Lets do a hypothetical here...

Say we have two identical fieros. Both 88GT's, one white, one black. Same wheels, tires, brakes, suspension, engine, transmission and aerodynamics. The only difference between the two is the total weight. Lets say the white one weighs 2000lbs and the black one weighs 4000lbs.

At a steady and again "identical" speed on the HWY which one gets better gas mileage? Does the black one with its added weight and therefor added momentum win? Or does the lighter white one win? Post your guess and reasoning behind it...

[This message has been edited by Fieroseverywhere (edited 06-25-2009).]

the light one wins everytime, except perhaps - maybe - on a downhill

that weight creates rolling resistance, and the faster you are going, the more this will show up.

edit: and, are you counting the accell up to HWY speed?
hmmm....and now...hmmm...does the momentum help break wind resistance...grrrrr

[This message has been edited by Pyrthian (edited 06-25-2009).]

Not counting acceleration. We all should know that the ligher car will always win in that department.

NO up or down hill. Everything being equal except the actual weight of the car. Everytime I think I come up with an answer I figure on something else that throws is all off. Does the added weight of the black one make the engine work harder because it has to move more weight? Or does it work less because the momentum requires less power to keep it rolling?

[This message has been edited by Fieroseverywhere (edited 06-25-2009).]

weight has little effect at steady highway speed compared to wind resistance, therefore both will perform about the same, with the slight advantage going to the lighter car due to rolling resistance, as already noted.

Interesting topic.I am gonna at the start of vehicals forward movement. with every thing the same from start. to me the heavier will lose ,cause more gas to lug more wt. to equal speed. what ever top speed maybe be. the bodys are the same so air resistance may be minimal. plus holding more wt. at speed seem harder over all. more gas. should be the 'Ole horsepower to weight song and dance. monentum only matters at the end stopping. And when gravity ( up/down) is involved as mentioned earlier, right?? never thought about it from this angle. i.e. straight line. wait heres a "i see what you mean moment" are there turns involved? momentum may play factor in this secnerio. what do you all think. Your turn going on the net for some facts. back later


HITTOKIRI BATOUSAI daniel

Ok. What if we throw another curveball into the mix. Say we adjust tire pressures to eliminate rolling resistance as a factor. Now which one gets better MPG?

The light one will win hands down. It is true that the heaver one will have a higher momentum, but it will take more energy to keep it up. I see this when I have an empty minivan and a fully loaded with family, and luggage.

The lighter car wins at just about everything except a head-on crash with the heavier car.

The black car will get less gas mileage because the sun will heat up the body panels & make them expand making the car bigger & have more wind resistence.
~ Paul
aka "Tha Driver"

Drunk chicks dig me!

Lets look at the factors.

Drag:
Assuming they they both have the same ride height, they have the same coefficent of drag by their aerodynamics. This is because their shape is exactly the same.

Friction:
All frictions would be identical, except for at the wheel bearings. There will be a bit more friction at the wheel bearings on the heavier car. Friction is calculated by multiplying the normal force (weight) and the coefficient of friction of each bearing. However, I suspect that the frictional forces of the wheel bearing would be negligible.

Springs, hills, and compressibles:
Springs store energy when compressed and release that energy when released, so there is no energy loss in springs. We can treat hills like springs in that it takes as much energy to fight gravity going up in elevation as gravity returns going down the same elevation. We'll assume that rubber bushings are negligible.

The shocks create energy loss (dampening) in the springs to keep them from reciprocating. There would be stronger springs on the heavier car, so there would be more energy to dissipate. This is probably negligible as well. I don't think that shocks would cause any real loss in mpg.

Conclusion:
There should be no difference in mpg between the two vehicles. They both have identical factors. There are a few exceptions, but I would suspect them to be similar. However, if climbing a mountain, there will be a noticeable difference.

Are you saying that an empty big rig and a full big rig get the same gas millage? I do not believe this to be true.

quote Originally posted by Fieroseverywhere: Say we adjust tire pressures to eliminate rolling resistance as a factor. Now which one gets better MPG?

