I Effect of moving cargo on vehicle fuel consumption

[WalterG]

I have to calculate how much fuel is wasted, when you do not fix a load in a vehicle.
So lets say there is a heavy box sitting in an almost empty truck. If this box moves how much fuel is lost? Or if you fix the box, how much fuel do you save?

 

[berkeman]

Welcome to PF.

Why do you think that a moving load would affect fuel economy? Are you taking the friction between the load bed and the load into account?

 

[jrmichler]

If the box is heavy enough, and shifts far enough, the question is moot because the truck will be uncontrollable. I once saw a stake body truck carrying a heavy electric motor that was not tied down. The motor shifted almost off the side when the driver took a corner too fast. The driver was barely able to get the truck back to the shop.

 

[WalterG]

Welcome to PF.

Why do you think that a moving load would affect fuel economy? Are you taking the friction between the load bed and the load into account?

Yes the friction creates thermal energy and that has to come from somewhere.

 

[berkeman]

Yes the friction creates thermal energy and that has to come from somewhere.

Agreed. Now we're getting somewhere!

That calculation should be pretty straightforward, no? It just involves how many cycles of the load sliding forward and backward are involved in a trip. Can you start that math for us so we can check it please?

 

[WalterG]

ok, thanks will need some time....

 

[WalterG]

Agreed. Now we're getting somewhere!

That calculation should be pretty straightforward, no? It just involves how many cycles of the load sliding forward and backward are involved in a trip. Can you start that math for us so we can check it please?

Let’s assume we have a Ute with the mass of 1600kg and a load of a 16 kg box with zero kinetic energy in our inertial reference frame. Now I bring this Ute from 0 to a velocity of 60km/h in 3 sec.

The box would move from behind the cabin to the end of the truck, let’s say 1m when the 60km/h are reached. I want to neglect the tire friction and air resistance.

As I don’t know the physics of the friction, because I don’t know the thermodynamic properties of the materials, I have to calculate where the energy comes from to move the box against the friction.

So the kinetic energy of the system after acceleration is E=1/2 m v2

My problem is that the kinetic energies before and after acceleration are the same. So where does the energy come from to warm up the box (and the Ute)?

 

[WalterG]

Let’s assume we have a Ute with the mass of 1600kg and a load of a 16 kg box with zero kinetic energy in our inertial reference frame. Now I bring this Ute from 0 to a velocity of 60km/h in 3 sec.

The box would move from behind the cabin to the end of the truck, let’s say 1m when the 60km/h are reached. I want to neglect the tire friction and air resistance.

As I don’t know the physics of the friction, because I don’t know the thermodynamic properties of the materials, I have to calculate where the energy comes from to move the box against the friction.

So the kinetic energy of the system after acceleration is E=1/2 m v2

My problem is that the kinetic energies before and after acceleration are the same. So where does the energy come from to warm up the box (and the Ute)?

Clarification: The kinetic energy of the system with moved box is the same as the kinetic energy of the system with a fixed box.

 

[jbriggs444]

Clarification: The kinetic energy of the system with moved box is the same as the kinetic energy of the system with a fixed box.

There is some math here that can be worked.

Are you using the same force to accelerate the truck in both cases? And achieving the same velocity for the COM?

Does this mean that the force is applied for the same or different times in the two cases?

One can calculate the acceleration of the truck (truck only, not truck plus crate) in the two cases. Is it the same?

You can then calculate the distance covered by the truck before the center of mass reaches the target velocity. Is that then the same in both cases?

Finally, you are in a position to consider the work done by the force. Is that the same in both cases?

I have to calculate where the energy comes from to move the box against the friction.

The energy does not move the box against friction. The energy moves the truck against the friction supplied by the box. The box just sits there and is acted upon by that same friction. And by the inelastic collision with the tail gate, of course.

 

[A.T.]

Yes the friction creates thermal energy and that has to come from somewhere.

If the cargo collides with the walls, it will dissipate energy too, via inelastic collision. And if you have cargo bouncing/sloshing around, the load on the engine/drive-train will be changing, making them less efficient.

 

[jrmichler]

I have to calculate how much fuel is wasted, when you do not fix a load in a vehicle.
So lets say there is a heavy box sitting in an almost empty truck. If this box moves how much fuel is lost? Or if you fix the box, how much fuel do you save?

In the physics world, where everything is 100% efficient and there is no friction unless specified, the energy to accelerate to a speed $v$ is $E = 0.5mv^2$. In the real world, it is necessary to add rolling friction, driveline losses, air drag, and engine efficiency. That's why it takes energy to maintain a constant speed on a level road.

