P&G does not save fuel
#81
I just gave your reputation a bump.
Glad to see another user that contributes on here with good stuff to make us all better.
I will plan to test this if I can find a decent test area.
I will use my SG for the data. I have found the accuracy to be remarkable.
Glad to see another user that contributes on here with good stuff to make us all better.
I will plan to test this if I can find a decent test area.
I will use my SG for the data. I have found the accuracy to be remarkable.
#83
I remember a little display at a local science museum. It had 2 grapefruit size balls each with an declined rail to roll down. Both started at the same level, ended at the same level, and the only acceleration was gravity. The difference was the right side was a hilly route, while the left side was a steady decline to the "finish line". The hilly route was significantly faster every time even realizing that it is going a slightly longer path.
P&G is similar as the gliding part uses next to nothing for fuel while the Pulse section would use a little more than a steady speed at that time, but is ultimately offset by the very low fuel usage during gliding.
P&G is similar as the gliding part uses next to nothing for fuel while the Pulse section would use a little more than a steady speed at that time, but is ultimately offset by the very low fuel usage during gliding.
#85
I don't know how to verify this, but here's a good assumption...
If you calculate the drag at 40 mph compared to 60 mph and 50 mph, you'll most likely see a big difference. Just because you hit 60 mph, doesn't mean the aerodynamic drag will take full effect because when the car slows down and reduces drag.
If the drag is significantly lower at the minimum speed compared with the average speed, the drag at lower speed will prevail to make up for accelerating the max speed.
I can't conclude anything at the moment until I see a P&G test from 60 mph to 80 mph, but there's nowhere safe around here where I can test it. I believe that it does work, but you just have to find the correct speed ranges vs the average speed, and that might vary from car to car.
If you calculate the drag at 40 mph compared to 60 mph and 50 mph, you'll most likely see a big difference. Just because you hit 60 mph, doesn't mean the aerodynamic drag will take full effect because when the car slows down and reduces drag.
If the drag is significantly lower at the minimum speed compared with the average speed, the drag at lower speed will prevail to make up for accelerating the max speed.
I can't conclude anything at the moment until I see a P&G test from 60 mph to 80 mph, but there's nowhere safe around here where I can test it. I believe that it does work, but you just have to find the correct speed ranges vs the average speed, and that might vary from car to car.
#86
drag equation is D = Cd * A * .5 * r * V^2
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
#87
[quote=Slovenian6474;341990]drag equation is D = Cd * A * .5 * r * V^2
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
Thank you. Couldn't have said it better.
Somewhere I remember hp required increases by the cube of velocity change. So if you double velocity the hp required is 8 times as much. Thats hp, not just drag.
Here's the old experiment: stick your hand safely out the window at 30 and at 60 mph. If you think thats about the same you think a Hummer is about the same as a FIT.
cheers.
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
Thank you. Couldn't have said it better.
Somewhere I remember hp required increases by the cube of velocity change. So if you double velocity the hp required is 8 times as much. Thats hp, not just drag.
Here's the old experiment: stick your hand safely out the window at 30 and at 60 mph. If you think thats about the same you think a Hummer is about the same as a FIT.
cheers.
#88
[quote=mahout;342051]
I'm going to try this...it looks safe enough...
Also, I'm going to try to compare how many seconds it takes from going from 65mph to 55mph on Neutral with going 40 mph to 30 mph on Neutral whenever I have a safe chance to. It'll give me a pretty good idea of how much force is pushing back on the car at certain speeds.
drag equation is D = Cd * A * .5 * r * V^2
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
Thank you. Couldn't have said it better.
Somewhere I remember hp required increases by the cube of velocity change. So if you double velocity the hp required is 8 times as much. Thats hp, not just drag.
Here's the old experiment: stick your hand safely out the window at 30 and at 60 mph. If you think thats about the same you think a Hummer is about the same as a FIT.
cheers.
D=drag
Cd= coefficient of drag (usually between .29 and .32 for cars)
A= frontal cross-sectional area
r= density of fluid (air in this case)
V= velocity
You can see velocity is the only squared variable. So you would have more drag if you doubled your speed than if you doubled the front end area of your car.
drag from increasing velocity goes up fast.
Thank you. Couldn't have said it better.
Somewhere I remember hp required increases by the cube of velocity change. So if you double velocity the hp required is 8 times as much. Thats hp, not just drag.
Here's the old experiment: stick your hand safely out the window at 30 and at 60 mph. If you think thats about the same you think a Hummer is about the same as a FIT.
cheers.
Also, I'm going to try to compare how many seconds it takes from going from 65mph to 55mph on Neutral with going 40 mph to 30 mph on Neutral whenever I have a safe chance to. It'll give me a pretty good idea of how much force is pushing back on the car at certain speeds.
#89
I'm not buying it. Lets say you can coast twice as far as it takes you to accelerate back up to speed. If you are getting 10mpg from 40-65mph and 100mpg 65-40mph, and using the distance of 10 miles accellerating and 20 miles coasting. . .the gas you would use accellerating would be 1 gal. Coasting would use .2gal for a total of 1.2gal in 30 miles. Thats 25mpg. I can get more then that at a steady 70mph! I call BS!
#90
I'm not buying it. Lets say you can coast twice as far as it takes you to accelerate back up to speed. If you are getting 10mpg from 40-65mph and 100mpg 65-40mph, and using the distance of 10 miles accellerating and 20 miles coasting. . .the gas you would use accellerating would be 1 gal. Coasting would use .2gal for a total of 1.2gal in 30 miles. Thats 25mpg. I can get more then that at a steady 70mph! I call BS!
