Here is a simple method of determining progressive drawer heights.

For our purposes, we will use the following criteria:
Total drawer cavity height will be 10 inches and we want to fit 5 drawers of progressively different heights in this cavity. I have purposely kept our example numerical data simple so that we can practically do the math in our heads as we progress through each step.

STEP 1
Divide cavity height by the number of drawers desired:

10 inches / 5 drawers = 2 inches / drawer

I’ll state the obvious. Our smallest drawer MUST be shorter in height than 2 inches if we wish to have drawers that increase progressively from top to bottom. For our criteria, the smallest drawer cannot be larger than 2 inches because 5 x (anything larger than 2 inches) cannot fit into a height of 10 inches. Therefore, choose a first drawer height that is less than 2 inches in height to start with.

Let’s say, for example, that we will choose a first drawer height of 1.5 inches in height. We know that 5 drawers x 1.5 inches will be less than 10 inches.

5 drawers x 1.5 inches = 7.5 inches. And 10 inches (total cavity height) – 7.5 inches = 2.5 inches.

This 2.5 inches in height is where we will get the variances in height that we will add to all drawers (except the first drawer) to make them vary in height from top to bottom. I will call this 2.5 inches, the Variance Gap (VG).

I will also call the variance between Drawer 1 and Drawer 2, delta ( Æ ). The variance between Drawer 2 and 3 will be different. However, it will be a multiple of this Æ value as you will soon see.

Let’s think about this a minute. What do we know so far about our drawers?

Drawer 1. We know that our first drawer will be 1.5 inches in height. This is fixed. Therefore, we will not need to add a Æ to this drawer.
Drawer 2. We know that this drawer must be greater in height than Drawer 1. Therefore, it will need to have a Æ added to it to make it greater in height than Drawer 1.
Drawer 3. All we know for certain is that this drawer will have to be greater in height than Drawer 2. And so on.

STEP 2
Finding Æ. How many Æ’s do we need for our configuration? Looking at the chart below:

Drawer Drawer Heights
Drawer 1 = 1.5
Drawer 2 = Drawer 1 + Æ = 1.5 + Æ
Drawer 3 = Drawer 2 + Æ = (1.5 + Æ) + Æ = 1.5 + 2 Æ
Drawer 4 = Drawer 3 + Æ = (1.5 + 2 Æ) + Æ = 1.5 + 3 Æ
Drawer 5 = Drawer 4 + Æ = (1.5 + 3 Æ) + Æ = 1.5 + 4 Æ
Now, adding up the Æ’s, we get 10 Æ’s

From this we now know that:
Drawer 2’s difference in height from Drawer 1 = Æ
Drawer 3’s difference in height from Drawer 2 = 2Æ
And so on.

Note 1. The number of Æ’s required will always be one less than the drawer number. For instance Drawer 1 will require zero Æ’s, while Drawer 4 will require 3 Æ’s. This is assuming you sequence your drawer numbers from 1 through n.

So, what is this magical Æ number?

We find Æ by dividing our VG by the total number of Æ’s we need. Therefore, VG / 10 = 2.5 inches / 10 = 0.25 inches. Æ = 2.5 inches.

Looking at the chart again:

Drawer 1 = 1.50
Drawer 2 = 1.5 + Æ = 1.5 + 0.25 = 1.75
Drawer 3 = 1.5 + 2 Æ = 1.5 + (2 x 0.25) = 1.5 + 0.50 = 2.00
Drawer 4 = 1.5 + 3 Æ = 1.5 + ( 3 x 0.25 ) = 1.5 + 0.75 = 2.25
Drawer 5 = 1.5 + 4 Æ = 1.5 + ( 4 x 0.25 ) = 1.5 + 1.0 = 2.50
Adding up our drawer heights we get 10.00 inches.

Note 2. If we want to increase our variances between drawers, we must increase our VG. This means that, for this example, our first drawer will have to be smaller in height than 1.5 inches.

Note 3. If you have uniform gaps between drawers, these will have to be subtracted from the configured drawer heights as calculated above. This may require adjusting the first drawer height until a solution is arrived at. Those drawer projects where only the drawer face is visible will still require a slight gap, say 1/16 inch between faces to keep them from binding. I have found that these are easily addressed when fitting and need not be re-calculated for as in the case of visible horizontal drawer supports.

My apologies for not being able to line up the chart data. Posting removes all the extra blank spaces. There may be an elegant way to do this, but I am not aware of it.

Regards,
Phillip Anthony