DIY, HOW TO DOSE NICOTINE IN A BASE?
Many of you ask the question so here is a little news for DIY enthusiasts who have a nicotine base that is a little too concentrated and who are looking to know how to mix it with a base to have a lower rate.
So you have a nicotine base with a high rate X (for example X = 36 mg/ml) and you wish to obtain an e-liquid with a rate of nicotine Y (for example Y = 12 mg/ml) by diluting it in a nicotine-free PG-VG base .
We will simplify the calculations without taking into account the addition of flavorings or additives, this can be neglected because the volumes are very small, just a few drops.
So for those in a hurry, here’s the formula to apply, we’ll explain later, but it’s for those who want to save time!
So if you have a base with a rate X and a PG-VG base without nicotine, and you want to have a base with a nicotine rate Y, you must apply the following dosage:
For a base volume at rate X you must put: X/Y – 1 base volume PG-VG without nicotine.
For example if you have a 36 mg/ml base and you want a 12 mg/ml liquid:
X/Y – 1 = 36/12 – 1= 2, It will be necessary to put 2 times more base without nicotine than base with strong nicotine.
If you put 1 ml of base at 36 mg/ml, you will have to put 36/12 – 1 = 2 ml of base without nicotine
If you put 3 ml of base at 36 mg/ml, you will have to put (36/12 – 1) x 3 = 6 ml of base without nicotine
But you can also do it by mixing 2 bases with different nicotine levels so more broadly:
If you want to get a total volume V of liquid at rate Y by mixing a base volume V1 at rate X and a base volume V2 at rate Z
It is necessary to put:
V1 = V.(YZ)/(XZ) base volume at rate X
V2 = V.(XY)/(XZ) base volume at rate Y
To get V, base volume at rate Z
For example if you have a 36 mg/ml base (X) and a 6 mg/ml base (Z) and you want 10 ml of a 12 mg/ml base (Y):
We have: V1 = 10 x (12-6)/(36-6) = 2 ch
V2 = 10 x (36-12)/(36-6) = 8 ch
It is therefore necessary to put 2 ml of the base at 36 mg/ml and 8 ml of the base at 6 mg/ml
Now here is the explanation for the curious:
By definition, the nicotine level of an e-liquid is equal to the number of mg of nicotine present in the liquid divided by the volume of the liquid.
Rate = Amount of nicotine in mg / Volume of liquid in ml
To know the rate it is therefore necessary to know on the one hand the quantity of nicotine and on the other the volume of the liquid.
For the calculation we set:
V1 = Volume of nicotine base at rate X (in ml)
V2 = Base volume PG-VG without nicotine (in ml)
Y = desired nicotine level (in mg/ml)
V = Total volume of liquid (in ml)
Q = amount of nicotine in the liquid (in mg)
And we are looking for V2 (the volume of the base without nicotine) that we must put to have a liquid with a Y rate.
By definition we have: Y = Q/V
The volume of the liquid is equal to the sum of the volumes of the liquids that you have mixed, so we have for our mixture: V = V1 + V2
The amount of nicotine present in our liquid is equal to the volume of liquid containing nicotine multiplied by the nicotine level of this liquid: Q = V1.X
So we have :
Y = Q/V = V1.X/(V1+V2) where Y is the rate we want to get.
And so after some operations we get V2:
V2 = V1.X/Y -1
Where V2 is the volume of our neutral base and V1 the volume of our high nicotine base.
For V1 basic volume at rate X, it will be necessary to put V1.X/Y – 1 basic volume without nicotine.
For 1 volume of nicotine, it will be necessary to put X/Y – 1 basic volume
Now we can also do the same calculation assuming that you have PG-VG base also containing nicotine, at the Z rate for the calculation. You therefore have 2 nicotine bases and you wish to obtain a 3rd base with a different rate.
We have the same notation by adding the rate Z to the base V2 and we seek to have a base at the rate Y.
V1 = Base volume at rate X (in ml)
V2 = Base volume at Z-rate (in ml)
Y = desired nicotine level (in mg/ml)
V = Total volume of liquid (in ml)
Q = amount of nicotine in the liquid (in mg)
And we are looking for V2 (the volume of the base at the Z rate) that we must put to have a liquid with a Y rate.
As above: Y = Q/V
V = V1+V2 same volume
But Q changes a bit because there is already nicotine in the PG-VG base
Q = V1.X + V2.Z (we add to Q the quantity of nicotine present in the base V2)
So we have: Y = (V1.X + V2.Z) / (V1+V2)
This gives us: V2 = V1. (XY)/(YZ)
For V1 volume of nicotine at rate X, it will be necessary to add V1. (XY)/(YZ) base volume at rate Z to obtain a liquid at rate Y.
If you have a base at the rate of 36 mg/ml (X) and a base at the rate of 6 mg/ml (Z), to obtain a liquid at the rate of 12 mg/ml (Y) you will have to put (XY)/(YZ) = (36-12)/(12-6) = 4 times more base at 6 mg/ml than base at 12 mg/ml.
For 1 ml of base at 36 mg/ml it will be necessary to put 4 ml of base at 6 mg to obtain a liquid at 12 mg/ml.
Now that you know the proportions, you have to know how much volume to put on each base to have a desired defined volume and there it is very simple.
If you want to have a total volume V of liquid at rate Y by mixing a base volume V1 at rate X and a base volume V2 at rate Z, we have:
V = V1+V2 the total volume is equal to the sum of the volumes.
And as seen above V2 = V1. (XY)/(YZ)
System of 2 equations with 2 unknowns, we obtain:
V1 = V.(YZ)/(XZ)
V2 = V.(XY)/(XZ)
Ultimately:
If you want to have a total volume V of liquid at rate Y by mixing a base volume V1 at rate X and a base volume V2 at rate Z
It is necessary to put:
V1 = V.(YZ)/(XZ) base volume at rate X
V2 = V.(XY)/(XZ) base volume at rate Y
For example if you have a 36 mg/ml base (X) and a 6 mg/ml base (Z) and you want 10 ml of a 12 mg/ml base (Y):
We have: V1 = 10 x (12-6)/(36-6) = 2 ch
V2 = 10 x (36-12)/(36-6) = 8 ch
It is therefore necessary to put 2 ml of the base at 36 mg/ml and 8 ml of the base at 6 mg/ml
Here, if you understood the system you can now dose everything you want, it also works for the design of a Mojito.
Feel free to leave comments if you have any questions or if it’s unclear.
Good DIY to all