The first consideration is the size of the lumber. Brickmolding and other smaller dimensional lumber would require the 1″ or 2″ long rods, fence posts normally take  2″ or 4″ rods and large logs require the 4″ or 3″ rods.

The second consideration is the volume of wood to be treated. Each rod contains a certain amount of material and has a Boric Acid Equivalent (BAE). In order to effectively protect wood, a loading of at least 2.0 Kg/m3 (2oz/cu ft) is recommended. The simplest way is to first calculate the the volume of wood you wish to protect. Then using the chart below divide by the appropriate size’s treatable wood volume to find the total number of rods required. In cases where existing decay is present the load rate should be tripled.

Rod Size

BAE

Volume of Wood Treatable

1″ – 1/3×1″

3.80g / 0.134oz

0.00141 m3 / 0.0498 ft3

2″ – 1/2×2″

13.83g / 0.488oz

0.00631 m3 / 0.223 ft3

3″ – 3/4×3″

53.30g / 1.880oz

0.0214 m3 / 0.754 ft3

4″ – 1/2×4″

26.93g / 0.950oz

0.0127 m3 / 0.447 ft3

i.e. To treat 3′ of a 4″ x 12″ beam with 2″ rods (4x12x36=1 cu ft / .223 = 4.48) would require 5 rods minimum.