Exploring the Relationship between Force & Power

[norrrbit]

I have a really fundamental problem with understanding relationship between force & power. Please help me with this example:

An electrically powered spaceship (e.g. 100% efficient ionic engine) is traveling through space. No external forces (like nearby stars) act on it. We turn on the engines which work with constant power: P - i.e. energy intake from the batteries is constant through time. To my intuition the ship should be now accelerating with constant accelaration: A. Then force acting on the ship should also be constant: F=ship_mass*A.

It seems then that F should be linearly proportional to P. But of course we know that it isn't, because Power is Energy/time and Energy=Force*Distance=0.5*F*A*time^2. So Power increases with time while Force is constant!
Is my intuition wrong?
If engines work with const Power then Accelaration decreases (in void)?

This problem bothers me since I wanted to calculate needed power (in terms of idealistic electrical, 100% efficient propulsion) to keep an object hover in air in Earth's gravitational field.Thanks in advance!

 

[rock.freak667]

It seems then that F should be linearly proportional to P. But of course we know that it isn't, because Power is Energy/time and Energy=Force*Distance=0.5*F*A*time^2. So Power increases with time while Force is constant!
Is my intuition wrong?
If engines work with const Power then Accelaration decreases (in void)?

P=dW/dt = d/dt(Fs)

For a constant force, P = F ds/dt = Fv

 

[K^2]

Your problem is that you are working with a space ship. With a space ship, there is the whole rocket-formula thing.

Basically, the energy you are spending now to accelerate the propellant you are carrying along with the rocket is going to be added to the power you are spending when you are using that propellant.

This is not something you want to look into until you have better understanding of power in simpler scenarios.

If you want to understand power and forces, it's better tot try and think of a car on a road, and just pretend there is no friction or drag.

Then you are pushing off the road that's going faster and faster relative to you all the time. So you need more and more work to do to push yourself at constant force, and therefore, constant acceleration.

Power required to push object traveling at V with force F is actually just F*V. If you want to consider direction of V and F as well, these are vectors, and multiplication is a dot product.

 

[norrrbit]

Thanks rock.freak667 for helping, but I know the formula. Please tell me though where do I make the mistake:
1. Idealistic engine is working with constant power P, by power I mean energy intake from batteries in 1 second.
2. Acceleration A is also constant, isn't it? So is force F too.

Thus P~=F.

 

[norrrbit]

K^2, I don't think that experiment with spaceship is problematic, since I have assumed space void with negligible gravitational forces from nearby cosmic objects. So no rocket formulas needed here.
I know that P=F*V. So when V increases (accelaration) P should also increase. But if my electric car's engine works with constant power it will accelarate at constant pace (no drag or friction)! So P is const. & F is const!

 

[K^2]

Rocket formula is needed any time you use propellant.

 

[Naty1]

To my intuition the ship should be now accelerating with constant accelaration: A.

I agree...
ok, so what is the "problem"? F = ma IS correct until you reach relativistic effects (very high velocities) then m becomes a variable based on v.

 

[rcgldr]

You seem to be ingoring the energy added to the propellent as it's expelled from the space ship. Using the rocket's initial velocity as a frame of reference, the total power is FxV for both rocket and propellent, so power = F (V_rocket - V_propellent), where V_propellent is the terminal velocity of the propellent, and it remains constant because as (+ V_rocket) increases, then (-V_propellent) deceases.

 

[K^2]

No, that's also wrong, because propellant is already traveling at velocity V, so it's final velocity is V-Vp.

Again, rocket formula exists for a reason.

Given a constant power of the drive, the thrust is constant. The mass of the rocket, however does decrease as you use up propellant, so acceleration is increasing. But yes, the thrust is constant.

Again, the extra power to apply thrust at higher and higher velocities comes from kinetic energy stored in the propellant.

For some mass m of propellant, you use $\frac{1}{2}mV_p^ 2$ of rocket drive's energy to accelerate propellant. The kinetic energy of the propellant, however, changes from $\frac{1}{2}mV^2$ to $\frac{1}{2}m(V-V_p)^2$.
That's a decrease in energy by $\frac{1}{2} m(2 V_p V - V_p^2)$. When you add to it the energy spent by the engine, the net energy increase of the rocket works out to $mVV_p$.