The process of finding fuel wasted by sloshing cargo starts by finding the fuel used without sloshing cargo. An example of how to do this, and the calculations, is in this thread: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html. Now that you have an estimate of the total power to maintain speed, the next step is to calculate the energy to slosh the cargo. Assume, as a sample calculation, that you have a 35 lb box that slides 5 feet every second. Ignore any bouncing off the sides of the truck. Assume a friction coefficient, $\mu$, of 0.5. Then the sliding force is $35 lbs * 0.5 = 17.5 lbs$, and the energy to slide the box is $17.5 lbs * 5 ft / 1 second = 87.5 ft-lbs / second$. The power is $87.5 / 550 = 0.16 hp$. Compare that to the total power to maintain speed, and you will have the amount of extra fuel needed. Note that sloshing a box that far that fast requires some seriously bad driving, which will in itself waste fuel.

 

[jrmichler]

If the cargo collides with the walls, it will dissipate energy too, via inelastic collision. And if you have cargo bouncing/sloshing around

If you have many pieces of cargo, say a few dozen rocks, bouncing around, the calculations are different. This is particle impact damping (search the term). It will also add to the power to drive the truck.

Particle impact dampers are used to control vibration. The basic principle is that anything that dissipates energy from the vibration of a vibrating system will reduce that vibration. The photo below shows a particle impact damper on the light bar of my hot rod snowblower:

The light bar had a bad torsional vibration about the vertical axis when the engine was at idle speed. Adding the particle impact damper reduced the vibration by a factor of several times. The damper is the 35 mm film container half filled with lead shot, wrapped in duct tape, and tie wrapped to the light bar at the arrow. If you want to see this snowblower in operation, search youtube hot rod snowblower hydrostatic. It has a hydrostatic transmission, so can be quickly run back and forth.

 

[berkeman]

The photo below shows a particle impact damper on the light bar of my hot rod snowblower:

Wait, what? Your hot rod snowblower doesn't appear to have any aero mods yet... WTH?

 

[jrmichler]

Well, um, err, actually I partly disassembled it to build an improved machine. This machine is clumsy to use, and only moves about 60% more snow than my 32" 10 hp Ariens. The limiter is impeller capacity, not power. My improved version will have two impellers and eliminate the auger to make a machine that moves even more snow, and will be physically smaller than my Ariens. That's if I ever get off my lazy butt and actually work on it.

 

[berkeman]

Well, it's the middle of the hot summer, so I can see how the motivation to work on the snowblower would be low right now.

 

[WalterG]

There is some math here that can be worked.

Are you using the same force to accelerate the truck in both cases? And achieving the same velocity for the COM?

Does this mean that the force is applied for the same or different times in the two cases?

One can calculate the acceleration of the truck (truck only, not truck plus crate) in the two cases. Is it the same?

You can then calculate the distance covered by the truck before the center of mass reaches the target velocity. Is that then the same in both cases?

Finally, you are in a position to consider the work done by the force. Is that the same in both cases?The energy does not move the box against friction. The energy moves the truck against the friction supplied by the box. The box just sits there and is acted upon by that same friction. And by the inelastic collision with the tail gate, of course.

Yes in both cases same final velocity = 60km/h. Acceleration shall be the same. So it may take longer time to accelerate to the same speed with the moving box. Hence more energy needed. May be

There is some math here that can be worked.

Are you using the same force to accelerate the truck in both cases? And achieving the same velocity for the COM?

Does this mean that the force is applied for the same or different times in the two cases?

One can calculate the acceleration of the truck (truck only, not truck plus crate) in the two cases. Is it the same?

You can then calculate the distance covered by the truck before the center of mass reaches the target velocity. Is that then the same in both cases?

Finally, you are in a position to consider the work done by the force. Is that the same in both cases?The energy does not move the box against friction. The energy moves the truck against the friction supplied by the box. The box just sits there and is acted upon by that same friction. And by the inelastic collision with the tail gate, of course.

Yes in both cases same final velocity = 60km/h. Acceleration shall be the same. So it may take longer time to accelerate to the same speed with the moving box. Hence more energy needed. May be all equations where the mass is involved assume that all the mass remains in one point? So I have to calculate the K.E. for all masses involved! I'll try that! Thanks for the help.

