The power output of a vehicle when descending hills

[martonhorvath]

TL;DR Summary

How to calculate power output at negative slopes?

By applying some simplifications, the power $P$ of a body moving at a speed of $v$ at an incline of $\alpha > 0^\circ$ can be expressed as:
$P = P_{acceleration} + P_{friction} + P_{gravity} + P_{air} = (ma + mgsin(\alpha) + C_{rr}mgcos(\alpha) + 0.5C_dA\rho v^2)\cdot v$

If we would apply this formula for a slope for which $\alpha < 0^\circ$, and assume the same power $P$ as for the example on the incline, would it be true that:
$$P = P_{acceleration} + P_{friction} + P_{air}$$?

According to my thought, in this case, power coming from the gravitational force is an "extra" and the body would start moving even with zero power produced by the body if the gravitational force were greater than the adhesion force.

The same would be my question for a case when the body is moving at $\alpha < 0^\circ$ and $a<0 \ \mathrm{m/s^2}$, does $P = P_{\text{friction}} + P_{\text{air}}$ in this case?

If my thoughts here are correct I would love to receive a proper explanation, if I was wrong, please correct me.

 

[russ_watters]

Power provided by gravity doesn't go away if the slope is negative, it just changes sign. If you drive a car down a steep hill and take your foot off the gas pedal, what happens?

 

[martonhorvath]

Power provided by gravity doesn't go away if the slope is negative, it just changes sign. If you drive a car down a steep hill and take your foot off the gas pedal, what happens?

I see, but that power is not produced by the car but by the hill. My point would be that in this case, the car has to overcome only three forces and there is an additional one which "helps" it. E.g., if the car would produce 1 kW under both conditions then I think that in the case of the downhill scenario, the net power would be more than 1 kW due to the gravitational force which comes on top of that 1 kW.

 

[russ_watters]

I see, but that power is not produced by the car but by the hill.

Right; the engine/car power is on the left side and all of the terms on the right side of the equation are power applied to the car by things other than the engine/car.

My point would be that in this case, the car has to overcome only three forces and there is an additional one which "helps" it.

Correct, so to properly account for it, you should leave it in the equation.

E.g., if the car would produce 1 kW under both conditions then I think that in the case of the downhill scenario, the net power would be more than 1 kW due to the gravitational force which comes on top of that 1 kW.

I'm not sure what you mean by "net power", but this equation (every conservation of energy equation) always sums to zero if you move all of the terms to the same side. Often when talking about forces summing to zero people will say a non-zero "net force" causes acceleration, but you already have a term for acceleration in your equation. In this case you've separated power applied by the car and power applied by other things on opposite sides of the equation.

Note, if you want to keep the car's speed constant on a hill, you're missing a term: power applied(absorbed) by the brakes.

 

[martonhorvath]

Right; the engine/car power is on the left side and all of the terms on the right side of the equation are power applied to the car by things other than the engine/car.

Correct, so to properly account for it, you should leave it in the equation.

I'm not sure what you mean by "net power", but this equation (every conservation of energy equation) always sums to zero if you move all of the terms to the same side. Often when talking about forces summing to zero people will say a non-zero "net force" causes acceleration, but you already have a term for acceleration in your equation. In this case you've separated power applied by the car and power applied by other things on opposite sides of the equation.

Note, if you want to keep the car's speed constant on a hill, you're missing a term: power applied(absorbed) by the brakes.

Maybe my question formulation was wrong here because I am starting to get a feeling that both of us are right here. So I will try to illustrate what I am trying to say here through another example, I think this one enlightens more about the problem I want to know:

let's assume that the car produces P power on a hill (i.e., $\alpha > 0^\circ$). Then the car gets to a downhill and outputs the same power P on the downhill but according to our previous agreement this time this P is spent only against friction and air if assuming a constant speed (i.e., the car is already accelerated). But if we look at the "net" power of the car based on that balance equation I wrote down, its power corresponding to its speed of travel should be greater than P because, in that speed, we have the effect of the gravitational force incorporated as well. But the car's power output is still P.

Does this make sense?

 

[Andy Resnick]

Right; the engine/car power is on the left side and all of the terms on the right side of the equation are power applied to the car by things other than the engine/car.

Correct, so to properly account for it, you should leave it in the equation.

I'm not sure what you mean by "net power", but this equation (every conservation of energy equation) always sums to zero if you move all of the terms to the same side. Often when talking about forces summing to zero people will say a non-zero "net force" causes acceleration, but you already have a term for acceleration in your equation. In this case you've separated power applied by the car and power applied by other things on opposite sides of the equation.