(To simplify things even further I'll also assume zero wind.) If we assume identical rolling resistance, the two cars will perform identically as long as the road is perfectly flat. But as soon as you introduce hills into the problem, however, the lighter car gains a slight advantage. The more hills and the steeper they are, the bigger the advantage that goes to the lighter vehicle. (Hint: The Second Law of Thermodynamics come into play when the road isn't perfectly flat.)

Assuming "all else is equal," wind, hills, and varying speed all favor the lighter vehicle. The extra energy required to accelerate, climb hills, or drive upwind will always be greater than the energy that can be "recovered" when decelerating, descending hills, or driving downwind.

[This message has been edited by Marvin McInnis (edited 06-25-2009).]

The hill problem does pose some good questions. We have the stored energy by gravity vs. the characteristics of loading engines.

This goes back to the idea of work. Work is force times distance. If it takes X Joules to go from A to B and we then introduce a hill between them, the equation says there is no difference. However, we know that there are other variables.

lets say the white car is going north and the wind is heading south at lets say 10mph...
and the black car is heading south with winds heading south at the same speed. does the black car get more mpg?
and what about air tempeture? hotter air VS colder air?

but i would assum the lighter car would have the upper hand. has less weight to around. engine has to work less. but i could be wrong.. blah! mind boggling!

Jeeze guys, this is begining to sound like the plane on a treadmill debate. YES the plane will takeoff - SIMPLE logic will tell you that (in spite of some pilots thinking otherwise: how smart do you have to be to be a pilot? - answer not very smart).
YES the lighter car will get better mileage because - even if you adjust the inflation of the tires to compensate for rolling resistence - you STILL have more friction to overcome in the bearings. Now, if you're going to start making multiple changes in the car to compensate for the weight, then yes both cars can get the same mileage ON A PERFECTLY FLAT ROAD.
~ Paul
aka "Tha Driver"

Remember 1/2 of the population is below average.

Forgetting small variations in friction and all that crap, at a constant speed, gaining no potential in terms of height, there is ONLY the force of wind resistance on the car. There is no term for vehicle weight in the finding of aero drag. Those forces will be the same whether the car weighs 2,700lbs or 27,000 (although that other crap that we forgot about earlier definitely starts coming into play).

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the heavier car wins.
this is with a flat road. no wind. rolling resistance from tires and bearings eliminated.
why?
simple. unless the speed was set exactly the same as in synchronized cruise control, any slight variation in speed will require more correction on the lighter car, where as the heavier car will maintain a more steady state. (conservation of energy, more momentum)

but the way this is going, it will be in a vacuum on a steel road with solid steel tires. at whic point both will get the same mpg, as once at speed neither will slow down.

in the real world, the lighter car wins

quote Originally posted by Tha Driver: Jeeze guys, this is begining to sound like the plane on a treadmill debate. YES the plane will takeoff - SIMPLE logic will tell you that (in spite of some pilots thinking otherwise: how smart do you have to be to be a pilot? - answer not very smart). YES the lighter car will get better mileage because - even if you adjust the inflation of the tires to compensate for rolling resistence - you STILL have more friction to overcome in the bearings. Now, if you're going to start making multiple changes in the car to compensate for the weight, then yes both cars can get the same mileage ON A PERFECTLY FLAT ROAD. ~ Paul aka "Tha Driver" Remember 1/2 of the population is below average.

Please forgive me for getting off topic, but a plane on a treadmill that matches the speed of the plane's wheels will NOT take off. No matter how fast the wheels are moving, if there's no wind to the wings (aka no forward fuselage movement), then there is no lift, and no lift = no takeoff.

EDIT: Looks like this is a whole can of worms that doesn't need to be opened... I apologize. *ducks for cover* I didn't realize this was such a big debate, but apparently it is...

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[This message has been edited by fieroboom (edited 06-26-2009).]

quote Originally posted by fieroboom: Please forgive me for getting off topic, but a plane on a treadmill that matches the speed of the plane's wheels will NOT take off. No matter how fast the wheels are moving, if there's no wind to the wings (aka no forward fuselage movement), then there is no lift, and no lift = no takeoff.

why would there not be wind to the wings, or forward movement of the fuselage?
the wheels do not power the plane - the propeller or jet engine does. the treadmill can go any speed it likes, and not affect the thrust of it engines. the treadmill can even suddenly STOP, and it will have ZERO affect on the plane. the wheels are free spinning.