#91
From actual testing where the coasting is far more than twice the acceleration distance. You picked unrealistic numbers when you say your Fit will only coast twice the distance that it takes to accelerate.
When you pick unrealistic models you always get unrealistic answers.
Let's try 500 ft for acceleration at 15 mpg and 2500 ft at 0 mpg per cycle.
Then the average is .0063 gal for 3000 ft or 90 mpg.
Worse, while actual testing didn't do as good as our model (only 20% better) our test vehicle wasn't a Fit either, but an all out comp car. Probably about like Jr's Chevy at MI.
When you pick unrealistic models you always get unrealistic answers.
Let's try 500 ft for acceleration at 15 mpg and 2500 ft at 0 mpg per cycle.
Then the average is .0063 gal for 3000 ft or 90 mpg.
Worse, while actual testing didn't do as good as our model (only 20% better) our test vehicle wasn't a Fit either, but an all out comp car. Probably about like Jr's Chevy at MI.
#92
From actual testing where the coasting is far more than twice the acceleration distance. You picked unrealistic numbers when you say your Fit will only coast twice the distance that it takes to accelerate.
When you pick unrealistic models you always get unrealistic answers.
Let's try 500 ft for acceleration at 15 mpg and 2500 ft at 0 mpg per cycle.
Then the average is .0063 gal for 3000 ft or 90 mpg.
Worse, while actual testing didn't do as good as our model (only 20% better) our test vehicle wasn't a Fit either, but an all out comp car. Probably about like Jr's Chevy at MI.
When you pick unrealistic models you always get unrealistic answers.
Let's try 500 ft for acceleration at 15 mpg and 2500 ft at 0 mpg per cycle.
Then the average is .0063 gal for 3000 ft or 90 mpg.
Worse, while actual testing didn't do as good as our model (only 20% better) our test vehicle wasn't a Fit either, but an all out comp car. Probably about like Jr's Chevy at MI.
#93
I'm pretty sure "0mpg" is a typo. 0mpg is idling at a stop light, not accelerating or coasting. Maybe he meant 0 gallons per mile (or approaching infinite mpg) which it would do if left in gear due to the fuel injection shutoff.
#94
Here's some help from my test a few pages back.
60 to 40mph
38.5 sec
I calc that to be 0.539 miles
neutral no wind
40 to 60
12.5 sec
I cal that to be 0.174 miles
5th gear flat footed
ratio is 3.1:1 GLIDE/PULSE
in my 2008 Fit Sport
have fun
60 to 40mph
38.5 sec
I calc that to be 0.539 miles
neutral no wind
40 to 60
12.5 sec
I cal that to be 0.174 miles
5th gear flat footed
ratio is 3.1:1 GLIDE/PULSE
in my 2008 Fit Sport
have fun
#95
Coasting does not get 0mpg! If you shut your engine off every 30 seconds you're going to kill the car. Even so. . .it wont matter. The Fit will not coast .5miles from 65-40mph on a level road any more then it will go from 40-65 in .1miles getting 15mpg. To be fair I will do an accurate test tonight and give distances. We can argue about MPG later.
#96
When you coast be sure to disengage the clutch and you may leave the engine idling. When you get down to 40 re-engage clutch and throttle to 60, then driop the clutch again and time the coast-down from 60 to 40. Unless you have a sail it will take longer than 30 sec on level ground. Even my truck took that long.
#97
Yes, based on injector flow rate at 40 psig and 40 ms open cycle on 4 cylinders but I would n't argue with 13 either. Even if you choose 10 mpg the o/a mpg is still 60+mpg.
#98
[quote=pcs0snq;342939]Here's some help from my test a few pages back.
60 to 40mph
38.5 sec
I calc that to be 0.539 miles
neutral no wind
40 to 60
12.5 sec
I cal that to be 0.174 miles
5th gear flat footed
What acceleration time 40 to 60 did you get in 3rd gear? 2nd?
60 to 40mph
38.5 sec
I calc that to be 0.539 miles
neutral no wind
40 to 60
12.5 sec
I cal that to be 0.174 miles
5th gear flat footed
What acceleration time 40 to 60 did you get in 3rd gear? 2nd?
#99
Ah! That's what I was trying to get at. Pulling numbers out of thin air and making comparisons with them is pretty much useless.
I'm pretty sure "0mpg" is a typo. 0mpg is idling at a stop light, not accelerating or coasting. Maybe he meant 0 gallons per mile (or approaching infinite mpg) which it would do if left in gear due to the fuel injection shutoff.
I'm pretty sure "0mpg" is a typo. 0mpg is idling at a stop light, not accelerating or coasting. Maybe he meant 0 gallons per mile (or approaching infinite mpg) which it would do if left in gear due to the fuel injection shutoff.
#100
[quote=mahout;343262]
.539mi + .174mi = .713mi SO 13mi/ga for .174 = .0133ga & at 100mpg for .539mi = .0053ga
Totals .713mi travel / .0186ga used = 38.33MPG
Thats based on the 13mpg during FLOORING it. ..(I dont have a scan gauge but I doubt you will get THAT)
SO I'll at least say under the right conditions P&G is plausable, but if it DOES work, its by such a small amount that it costs you more in wear and tear on your car. But personally, having seen MPG read outs in rental cars. . .you dont get double digit MPG by flooring it.
Totals .713mi travel / .0186ga used = 38.33MPG
Thats based on the 13mpg during FLOORING it. ..(I dont have a scan gauge but I doubt you will get THAT)
SO I'll at least say under the right conditions P&G is plausable, but if it DOES work, its by such a small amount that it costs you more in wear and tear on your car. But personally, having seen MPG read outs in rental cars. . .you dont get double digit MPG by flooring it.