Note, if you want to keep the car's speed constant on a hill, you're missing a term: power applied(absorbed) by the brakes.

This is an interesting discussion and I want to display my ignorance: I drive a manual transmission and can use 'engine braking' to maintain constant speed (or even speed decreases) on downhills. What's that all about?

 

[russ_watters]

let's assume that the car produces P power on a hill (i.e., $\alpha > 0^\circ$). Then the car gets to a downhill and outputs the same power P on the downhill but according to our previous agreement this time this P is spent only against friction and air if assuming a constant speed (i.e., the car is already accelerated). But if we look at the "net" power of the car based on that balance equation I wrote down, its power corresponding to its speed of travel should be greater than P because, in that speed, we have the effect of the gravitational force incorporated as well. But the car's power output is still P.

You've created an inequity. If the car/engine is still outputting P power when the slope goes from positive to negative, the car must accelerate. That's what actually happens.

 

[russ_watters]

This is an interesting discussion and I want to display my ignorance: I drive a manual transmission and can use 'engine braking' to maintain constant speed (or even speed decreases) on downhills. What's that all about?

I'm not quite sure what you are asking, but: When you use engine braking the engine's power output goes negative: it stops burning fuel and its internal friction - which is substantial - opposes the car's motion.

Incidentally, my automatic transmission car has several selectable modes for this:

1. Disengage "clutch"
2. Stay in gear
3. Stay in gear but downshift

The manual claims option 1 is the most energy efficient, but I suspect that is only true on gentle slopes. On steeper slopes, in neutral, you are powering the engine at idle while using the brakes to dissipate the GPE.

 

[Steve4Physics]

This might help, though it's a bit long/wordy...

Power is energy transferred per second – but confusion can arise if it is not clear where the energy has come from and where it is going.

We can construct an equation relating the instantaneous values of various powers, all measured at the same moment in time, $t$.

Employ a sign-convention in which positive power values indicate the car is receiving energy and negative power values indicate that the car is transferring energy elsewhere.

$P_{\text{fuel to engine}}$ is the rate of energy transfer from fuel to the engine. (This will depend on the rate at which fuel is being supplied at time $t$ and the efficiency.) It is positive or zero. (It could be negative if you have an electric vehicle with regenerative braking!)

$P_{\text{friction}}$ is the rate at which the car loses energy due to frictional losses, e.g. due to tyre rolling-resistance, friction in bearings. It has a negative value if the car is moving.

$P_{\text{air}}$ is the rate at which the car loses energy due to air-resistance. It will usually be negative but could be positive if a wind blows in the direction of the car’s motion faster than the car’s speed.

$P_{\text {gravity}}$ is minus the rate of change of the car’s gravitational potential energy(GPE).
If the car is going downhill, GPE decreases and $P_{\text {gravity}}$ will be positive (the car’s weight is doing work on the car).
If the car is going uphill, GPE increases and $P_{\text {gravity}}$ will be negative.

From conservation of energy, the net power being delivered to the car at time $t$ is the sum of the above:
$P_{\text {fuel to engine}} +P_{\text {friction}} +P_{\text{air}} +P_{\text{gravity}}$
(remembering the correct sign is need for the value of each term).

The overall effect is that the car’s speed increases, decreases or remains constant.
If $P_{kinetic}$ is the rate of change of the car's kinetic energy then;
$P_{kinetic} = P_{\text {fuel to engine}} +P_{\text{friction}} +P_{\text{air}} +P_{\text {gravity}}$

 

[martonhorvath]

You've created an inequity. If the car/engine is still outputting P power when the slope goes from positive to negative, the car must accelerate. That's what actually happens.

Okay, you are right concerning acceleration but can you see what I mean under 'net power' now? If we take a look at the power balance equation and want to calculate the car's contribution to the 'net power' (i.e., power corresponding to the speed of travel) at an incline less than zero, it doesn't make sense to have a negative gravity component in there, we should rather take $P= P_{acceleration} + P_{friction} + P_{air}$. At least, this is my thought here.

 

[russ_watters]

Okay, you are right concerning acceleration but can you see what I mean under 'net power' now? If we take a look at the power balance equation and want to calculate the car's contribution to the 'net power' (i.e., power corresponding to the speed of travel) at an incline less than zero, it doesn't make sense to have a negative gravity component in there, we should rather take $P= P_{acceleration} + P_{friction} + P_{air}$. At least, this is my thought here.

I don't see it. How can you account for the influence of gravity if you remove it from the equation? The two sides are now no longer equal.

 

[martonhorvath]

This might help, though it's a bit long/wordy...

Power is energy transferred per second – but confusion can arise if it is not clear where the energy has come from and where it is going.

