# Combining solutions with multiple concentrations - Mixing NPK fertilizer ratios

I'm trying to mix my own fertilizer to grow some peppers.

Fertilizers are labeled with NPK ratios, so mixing two volumes and finding the result is easy: Just multiply the concentration of each component by the ratio of the solution over the total volume. For example, $$\pu{1L}$$ of a $$3:4:6$$ solution plus $$\pu{1L}$$ of a $$5:2:6$$ solution yields $$\pu{2L}$$ of an $$4:3:6$$ solution.

Something like:

$$NPK_\mathrm{final} = \left(NPK_{s1} \times \frac{V_{s1}} {V_{s1} + V_{s2}} \right)+ \left(NPK_{s2} \times \frac{V_{s2}} {V_{s1} + V_{s2}}\right)$$

However, if I have a desired ratio in mind, how can I solve for how much of each of 2 solutions to add? I can plug in different values till I get close, but I'm curious if there's a general formula. I figured there's probably a way to get there with linear algebra (but I never took linear algebra!).

• Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice. You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources. Jan 31, 2021 at 18:03
• Yes- mixing 3 fertilizers is the only way to alter the ratios of all 3 components. For simplicity's sake (and because I didn't want the example equation to be too long) I figure you could do a two step process. Mix two fertilizers, then mix the third into the resulting solution. Jan 31, 2021 at 18:17
• Yes, I was going to add exactly a note like this to my answer . Jan 31, 2021 at 18:18

Let consider ratio of N/P.

Let fertilizer A has this ratio N/P=a.
Let fertilizer B has this ratio N/P=b.

Let you want the ratio c, where a < c < b.

Then you need to mix A : B in ratio $$\frac{ b - c}{c-a}$$

In the Central Europe with German scientific influence, it is called the "mixing cross rule".

Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice for the other two elements.

You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources.

A graphical chart method(*) may be more practical than a numerical method.

One can draw a equal-side triangular ternary chart. Each each corner of the triangular ternary plot represents formally a fertilizer with the solely N ( or P or K ) component.

If we take e.g. the N corner, than lines parallel to the side opposite to N corner represent N relative content to the sum N + P + K. The same for other 2.

Within this triangle, one can draw 3 points, each representing 3 different particular fertilizers.

Then one can draw similar chart between these 3 points for these fertilizers.

Available ratios than fall into this smaller triangle.

Then one can on the big chart choose N:P:K ratios and than read from the small triangle chart the ratios of all 3 fertilizers.

There is small catch in all this. The amounts of fertilizers to be mixed should be normalized for their total NPK content.

So mixing fertilizers A, B, C would not affect the total NPK amount per volume/mass, only the component ratio.

(*) It is also called a ternary plot on Wikipedia. See also the listed referenced there:

Or you can used PDF chart template to be printed, provided by @MaxW.

• @Maurice - The gist is that you plot ($\frac{N}{N+P+K}, \frac{P}{N+P+K}, \frac{K}{N+P+K}$) on ternary scale for each type of fertilizer. With two types you can draw a line between them. You can make any mix along the line. With three different types you get a triangle when joining the three with line segments. You can only make mixes which would be inside the triangle.