Need some help with my math, how much HHO needed?

[Dr. Jerryrigger]

okay, lost of posts on this topic, and I'm going to ignore most of them and reply mostly to the start.

dtvg,
First of all I'd like to show you the math, but with a different approach.

from your gas log I think it is fare to say that on a good day on the highway you can get about 57 MPG. so lets round up and make the assumption that your car uses 1 gallon of gas to go 60 miles at 60mph therefor running for 60 min on one gal of gas.
gas had about 125,000 BTU per gal. so 125,000/1= 125,000 BTU per hour

Okay now the HHO side of things.
And the amount of HHO we get is directly related to the electrical energy put in to it, so lets take a look at the necessary electrical system

125,000 btu * .293= 36,635W
Okay lets say you have a really good HHO set up with only about 10% loss. so you would need 1.1 X 36,635 =~40,000W
okay so at 14 V 40,000/14= 2857A
okay so a DF8D car battery (really big one) holds 200 Amp Hours (when drained slowly) and has a weight of 150lbs.
So if drained in an hour lets say it will give 180 Ah (generous)
2857A/180Ah = 204 batteries at 150 lb each
that's 30,600 lbs of batteries in a 1700 lb car, just to go for one hour.

or you could just go with an electric swap...

HHO is a waist of electricity, because ICE's are huge energy holes that give very little back.

That said small amounts of HHO can improve FE by helping gasoline combustion. Small amounts of propane can do the same thing.

If you do what to know how much HHO that is by volume find out what 36,635W is equal to in HHO and dived by 60 go get a min.