We can construct an equation relating the instantaneous values of various powers, all measured at the same moment in time, $t$.

Employ a sign-convention in which positive power values indicate the car is receiving energy and negative power values indicate that the car is transferring energy elsewhere.

$P_{\text{fuel to engine}}$ is the rate of energy transfer from fuel to the engine. (This will depend on the rate at which fuel is being supplied at time $t$ and the efficiency.) It is positive or zero. (It could be negative if you have an electric vehicle with regenerative braking!)

$P_{\text{friction}}$ is the rate at which the car loses energy due to frictional losses, e.g. due to tyre rolling-resistance, friction in bearings. It has a negative value if the car is moving.

$P_{\text{air}}$ is the rate at which the car loses energy due to air-resistance. It will usually be negative but could be positive if a wind blows in the direction of the car’s motion faster than the car’s speed.

$P_{\text {gravity}}$ is minus the rate of change of the car’s gravitational potential energy(GPE).
If the car is going downhill, GPE decreases and $P_{\text {gravity}}$ will be positive (the car’s weight is doing work on the car).
If the car is going uphill, GPE increases and $P_{\text {gravity}}$ will be negative.

From conservation of energy, the net power being delivered to the car at time $t$ is the sum of the above:
$P_{\text {fuel to engine}} +P_{\text {friction}} +P_{\text{air}} +P_{\text{gravity}}$
(remembering the correct sign is need for the value of each term).

The overall effect is that the car’s speed increases, decreases or remains constant.
If $P_{kinetic}$ is the rate of change of the car's kinetic energy then;
$P_{kinetic} = P_{\text {fuel to engine}} +P_{\text{friction}} +P_{\text{air}} +P_{\text {gravity}}$

Thank you for your detailed reply, I think your answer reassures me that I do not think wrongly about this. If I understand you correctly, you described $P_{gravity}$ from the perspective of the gravitational force, thus in your thread of thoughts if gravity does work on the car, $P_{gravity} >0$ and vice versa

I don't see it. How can you account for the influence of gravity if you remove it from the equation? The two sides are now no longer equal.

Let's remove acceleration thus assuming $v=const.$ and use a new example. What I am particularly interested in is the following:

A vehicle produces e.g., P=200 W on an uphill, in this case, the energy is spent against gravity, friction and air resistance (we assumed $a=0~m/s^2$). Then we take this vehicle and the same P=200 W power output on a downhill but this time, thus P should equal only power against friction and air resistance because the body is not working against gravity anymore but gravity does work on the body (and accelerates it in the beginning until terminal velocity is reached but in this case I assumed that the body is already travelling at the terminal velocity).

Is this correct or not?

 

[russ_watters]

A vehicle produces e.g., P=200 W on an uphill, in this case, the energy is spent against gravity, friction and air resistance (we assumed $a=0~m/s^2$). Then we take this vehicle and the same P=200 W power output on a downhill but this time, thus P should equal only power against friction and air resistance because the body is not working against gravity anymore but gravity does work on the body (and accelerates it in the beginning until terminal velocity is reached but in this case I assumed that the body is already travelling at the terminal velocity).

Is this correct or not?

Not. Let's assume on the uphill you have 100 W going to gravitational potential energy, 50W each going to friction and air resistance (and no acceleration term): 200W = 100W + 50W + 50W

The car crests the hill and nothing changes except you have removed the GPE term. Now you have: 200W = 50W + 50W. A broken equation.

 

[Steve4Physics]

@martonhorvath, a car doesn’t actually have a power, its engine does. So by $P$ you could mean:

- the power delivered by the engine (rate of energy conversion from chemical energy in the fuel to mechanical energy);

- the rate of change of the car’s kinetic energy (which is the sum of all the other powers combined as explained in Post #9);

- something else.

So it would help if you clarify what you mean by$P$.

 

[russ_watters]

So it would help if you clarify what you mean by$P$.

I've been assuming he means engine for the "car power" (also ignoring brakes).

 

[PeroK]

Thank you for your detailed reply, I think your answer reassures me that I do not think wrongly about this. If I understand you correctly, you described $P_{gravity}$ from the perspective of the gravitational force, thus in your thread of thoughts if gravity does work on the car, $P_{gravity} >0$ and vice versa

Let's remove acceleration thus assuming $v=const.$ and use a new example. What I am particularly interested in is the following:

A vehicle produces e.g., P=200 W on an uphill, in this case, the energy is spent against gravity, friction and air resistance (we assumed $a=0~m/s^2$). Then we take this vehicle and the same P=200 W power output on a downhill but this time, thus P should equal only power against friction and air resistance because the body is not working against gravity anymore but gravity does work on the body (and accelerates it in the beginning until terminal velocity is reached but in this case I assumed that the body is already travelling at the terminal velocity).