 

[jbriggs444]

Yes in both cases same final velocity = 60km/h. Acceleration shall be the same. So it may take longer time to accelerate to the same speed with the moving box. Hence more energy needed. May be

That is not the logic that I expected you to follow. The reasoning that I expected was:

1. The total mass for each truck plus box is identical.
2. The final velocities of the two centers of mass are identical.
3. It follows that the change in momenta of the two trucks getting to the final COM velocity are identical.
4. The applied forces on the two trucks are identical.
5. It follows that it takes exactly the same amount of time to accelerate both centers of mass to the final speed.
6. The truck with the movable box accelerates faster than its carried box.
7. It follows that the truck with the movable box accelerates faster than the center of mass of that truck+box.
8. The truck with the fixed box accelerates at the same pace as the center of mass of that truck+box.
9. It follows that the truck with the movable box accelerates more rapidly than the truck with the fixed box.
10. It follows that, over the course of the acceleration to the final COM speed, the truck with the movable box traverses a greater distance.
11. It follows that, over the course of that acceleration, more work is done on the truck with the movable box.

That is where the extra energy required to give the box a velocity relative to its truck comes from.

Of course, this description is for the idealized world of a smooth road, no air resistance and a truck bed long enough that the box is still sliding when final COM speed is achieved. For a bumpy road it gets trickier.

 

[WalterG]

That is not the logic that I expected you to follow. The reasoning that I expected was:

1. The total mass for each truck plus box is identical.
2. The final velocities of the two centers of mass are identical.
3. It follows that the change in momenta of the two trucks getting to the final COM velocity are identical.
4. The applied forces on the two trucks are identical.
5. It follows that it takes exactly the same amount of time to accelerate both centers of mass to the final speed.
6. The truck with the movable box accelerates faster than its carried box.
7. It follows that the truck with the movable box accelerates faster than the center of mass of that truck+box.
8. The truck with the fixed box accelerates at the same pace as the center of mass of that truck+box.
9. It follows that the truck with the movable box accelerates more rapidly than the truck with the fixed box.
10. It follows that, over the course of the acceleration to the final COM speed, the truck with the movable box traverses a greater distance.
11. It follows that, over the course of that acceleration, more work is done on the truck with the movable box.

That is where the extra energy required to give the box a velocity relative to its truck comes from.

Of course, this description is for the idealized world of a smooth road, no air resistance and a truck bed long enough that the box is still sliding when final COM speed is achieved. For a bumpy road it gets trickier.

 

[WalterG]

Well, if 9. is true, then 10. is certainly false. Because if the truck accelerates more rapidly it should reach the 60km/h earlier, hence less distance travelled.

I was more thinking about the fact, that the center of mass is moving backwards during the acceleration for the truck with moving box. So the center of mass of the truck with the moving box reaches the 60km/h a bit later, because it moves backwards. Hence more energy needed.

 

[jbriggs444]

Well, if 9. is true, then 10. is certainly false. Because if the truck accelerates more rapidly it should reach the 60km/h earlier, hence less distance travelled.

Not so. Note the distinction between truck speed and center of mass speed. We've already decided that both accelerations (of the center of mass) take the same amount of time. The truck (alone) travels further than the center of mass.

I was more thinking about the fact, that the center of mass is moving backwards during the acceleration for the truck with moving box.

The center of mass is moving forward. Do not frame jump. Strive to avoid accelerating frames if you are using a conservation of energy argument. [Or look out for an associated potential term].

So the center of mass of the truck with the moving box reaches the 60km/h a bit later, because it moves backwards. Hence more energy needed.

Same force = same acceleration of the center of mass. The center of mass does not lag. The truck leads.

 

[PeroK]

Well, if 9. is true, then 10. is certainly false. Because if the truck accelerates more rapidly it should reach the 60km/h earlier, hence less distance travelled.

I was more thinking about the fact, that the center of mass is moving backwards during the acceleration for the truck with moving box. So the center of mass of the truck with the moving box reaches the 60km/h a bit later, because it moves backwards. Hence more energy needed.

I doubt you'll get a reliable answer without specific data about the load and what it is doing. That said, I can't see a 16kg load moving around making much difference to the fuel consumption of a 1600kg vehicle, whether the load is sliding about or not. Especially if most of the journey is at constant speed.

PS It would be interesting to measure the effect of, say, three passengers against none (driver only). How much difference would the extra 200-300kg make? That would make no difference to the air resistance, for example.

 

[WalterG]

Not so. Note the distinction between truck speed and center of mass speed. We've already decided that both accelerations (of the center of mass) take the same amount of time. The truck (alone) travels further than the center of mass.The center of mass is moving forward. Do not frame jump. Strive to avoid accelerating frames if you are using a conservation of energy argument. [Or look out for an associated potential term].

Same force = same acceleration of the center of mass. The center of mass does not lag. The truck leads.

Agreed! The Truck with the fixed load would lead. Maybe a fraction of a millimeter :); haven't calculated it yet.

