane exit geometry no longer matches the volute and the real pump falls below what these relations promise. And note the difference in kind between the two levers: speed is reversible and can follow the load hour by hour, while a trim is permanent, free to run, and exactly right for a pump that was simply bought too big.