Is this correct or not?

Terminal velocity has a different value uphill and downhill.

 

[martonhorvath]

Not. Let's assume on the uphill you have 100 W going to gravitational potential energy, 50W each going to friction and air resistance (and no acceleration term): 200W = 100W + 50W + 50W

The car crests the hill and nothing changes except you have removed the GPE term. Now you have: 200W = 50W + 50W. A broken equation.

The 200 W is a capacity in the example, it is spent regardless of incline. That would be the point of my question. To have the same work rate from the engine both uphill and downhill, so from 200 = 100 + 50 + 50, it would be e.g., 200 = 120 + 80, I want to keep this 200W which is produced by the engine the same for the two conditions. And my thought was that gravitation 'adds' to this power...

 

[Steve4Physics]

@martonhorvath, how about this:

Car going down a hill. At a particular moment in time:
Power output of engine = 400W
Power lost due to air resistance = 200W
Power lost due to rolling resistance and internal friction = 100W
Power gained due to gravity (loss of potential energy) = 200W
Net power given to car = 400 – 200 – 100 + 200 =300W
This ‘net power’ is the rate of change of the car’s kinetic energy; it is positive (300W) so the car I speeding up at this moment.

Car going up a hill. At a particular moment in time:
Power output of engine = 400W
Power lost due to air resistance = 200W
Power lost due to rolling resistance and internal friction = 100W
Power lost due to gravity (gain of potential energy) = 200W
Net power given to car = 400 – 200 – 100 – 200 =-100W
This ‘net power’ is the rate of change of the car’s kinetic energy; it is negative (-100W) so the car is slowing down at this moment.

 

[russ_watters]

The 200 W is a capacity in the example, it is spent regardless of incline. That would be the point of my question.

Usually the word capacity is used to indicate the amount available, not the amount actually delivered. For example if the engine was capable of 400 watts but delivering 200 Watts we would say its capacity is 400 watts, but its actual output is 200 w. I'm not sure if that distinction is critical here because I think I understand you are talking about the actual power delivered being 200 w. You want to keep that the same in both cases. Correct?

To have the same work rate from the engine both uphill and downhill, so from 200 = 100 + 50 + 50, it would be e.g., 200 = 120 + 80, I want to keep this 200W which is produced by the engine the same for the two conditions.

If friction and wind resistance were 50 and 50 before cresting the hill and the speed stays the same they have to still be 50 and 50. I don't understand how or why you would think they just change.

And my thought was that gravitation 'adds' to this power...

I don't understand what you mean by that. Where does that show up in the equation?

 

[martonhorvath]

Terminal velocity has a different value uphill and downhill.

If friction and wind resistance were 50 and 50 before cresting the hill and the speed stays the same they have to still be 50 and 50. I don't understand how or why you would think they just change.

The speed does not stay the same, I am sorry if I have been unclear about this before.

You want to keep that the same in both cases. Correct?

That is correct, indeed.

I don't understand what you mean by that. Where does that show up in the equation?

If we want to look at an equation, we could write something like $P_{downhill} = P + P_{gravity} - (P_{friction} + P_{air})$, where P is the same 200 W as in the uphill example.

 

[martonhorvath]

@martonhorvath, how about this:

Car going down a hill. At a particular moment in time:
Power output of engine = 400W
Power lost due to air resistance = 200W
Power lost due to rolling resistance and internal friction = 100W
Power gained due to gravity (loss of potential energy) = 200W
Net power given to car = 400 – 200 – 100 + 200 =300W
This ‘net power’ is the rate of change of the car’s kinetic energy; it is positive (300W) so the car I speeding up at this moment.

Car going up a hill. At a particular moment in time:
Power output of engine = 400W
Power lost due to air resistance = 200W
Power lost due to rolling resistance and internal friction = 100W
Power lost due to gravity (gain of potential energy) = 200W
Net power given to car = 400 – 200 – 100 – 200 =-100W
This ‘net power’ is the rate of change of the car’s kinetic energy; it is negative (-100W) so the car is slowing down at this moment.

I agree with this, as I said earlier, I think we have been talking about the same thing, just you brought kinetic energy into the conversation and I was trying to explain my thoughts through constant velocity examples after a point.

 

[Steve4Physics]

... you brought kinetic energy into the conversation and I was trying to explain my thoughts through constant velocity examples after a point.

OK. But it may be worth adding this in case it's not already clear...

$P_{downhill}$ is the car's rate of change of kinetic energy when going downwhill.

