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Question about the ratio between spice and PG
- Thread starter Mr Parcan
- Start date
I use zero VG.
I noticed no improvement in taste with VG. It made only makes DMT and its salt less soluble.
No downsides for me. I found the reports of VG improving flavor with nicotine mixes not applying to DMT my experience.
I noticed no improvement in taste with VG. It made only makes DMT and its salt less soluble.
No downsides for me. I found the reports of VG improving flavor with nicotine mixes not applying to DMT my experience.
Mr Parcan
Rising Star
Thanks for your coments, The reason for using VG is for convenience. Easier to keep in a bottle with a decent mod and dedicated RDA or maybe a tank.
I wouldnt put nicotine with it, dont know about flavour either as not really needed. However experimenting makes life interesting.
I wouldnt put nicotine with it, dont know about flavour either as not really needed. However experimenting makes life interesting.
To clarify, I mentioned nicotine since we followed in the footsteps of that technology.
In the nicotine space, VG is recommended to improve vapor smoothness, flavor, feel, cloud production, and to avoid dryness when vaping heavily (as pure PG can be slightly desiccant).
But after testing, it turns out that none of these VG benefits apply to DMT juices as far as I know. For example, the amount of vapor needed is nowhere close to feeling a desiccant effect from pure PG. We are not chasing big clouds either.
Other nicotine technologies do apply, such as salting part or all of the freebase to make the pH closer to (or in) the physiological range. This makes the vaping experience more pleasant in both cases. Arguably, salting has is a bigger positive effect with DMT because the concentration of alkaloid is much higher.
In the nicotine space, VG is recommended to improve vapor smoothness, flavor, feel, cloud production, and to avoid dryness when vaping heavily (as pure PG can be slightly desiccant).
But after testing, it turns out that none of these VG benefits apply to DMT juices as far as I know. For example, the amount of vapor needed is nowhere close to feeling a desiccant effect from pure PG. We are not chasing big clouds either.
Other nicotine technologies do apply, such as salting part or all of the freebase to make the pH closer to (or in) the physiological range. This makes the vaping experience more pleasant in both cases. Arguably, salting has is a bigger positive effect with DMT because the concentration of alkaloid is much higher.
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U r welcome, but I don't have YouTube videos
Mr Parcan
Rising Star
must be someone with a similar name. not doing very well on my 1st day
All good, no worriesmust be someone with a similar name. not doing very well on my 1st day
Timmyt
Timmyt
I've been using vapes for a while. But what's the watts u use?
pantostao
Established member
The wattage needed depends on the resistance of the atomizer.
If you're using a cart, those are around 1 to 1.2 ohms. You should use a low wattage (3-10W).
If you're using a "modern" tank, using a sub-ohm resistance (around 0.3 ohms usually), you should use higher wattage. Maybe 20 to 70W.
The number itself is not important, the interplay between all the components is.
fractals4life
Titanium Teammate
It's the voltage needed that depends on the coil resistance ... wattage mode should be automatically varying the voltage to produce however many watts are needed in the coil. Subject to the maximum voltage that can be provided by the battery unit.
Lower resistance needs lower volts to drive the same wattage, so sub ohm coils can do higher wattages given the same maximum voltage available ... provided the current needed can be delivered by the battery circuitry ... it is indeed all a big interplay, with any attempt to reduce settings to just one of the numbers doomed to failure unless all other factors are taken into account.
Ultimately, Watts are what heat the juice, and the wattage needed depends on how fast juice can be supplied to get boiled, too many and you get dry wick and burnt DMT, not enough and you don't get enough vapour to be effective.
pantostao
Established member
This is right, I'm definitely not an electrician and got confused by the units. Thanks
It's really quite simple:
P=V×I V=I×R I=√(P/R)
V=P/I I=V/R P=V²/R P=I²R
I=P/V R=V/I V=√(P×R)
where 'P' is power (in Watts, W), 'V' is voltage (in, you've guessed it, Volts, V), 'I' is current (in Ampères, A) and 'R' is resistance (in Ohms, Ω).
A lovely collection of dead white men to play with
There's also the relative resistance of your coil material in Ωm⁻¹ to be calculating the resistance with, as well as its temperature coefficient in Ωm⁻¹K⁻¹ if you really want the fun to start, but I'm not an electrician either.
P=V×I V=I×R I=√(P/R)
V=P/I I=V/R P=V²/R P=I²R
I=P/V R=V/I V=√(P×R)
where 'P' is power (in Watts, W), 'V' is voltage (in, you've guessed it, Volts, V), 'I' is current (in Ampères, A) and 'R' is resistance (in Ohms, Ω).
A lovely collection of dead white men to play with
There's also the relative resistance of your coil material in Ωm⁻¹ to be calculating the resistance with, as well as its temperature coefficient in Ωm⁻¹K⁻¹ if you really want the fun to start, but I'm not an electrician either.