 

[jbriggs444]

Agreed! The Truck with the fixed load would lead. Maybe a fraction of a millimeter :); haven't calculated it yet.

No. The truck with the loose load leads.

 

[WalterG]

No. The truck with the loose load leads.

Please explain: If the truck with the loose load would lead, it could make more distance, hence use less energy to reach destination. That can't be right. Remember, that I assume same acceleration in both cases.

 

[jbriggs444]

Please explain: If the truck with the loose load would lead, it could make more distance, hence use less energy to reach destination. That can't be right. Remember, that I assume same acceleration in both cases.

The load is in the tail of the truck. The truck is farther ahead as a result.

Recall that the assumptions that I am working under are a smooth road, no air resistance, a truck body long enough that the load attains the speed of the truck before it hits the tail gate and an initial acceleration that persists only long enough so that the center of mass of the truck plus load reaches the designated cruising speed. The only energy loss to friction is during the initial acceleration phase. After that, the truck coasts all the way to the finish line.

 

[WalterG]

The load is in the tail of the truck. The truck is farther ahead as a result.

Recall that the assumptions that I am working under are a smooth road, no air resistance, a truck body long enough that the load attains the speed of the truck before it hits the tail gate and an initial acceleration that persists only long enough so that the center of mass of the truck plus load reaches the designated cruising speed. The only energy loss to friction is during the initial acceleration phase. After that, the truck coasts all the way to the finish line.

Yes, right. But what about the initial acceleration? Would you want to increase it according to the movement of the center of mass during acceleration? So that it can keep up with the fixed truck? Or do we want to leave the acceleration constant, meaning that we put a constant amount of energy per time into the system?

 

[jbriggs444]

Yes, right. But what about the initial acceleration? Would you want to increase it according to the movement of the center of mass during acceleration? So that it can keep up with the fixed truck? Or do we want to leave the acceleration constant, meaning that we put a constant amount of energy per time into the system?

You seem to be confused.

The centers of mass already keep up with each other. The force is fixed, so the centers of mass accelerate identically.

Under a constant (over time) acceleration, the energy per unit time going into the system is increasing because velocity is increasing. Power (energy per unit time) goes as force times velocity.

Under a constant (between the two scenarios) acceleration of the center of mass, the truck with the movable box will be accelerating more rapidly and will require an increased amount of work. Increased acceleration means increased velocity. Power (energy per unit time) goes as force times velocity.

 

[WalterG]

You seem to be confused.

The centers of mass already keep up with each other. The force is fixed, so the centers of mass accelerate identically.

Under a constant (over time) acceleration, the energy per unit time going into the system is increasing because velocity is increasing. Power (energy per unit time) goes as force times velocity.

Under a constant (between the two scenarios) acceleration of the center of mass, the truck with the movable box will be accelerating more rapidly and will require an increased amount of work. Increased acceleration means increased velocity. Power (energy per unit time) goes as force times velocity.

It took me a while, but I think I understand your reasoning now. If we feed a constant amount of energy E per time slice delta t (dt) into both trucks, the fixed-load truck-A would reach the final velocity Vf earlier than the moving load truck-B. Hence truck-B would go a further distance until reaching Vf. If we now measure that difference (the difference between the two distances traveled) we can deduct how much more energy we put into truck-B. (because E/dt is constant) Thanks again.

 

[sophiecentaur]

I can't find any mention of Work Done in shifting the box against friction (ignore the reference frame niggle). This is an alternative approach.

You really need to ignore the case when the box hits the side, front or back because that would be totally scary and fuel economy would be the furthest thing from your mind! Also, PF doesn't encourage dangerous pastimes.

I saw a suggested wood on wood coefficient of static friction of 0.6. That would suggest you'd be needing 0.6g acceleration to make a wood box slide on a wood floor. Perhaps a bit of a bump in the road when braking (0.8g max that can be relied on) could get the box sliding. This is not going to happen very often and the quoted best 0to60 performance is less than 1g. It would be driving for worst fuel economy to have the box moving frequently (boring practicality).

The energy to move the box against friction by, say 3m would be force times distance. Force would be weight of box (mg = 160N) X 0.6
Work done (for and aft shifting) would be about 288J. That's equivalent to raising the box (going uphill) by 1.8m. And 1l of petrol (gas) stores about 36MJ. We're almost down in the noise.

The box in the model is 1% of the truck mass and the Energy dissipated by the brakes ( and then fuel to get it up to speed) will be significantly more than that. So driving in a non-flashy way could make far more difference to fuel consumption than anything that box can do.

If I've missed out a zero here or there, please put me right. But I bet my suggested method should give a realistic answer.