If the speed is constant, then $P_{downhill} = 0$.
If the speed is increasing, then $P_{downhill} > 0$.
If the speed is decreasing, then $P_{downhill} < 0$.

 

[martonhorvath]

OK. But it may be worth adding this in case it's not already clear...

$P_{downhill}$ is the car's rate of change of kinetic energy when going downwhill.

If the speed is constant, then $P_{downhill} = 0$.
If the speed is increasing, then $P_{downhill} > 0$.
If the speed is decreasing, then $P_{downhill} < 0$.

Thank you, it is clear.

 

[russ_watters]

The speed does not stay the same, I am sorry if I have been unclear about this before.

If we want to look at an equation, we could write something like $P_{downhill} = P + P_{gravity} - (P_{friction} + P_{air})$, where P is the same 200 W as in the uphill example.

Ok, that will be fine. Just note that:

1. You've changed the name of the acceleration term - there's no need to do that.
2. Previously you were talking about removing the GPE term. Now it's back. That's good.
3. You've rearranged the terms in a way that isn't necessary and changed signs when not necessary. The terms and equation can all stay the same in the uphill and downhill cases. The original equation worked fine and does not need to change as you crest the hill. All that changes is the value of the numbers associated with the terms. The GPE term's value changes from positive to negative, but the way it is expressed as a variable in the equation does not need to change (and really shouldn't - you'll make a physicist's eyes bleed).

 

[Vanadium 50]

Incidentally, my automatic transmission car has several selectable modes for this:

My hybrid will "engine brake" on hills by using the gravitational energy to charge the battery. It takes a little practice (the Penna. Turnpike, eg) but can be surprisingly efficient. I was getting maybe 35 mpg as opposed to 40-45 on flat land like Ohio.

I bring this up because now that you're all nodding your heads, the power of the combined battery+engine system clearly goes negative downhill: the battery is charging. What if I don't have a battery? It's still got to go negative, so the power must go to engine losses.

 

[jbriggs444]

What if I don't have a battery? It's still got to go negative, so the power must go to engine losses.

If you do not have a battery to charge and if engine braking is inadequate to maintain a constant speed on a downhill then speed increases.

No bananas in Scranton PA.

 

[Vanadium 50]

The point is that the engine-battery system can be analyzed, and the result interpolated to an engine-only system. There are two parts - a part that consumes it.

"What if one part overwhelms the other?" is not the question being asked. But the answer should be unaupriainf - the car either stops or goes very fast.

 

[Vanadium 50]

And yoy know what else I am sick of? Being nearly blind, and having PF think its a great big joke. I know my typing sucks and I an doing the best I can.

Not a day goes by when I am not reminded that I an blind, crippled and of negative value to society. PF can pile on too. Plenty of company,

 

[Andy Resnick]

I'm not quite sure what you are asking, but: When you use engine braking the engine's power output goes negative: it stops burning fuel and its internal friction - which is substantial - opposes the car's motion.

Incidentally, my automatic transmission car has several selectable modes for this:

1. Disengage "clutch"
2. Stay in gear
3. Stay in gear but downshift

The manual claims option 1 is the most energy efficient, but I suspect that is only true on gentle slopes. On steeper slopes, in neutral, you are powering the engine at idle while using the brakes to dissipate the GPE.

That's actually quite a helpful answer- thanks! I drive a manual transmission and often use an engine brake (clutch engaged after downshifting to a lower gear) but wasn't sure what was going on re: spark plugs and fuel injector. Conceptually, is the engine essentially running backwards? Not in reverse, but backwards in the sense of a reversible engine...

I should ask my PHY I class (during our thermo/engine lectures) if they think I am generating gasoline when I do this..... :)

 

[jbriggs444]

Not in reverse, but backwards in the sense of a reversible engine...

In a classical four stroke engine (Otto Cycle) there is throttle plate in the carburetor. On a down slope, the throttle will be almost completely closed. On the intake stroke, the engine will be inhaling a partial vacuum as a result of the throttle. This will tend to slow the engine down.

The compression and power strokes will net out to almost nothing without a good charge of combustible vapors to work with. If one is running with electronic fuel injection, the fuel supply may be completely cut off in this situation.

When the exhaust stroke begins, the exhaust valves will open into a cylinder still containing a partial vacuum, allowing ambient pressure air to enter. The remainder of the exhaust stroke will expel this air.

If you think about it, it is the difference between inhaling a partial vacuum and expelling ambient pressure air that is responsible for the engine braking effect. The throttle plate is where energy is lost.

Plus the friction of the piston rings, the crank shaft, etc, etc.