Wow, and now that I think about it a second time, I was incorrect, because the engines provide thrust against the relative air around the plane, not the treadmill, so the treadmill is irrelevant... What made me think about it was a seaplane... If I had been correct before, then seaplanes attempting to take off against water current would be screwed, but they aren't. That's a good one; can't believe I missed that! Again, my apologies.
-Paul

If you throw a conveyor belt under them, then which one wins? (runs and hides )

Jim

I think given everything but wieght was equal the lighter of the two would win. As said, not by much.

Real world there are a bazillion variables that come into play though. Type of road (condition of road), raining? Speed? I think the heavier unit might actually beat out the lighter one in certain conditions, if the speed was just right along with the type of road it was on and such.

If everything is the exact same with the exception of weight, then the sole factor between them will be rolling resistance which happens at the bearings, brake pad drag, and tire contact patch (where the tire flattens out). The heavier car will have a slightly larger contact patch so more energy will be used to bend the round tire into the larger flat contact patch.


Once transients come into play (starting/stopping/climbing/turning) the benefit of the lighter car becomes much more pronounced...

[This message has been edited by fieroguru (edited 06-26-2009).]

Anyone with a very large building that is temp and humidity controlled that we can drive a car around inside?

At speed - now that I think about it, they should be equal, however, based on the total drive (stop/starting/etc) the lighter car would win as it's average fuel consumption should be less.

quote Originally posted by fieroguru: If everything is the exact same with the exception of weight, then the sole factor between them will be rolling resistance which happens at the bearings, brake pad drag, and tire contact patch (where the tire flattens out). The heavier car will have a slightly larger contact patch so more energy will be used to bend the round tire into the larger flat contact patch. Once transients come into play (starting/stopping/climbing/turning) the benefit of the lighter car becomes much more pronounced...

Bingo.

Interesting. I expected to have a definate answer by this point.

Linear momentum:
Mechanics. a quantity expressing the motion of a body or system, equal to the product of the mass of a body and its velocity, and for a system equal to the vector sum of the products of mass and velocity of each particle in the system.

Resistance:
the opposition offered by one thing, force, etc., to another. (in this case wind resistance)

This is figuring on adjusted tire pressures to eliminate rolling resistance from the equation (tire and bearing). Same contact patch with the road for both vehicles.

The wind resistance is the same between the two objects in this equation. The differernce being it is easier (takes less force) to push a lighter object. This would lead me to believe that the heavier object would have an advantage here by its increased momentum. Think of a freight train. It takes an enormous amount of energy to move it. It takes very little, relativly, to maintain a constant speed because the momentum keeps it moving. I just don't know either way.

What was never introduced into this equation is either Time or Distance. Would those also change the answer? What about the difference in weight? Say we're comparing 2000lbs to 20000lbs.

quote Originally posted by Fieroseverywhere: Interesting. I expected to have a definate answer by this point.

????

quote Originally posted by FastIndyFiero: Forgetting small variations in friction and all that crap, at a constant speed, gaining no potential in terms of height, there is ONLY the force of wind resistance on the car. There is no term for vehicle weight in the finding of aero drag. Those forces will be the same whether the car weighs 2,700lbs or 27,000 (although that other crap that we forgot about earlier definitely starts coming into play).

Physics has spoken...

quote Originally posted by Fieroseverywhere: Linear momentum: Mechanics. a quantity expressing the motion of a body or system, equal to the product of the mass of a body and its velocity, and for a system equal to the vector sum of the products of mass and velocity of each particle in the system. Resistance: the opposition offered by one thing, force, etc., to another. (in this case wind resistance) This is figuring on adjusted tire pressures to eliminate rolling resistance from the equation (tire and bearing). Same contact patch with the road for both vehicles. The wind resistance is the same between the two objects in this equation. The differernce being it is easier (takes less force) to push a lighter object. This would lead me to believe that the heavier object would have an advantage here by its increased momentum. Think of a freight train. It takes an enormous amount of energy to move it. It takes very little, relativly, to maintain a constant speed because the momentum keeps it moving. I just don't know either way. What was never introduced into this equation is either Time or Distance. Would those also change the answer? What about the difference in weight? Say we're comparing 2000lbs to 20000lbs.

