bang guy
Moderator
Originally Posted by digitydash
http:///forum/post/2600450
Here is my reply I got from him.
Here's how we did it. We filled up our tank and let the pump shut off automatically. We then put it on the dyno with a starting milage on the odometer of 125,691. We ran it 20 minutes at 55mph and put 21 miles on. We then topped the tank off, and put 2.216 gallons in our tank. Divde 21 by 2.216 and you get 9.4.
We then repeated the test a month later. We filled up, and started the dyno at 126,566. We ran the machine for 20 minutes at 55mph. We put twenty miles on. We then filled up again, and put .862 gallons in our tank. Divide 20 by .862 and you get 23.2mpg.
Now take the difference between the amount of gas that we put into each tank. 2.216 minus .862 equals 1.3.
Take the 1.3 and divide it 2.216, which was the original amount of gas we put into the tank. That equals 61-percent.
If you think this math doesn't work, I'd more than happy to hear back from you.
"2.216 minus .862 equals 1.3.
Take the 1.3 and divide it 2.216,"
This isn't the correct formula. The correct way to determine % increase is (2.216/.862) * 100.
The test is flawed because they didn't measure how much fuel was used. They used the automatic stop on the fuel pump which is only accurate to a GALLON or so. If they repeated each of the tests 100 times they might have numbers with statistical significance.
http:///forum/post/2600450
Here is my reply I got from him.
Here's how we did it. We filled up our tank and let the pump shut off automatically. We then put it on the dyno with a starting milage on the odometer of 125,691. We ran it 20 minutes at 55mph and put 21 miles on. We then topped the tank off, and put 2.216 gallons in our tank. Divde 21 by 2.216 and you get 9.4.
We then repeated the test a month later. We filled up, and started the dyno at 126,566. We ran the machine for 20 minutes at 55mph. We put twenty miles on. We then filled up again, and put .862 gallons in our tank. Divide 20 by .862 and you get 23.2mpg.
Now take the difference between the amount of gas that we put into each tank. 2.216 minus .862 equals 1.3.
Take the 1.3 and divide it 2.216, which was the original amount of gas we put into the tank. That equals 61-percent.
If you think this math doesn't work, I'd more than happy to hear back from you.
"2.216 minus .862 equals 1.3.
Take the 1.3 and divide it 2.216,"
This isn't the correct formula. The correct way to determine % increase is (2.216/.862) * 100.
The test is flawed because they didn't measure how much fuel was used. They used the automatic stop on the fuel pump which is only accurate to a GALLON or so. If they repeated each of the tests 100 times they might have numbers with statistical significance.