Lets say you expel propellant at some rate dm/dt. Then the energy per dm is $dE = VV_p dm$. And the power $P = \frac{dE}{dt} = VV_p \frac{dm}{dt}$. That gives us thrust $F = P/V = V_p \frac{dm}{dt}$. Which is exactly the impulse per dm divided by dt.

All of this is the reason why rocket formula exists and why it works.

 

[rcgldr]

You seem to be ingoring the energy added to the propellent as it's expelled from the space ship. Using the rocket's initial velocity as a frame of reference, the total power is FxV for both rocket and propellent, so power = F (V_rocket - V_propellent), where V_propellent is the terminal velocity of the propellent, and it remains constant because as (+ V_rocket) increases, then (-V_propellent) deceases.

No, that's also wrong, because propellant is already traveling at velocity V, so it's final velocity is V-Vp.

You missed the part that I stated these velocities were using the rocket's initial velocity as a frame of reference, so the (V-Vp) aspect was already covered due to the frame of reference. My point was that V_rocket - V_propellent will be a constant (assuming a contant throttle application), with any inertial frame of reference.

 

[Gerenuk]

Is my intuition wrong?

Yes, why would you think power and force are constant?
And also cars with a given power cannot accelerate forever, right?

Why should acceleration be constant? I suppose you believe that both a some kind of "drive" to a vehicle, but they are again different and you have to look at the details.

As you know P=F*v. So provided all the details allow to apply this equation, then you would have F=P/v which means that the acceleration decreases as velocity increases.

This problem bothers me since I wanted to calculate needed power (in terms of idealistic electrical, 100% efficient propulsion) to keep an object hover in air in Earth's gravitational field.

You cannot look up a random equation from the formula book and plug in some value for some letter. I mean you have to look at the details of the problem, and then pick the right equation. For a rocket its completely different than for a helicopter. And the cup on the table doesn't need any power at all to not fall.

@All:
The rocket might be an easy case.
What about a helicopter in air? How much power does it need?

 

[jack action]

Lots of words for something that appears very basic:

P = Fv and F = ma (so P = mav), dv = adt. P and m are constant so:

av = constant

So a is always positive causing v to always increase, which means a is always decreasing without ever reaching zero (so v will increase forever).

Even considering relativity, this will always be true (although calculations will be more complex).

 

[Gerenuk]

Lots of words for something that appears very basic:

As a rule of thumb, when someone shouts out something is very easy, he usually doesn't have a clue what the question is. At least that's my experience in this forum.

The calculation you do is useful under some circumstances, but it still misses the point that the particular method of levitation is important. Your estimate works for neither rockets nor helicopters.

 

[K^2]

You missed the part that I stated these velocities were using the rocket's initial velocity as a frame of reference, so the (V-Vp) aspect was already covered due to the frame of reference. My point was that V_rocket - V_propellent will be a constant (assuming a contant throttle application), with any inertial frame of reference.

Then you're dealing with accelerated reference frames. Though, in this case, it's not making a difference, so fair enough.

What about a helicopter in air? How much power does it need?

For what? Hover? Climb? Cruise?

Besides, the helicopter question can be as complicated as you want to make it, down to considering effects of vibrations of the blades on overall efficiency.

 

[jack action]

As a rule of thumb, when someone shouts out something is very easy, he usually doesn't have a clue what the question is. At least that's my experience in this forum.

The calculation you do is useful under some circumstances, but it still misses the point that the particular method of levitation is important. Your estimate works for neither rockets nor helicopters.

Either you misunderstood the question in the OP or I did.

From my understanding, the OP asks a very simple question about the relationship between power, force and acceleration ("I have a really fundamental problem with understanding relationship between force & power."). The OP gives the example of a ship in space just to illustrate what he/she doesn't understand. The assumption is that there is no power loss from drag, friction or even propellant velocity. All the power is used to accelerate the mass, which is also constant. Those are the assumptions stated in the problem, so no need to complicate everything.

 

[K^2]

All the power is used to accelerate the mass, which is also constant. Those are the assumptions stated in the problem, so no need to complicate everything.