Momentum does not have anything to do with the problem you described, because it is a static case, not a dynamic one. If you say that there is the same frictional loss acting on both cars (summarized into a single force), and they are moving at the same speed (with the aerodynamic drag summarized into a single force), the engine ONLY ONLY has to react these forces to maintain the vehicles speed.

Consider summing the forces in the direction the car is travelling in. Denote drag by D, friction by F_f, acceleration by a, vehicle mass by m, and engine force output by P. P is found to be the sum of D and F_f, except acting in the opposite direction. Acceleration a is 0.

D + F_f - P = Force = m*a (but a is DEFINED to be zero) => D + F_f - P = 0 => D + F_f = P

So there is NO net force acting on the car. One vehicle can weigh 200 pounds and the other 200,000. They can both be travelling at Mach 2. But as long as both have the same forces opposing them (friction, drag) it will take the same engine output to keep them at a constant speed.

[This message has been edited by FastIndyFiero (edited 06-26-2009).]

Most likely the black one will break down, and AAA will come and tow it to the destination therefor it will win because it used less gas.

In a very practical sense, identical cars will have identical tires, correct? If that's the case, the 4000lb car would have a larger contact patch, as well as more tire deflection. Rolling resistance of tires is proportionate to the weight on the tire, and while you can minimize that effect (think 120PSI in truck tires), the rubber itself would deflect more (being that it's the same rubber as on the 2000lb car), thus creating more wasted heat. This means that the lighter car would still get better MPG. Perhaps negligible, but better none the less.

As far as the shock absorbers go, even if the road was fairly flat, any time wheel and body motion needs to be converted through the shocks to heat, that's wasted energy as well. This is why race bicycles don't have suspensions in the first place. If a particular shock has to control twice the dynamic weight (of an undulating car), it will no doubt waste more energy.

It's not the WEIGHT of the vehicle that makes it less efficient, but the movement and control of the weight that converts energy to heat.
Hell, if you had no shocks, and steel wheels, it'd be scary as piss to drive but you'd definitely get better mileage..


Edit: That's another reason trains use steel wheels (besides just being cheap and strong) and no shocks, just springs.

[This message has been edited by aaronkoch (edited 06-26-2009).]

quote Originally posted by Fieroseverywhere: Interesting. I expected to have a definate answer by this point.

You do - in the very first response to your question. "that weight creates rolling resistance, and the faster you are going, the more this will show up."
In EQUAL cars the heavier will use more gas. It's not up for debate. If you CHANGE the parameters to compensate for the weight you can equal them out or get better mileage with a heavier car; but then they're no longer EQUAL cars.
~ Paul
aka "Tha Driver"

Maybe it's more than half the population that's below average?

FastIndyFiero finally spells it out. The original question was related to momentum and the given circumstances do not allow any change in momentum therefor no difference in mileage. (this is of course assuming that friction is not allowed to be considered. Otherwise the heavy car get less mileage.)

Now, Paul, aka "ThaDriver", I would normally take your comments about Pilots being dumb personal but truth be told. We ain't all the sharpest tools in the shed!

One of the reasons that people get all up in arms about the airplane and the treadmill controversy is that pilots look at it one way while others seem to see it another way. Fact of the matter is, no air over the airfoil equal no lift. (duhh) But, others see it as Pyrthian described it, if the airplane is allowed to accelerate (think of an endless treadmill) then he would be absolutely correct. You get enough relative wind and you fly away.

By the way, its a darn good thing pilots don't have to be smart...

Charlie

If the black car leaves Chicago at 12 noon & the while car leaves LA at 10 am, where will they be at 6pm?

