kcoletisoppO

# Math is Fun! E30 Edition

No, not that kind of E30. Though that one is likely more fun than this conversation. I’m talking about an E30 gasoline, or put another way - a 30% mixture of ethanol.

E30 can provide a nice boost in power, particularly for turbocharged cars. Some cars pick up over 50hp (with a tune) by switching to E30. This comes at a slight sacrifice in fuel economy, but you take the good with the bad.

Some gas stations carry E30 gasoline at the pump! Something Georgia, a state that is unpleasant in many ways, does not seem to have at any of its gas stations. If you live in an area without E30, but you DO have E85, there is an easy solution! One could mix E85 into a tank that is partially filled with “standard” gasoline.

Now, most people who take the gas mixture approach keep it relatively simple and fill up with about 25% E85 and 75% 93 octane. I don’t like that approach - in part because it’s not true E30 - but also because I like to be precise!

Instead, lets do some math to create a formula to find how many gallons of E85 you need in a full tank of gas to get to an E30 mixture. In our fun math problem, we are not going to assume that 93 contains maximum ethanol % at most stations (10%) or that E85 is truly 85% ethanol, because sadly it rarely is. To find the true ethanol percent, you will have to use an ethanol content tester (most are junk unfortunately).

I did not check to see if anyone has published other formulas you can use. Mine is likely convoluted or a repeat of what someone else has done. Is anything original? It’s also honestly more work than it’s worth towards the end. But here’s what I scribbled out a while ago and what I often do.

MATH TIME!

We will assume:
A = Tank size (Gallons)
B = Gallons of “E85"
C = % ethanol in 93 octane
D= true % ethanol in E85

Gallons of E85 in a full tank of gas (solve for B):

B*D + (A-B)*(C) = A*.3
BD + AC-BC = .3A
BD-BC = .3A-AC
B*(D-C) = A*(.3-C)
B = (A*(.3-C))/(D-C)

Let us test it. If our tank is 20 gallons and we have true E85 and our 93 octane is 10% ethanol, then:

A = 20
B = ?
C = .85
D = .1

If we solve for B, we will find we need 5.33333 gallons of E85 in a full tank of gas (and 14.666 gallons of 93 octane, of course).

So now we have a full tank of E30. Congrats! We’re done, right? Not exactly... We’ve been driving around with E30 and now we need to fill up. But we can’t add 5.3333 gallons of E85 unless we have a COMPLETELY empty tank. Anything left in there and we will throw off our ratio. We also don’t know EXACTLY how much gas is left, so we can’t just do a percentage of the 5.3333 based on how much we need to fill up again, and we want to be precise, right?

So let’s find how many gallons of 93 octane we would add for each gallon of E85 we add to the tank instead!

We will assume:
A = gallons of 93 octane
B = % ethanol in 93 octane
C = true % ethanol in E85

Find: How many gallons of 93 per 1 gallon of E85 (solve for A)

1*C + A*B = (1+A)*(.3)
C + AB = .3+.3A
AB - .3A = .3 - C
A*(B - .3) = .3 - C
A = (.3 - C)/(B - .3)

So, if our 93 octane is 10% ethanol and our E85 is truly 85% ethanol, then:
A = ?
B = .1
C = .85

If we solve for A, we will find that we need 2.75 gallons of 93 octane per each gallon of E85. If we have roughly a 1/4 tank for a 20 gallon tank, we could add 3 gallons of E85 and 8.25 gallons of 93 octane. We won’t be full, but we will know we have maintained an E30 mixture.

The inverse is true also. To determine how many gallons of E85 per 1 gallon of 93 octane, we use:

A = (B - .3)/(.3 - C)

That about covers the formulas I use when trying to come up with a relatively precise mixture of E30.