Confirm or deny?

[emm-dee]

Wanna check those numbers? 14000 pound MH / 2800 = 5. *35 = 175 mpg. ?????

[dustman_stx]

Originally Posted by 270S

I have a 2017 that has the V10. With the $500 5Star ODBC upgrade, there is NOTHING wrong with the Engine or Transmission at all.


Better mileage would be nice, but that is just physics and software can not help that. MPG = (weight / 2800) * 35. It is that simple! Use any of your gas vehicles, put in its weight and the result will be extremely close from my real world tests. I always heard different vehicles, designs, manufacturers, … are different, but reality is that for every gallon of fuel (regardless of type), it yields XX quantity of energy. The higher the mass, the more energy (fuel consumption) it takes to move it. Assuming gearing (Tx) is correct, a smaller engine at higher RPMs will consume the exact same amount of fuel as a larger one at lower RPMs - just physics.


Unless there is a breakthrough in designs (something like Fuel Injection was), all current engines are fairly equal in efficiency regardless of manufacturer. They all use the same design technology in hardware and software and most vehicles shapes are based on some form of wind tunnel input.

I'm not sure I agree completely with that. I think you've attempted to seriously oversimplify. For example, an F150 with the 6.2 V8 tends to get maybe 17-18 MPG hwy. You take the same truck with the 2.7L Ecoboost and it'll get 25ish MPG hwy. Something as simple as changing the tires to a more aggressive tread has in my experience had as much as a 10% effect on mileage (no size or LR change- just tread). I regularly see sports cars in Motor Trend that are light and extremely aerodynamic that get less than 20mpg. They'll have 700+HP under the hood, but they'd blow your formula up completely.

[Cowracer]

Originally Posted by Paulie1138

I do know that Ford is about to release a 7.3 gas engine
(oh the sacrilege!....LOL)

Interesting to see how that works out. Too bad they don’t re-release the 7.3 diesel!

After driving an Ecoboost 3.5L v6, I have held the firm opinion that if ford made a v-8 motor with all the ecoboost technology in the 5.5 to 6 liter range, they would probably never sell another diesel.

Shame they are going with the old low-speed, big cube, pushrod 2 valve engine for the new 7.3. Seems like a step backwards.

Tim

[dustman_stx]

Originally Posted by Cowracer

After driving an Ecoboost 3.5L v6, I have held the firm opinion that if ford made a v-8 motor with all the ecoboost technology in the 5.5 to 6 liter range, they would probably never sell another diesel.

Could it be we are looking at a 7.3 Ecoboost gasser? That would be a MONSTER rv engine.

Tim

The 7.3L is not forced induction, hence no "boost".

[Cowracer]

Originally Posted by dustman_stx

The 7.3L is not forced induction, hence no "boost".

Yeah. I read up on it and edited my post.

Tim

[dustman_stx]

Originally Posted by Cowracer

Yeah. I read up on it and edited my post.

Tim

I was hoping for a smaller boosted V8 for the unloaded economy and low end torque, but the 7.3L does sound promising.

[270S]

Originally Posted by emm-dee

Wanna check those numbers? 14000 pound MH / 2800 = 5. *35 = 175 mpg. ?????

Thanks - Typo, meant to type (2800 / Weight) *35
2800/14000 = .2 * 35 = 7mpg which is fairly typical for comparable Highway speeds. If going slower (55 group), the numbers will be apx 20-30% higher in both vehicles.

Again for those that always try to look for an argument ===> This was not presented as an absolute rule. It is a close approximation for NORMAL VEHICLES. If you have a blown dragster, this isn't applicable for extremes, too many other factors then enter the equation, which doesn't actually change the physics. F = M * A, unless you want to argue with Newton, Einstein, …. Without getting too technical, the A (acceleration) ends up translating into a constant value (Velocity) so the Force required is directly proportional to the mass. The force required is generated by the combustion of Fuel. So again, unless we are dealing with extremes that need additional equation variables to be entered, the MPG is closely directly proportional to weight (mass) alone.

[dustman_stx]

Originally Posted by 270S

Thanks - Typo, meant to type (2800 / Weight) *35
2800/14000 = .2 * 35 = 7mpg which is fairly typical for comparable Highway speeds. If going slower (55 group), the numbers will be apx 20-30% higher in both vehicles.

Again for those that always try to look for an argument ===> This was not presented as an absolute rule. It is a close approximation for NORMAL VEHICLES. If you have a blown dragster, this isn't applicable for extremes, too many other factors then enter the equation, which doesn't actually change the physics.

I would say it probably works out pretty well for naturally aspirated pickups being driven reasonably. But, since it doesn't take a rocket scientist to figure our you're referring to my post- the example I gave of the same truck having a 6.2V8 versus a 2.7L boosted V6 is light years from being a blown dragster. Yet their MPG numbers are quite different in reality whereas your formula would show them being almost identical.