Tire deflection will also be greater with heavier vehicle, thereby adding to the rolling resistance and decreasing mileage.
Actually, the difference in the tire performance is one of the main contributors to lower gas mileage.

let's correct the question here. HOW MUCH better will the lighter car's mileage be? And we can't rule out getting the car out of the gas station and up the on ramp to the interstate,unless we use a separate, one or two gallon fuel cell for that part of the trip. But we can find 300 miles of interstate with no hills whatsoever. We can make the rolling drag equal, and the wind drag. That's realistic. We can even have computer-controlled cruise control to halfway rule out the variance of the fool who's also holding the steering wheel. ALL else truly equal. Now, 1 mpg? 1%? HOW MUCH?

5 to 10% depending on the type of identical tires on both cars.

[This message has been edited by 30+mpg (edited 06-26-2009).]

quote Originally posted by cptsnoopy: FastIndyFiero finally spells it out. The original question was related to momentum and the given circumstances do not allow any change in momentum therefor no difference in mileage. (this is of course assuming that friction is not allowed to be considered. Otherwise the heavy car get less mileage.) Now, Paul, aka "ThaDriver", I would normally take your comments about Pilots being dumb personal but truth be told. We ain't all the sharpest tools in the shed! One of the reasons that people get all up in arms about the airplane and the treadmill controversy is that pilots look at it one way while others seem to see it another way. Fact of the matter is, no air over the airfoil equal no lift. (duhh) But, others see it as Pyrthian described it, if the airplane is allowed to accelerate (think of an endless treadmill) then he would be absolutely correct. You get enough relative wind and you fly away. By the way, its a darn good thing pilots don't have to be smart... Charlie

LOL! Just to be clear, I didn't say all pilots were not smart. I just said you didn't have to be smart to be one.
The fact is, Mythbusters proved it. A plane on a treadmill DID take off just like I knew it would. The wheels turning have nothing to do with forward thrust & therefore the plane pulls forward (yes even on a treadmill going the liftoff speed of the plane) & takes off.
~ Paul
aka "Tha Driver"

"The police said he commited suicide: he shot himself in the head - twice!" (real news story)

real quick about the treadmill. is this treadmill powered. or a old school like a hamster wheel. I know this one is a stupid ?? but its a questionable topic any way right?? hahaha. sorry its been a looong day. later..

As has been said, the plane takes off.

People are used to thinking about cars. A car with wings would certainly be stymied by the treadmill. After all, the car pushes on the ground.

The airplane, however, pushes on the AIR, and can thus accelerate even with the treadmill going the opposite direction. The airplane would accelerate to its takeoff speed and takeoff. Since the treadmill would be moving the opposite direction at that speed, the wheels would be turning twice the takeoff speed.

Here's one for the aerospace engineers: Could an airplane with a 90 knot approach speed hover onto the deck of a hydrofoil aircraft carrier that could cover the water at 90 knots?

[This message has been edited by Will (edited 06-26-2009).]

quote Originally posted by Will: Could an airplane with a 90 knot approach speed hover onto the deck of a hydrofoil aircraft carrier that could cover the water at 90 knots?

Absolutely! I've almost done the same thing when landing a lightly loaded airplane with a published stall speed of less than 60 knots into a 40+ knot wind aligned straight down the runway. It does require airspeed and pitch attitude discipline, though, and you use power to modulate the descent rate; diving for the runway is counterproductive. The feeling is very much one of an almost vertical descent ... crossing the runway threshold at 100 feet or higher and still touching down on the numbers. After touchdown the ground roll is very short, and you have to add a lot more power than usual to taxi upwind.

[This message has been edited by Marvin McInnis (edited 06-28-2009).]

quote Originally posted by Will: As has been said, the plane takes off. People are used to thinking about cars. A car with wings would certainly be stymied by the treadmill. After all, the car pushes on the ground. The airplane, however, pushes on the AIR, and can thus accelerate even with the treadmill going the opposite direction. The airplane would accelerate to its takeoff speed and takeoff. Since the treadmill would be moving the opposite direction at that speed, the wheels would be turning twice the takeoff speed. Here's one for the aerospace engineers: Could an airplane with a 90 knot approach speed hover onto the deck of a hydrofoil aircraft carrier that could cover the water at 90 knots?

As Marvin said, yes. The only minor issue to deal with is the airflow around the hover craft would create a little turbulence but it should not present any real problem.