That's the problem. A ship with constant mass cannot accelerate. It has to eject mass. The kinetic energy change of the ejected mass then goes into power consideration.

The net power driving the spaceship is generally much greater than net power of its engine.

That's why the engine can provide constant thrust under constant power under increasing velocity, and it is not something you can disregard by assuming constant mass. That assumption in itself is unphysical.

 

[jack action]

That's the problem. A ship with constant mass cannot accelerate. It has to eject mass. The kinetic energy change of the ejected mass then goes into power consideration.

The net power driving the spaceship is generally much greater than net power of its engine.

That's why the engine can provide constant thrust under constant power under increasing velocity, and it is not something you can disregard by assuming constant mass. That assumption in itself is unphysical.

Who said that it needed to be physical and possible? The OP also states that it has a 100% ionic engine: Are you saying you cannot answer the question because there is no such thing as a 100% efficiency?

It is a thought experiment to help understand basic concepts. Whatever is assumed must be taken as true.

If I ask you how much force is needed to create an acceleration of 2 m/s² for a block of 10 kg sliding on a surface, assuming no friction:

Are you going to argue for hours that there is no such thing as a frictionless surface, try to guess what would be an adequate coefficient of friction and when you're not going to find one, you will claim the problem is unsolvable;

or are going to say: F = ma = 10 X 2 = 20 N?

I don't think norrrbit is going to use the info found here to built the next space shuttle for NASA.

Anyway, that's how I read the OP.

 

[Gerenuk]

For what? Hover? Climb? Cruise?

Let's say hover.

Maybe one can make an estimate by saying the helicopter accelerates air from stand-still and it has to accelerate enough air to make up for the momentum?

My late-night calculation gives
$$P=\frac{mgv}{2}$$
where v is the velocity of the air. Hmm...

 

[Gerenuk]

The OP gives the example of a ship in space just to illustrate what he/she doesn't understand. The assumption is that there is no power loss from drag, friction or even propellant velocity. All the power is used to accelerate the mass, which is also constant. Those are the assumptions stated in the problem, so no need to complicate everything.

You shouldn't do what I call alphabet physics, when some students do it. They pick an equation and plug in just any power, any force and any velocity for any equation which has P, F and v in them. What you do should make sense. As the others have explained the problem with the rocket is not about losses or friction.
I mean you are right that if all the power goes into kinetic energy, then P=mav. But you haven't understood what the others were discussing, so it's not right to claim they make something simple overly complicated. They were explaining that any real world application for flight is different and not due to losses but since its different physics.

 

[Gerenuk]

Who said that it needed to be physical and possible? The OP also states that it has a 100% ionic engine: Are you saying you cannot answer the question because there is no such thing as a 100% efficiency?
...
Are you going to argue for hours that there is no such thing as a frictionless surface, try to guess what would be an adequate coefficient of friction and when you're not going to find one, you will claim the problem is unsolvable;

I advice you to look up the equation for a rocket and think about it. You are completely missing the point here (and making a fool of yourself if you play cocky).
It's not about efficiency and perfectness. To propell a spaceship you need to eject mass. There is no other way. This mass carries away energy. To accelerate to a given speed you need to eject some fixed amount of mass (at a given velocity). This mass will have a well defined amount of energy that is "wasted".

 

[K^2]

Who said that it needed to be physical and possible? The OP also states that it has a 100% ionic engine: Are you saying you cannot answer the question because there is no such thing as a 100% efficiency?

No, I'm saying there is no such thing as an ion thruster that violates conservation of momentum, and there never can be. It's one thing to have a problem which has unachievable efficiency or something like that, and another to have setup violate something as fundamental as conservation of momentum.

If you start out with rocket that violates conservation of momentum, nobody knows what it's going to do. It's an impossible question. We have to have some fundamental properties preserved, and there is nothing more fundamental.

That's why I suggested a simpler case. Car on a road. We can still ignore all sources of friction, but use road with "infinite" mass to push off of.

@Gerenuk:

Power will depend on loading factor. That is weight/area of the rotor. I don't see area in your equation, so I can tell you right away that it's wrong.

Keep in mind, though, that rotor isn't 100% efficient at pushing air, and some of the air it does push, will loop around over the blade tips and re-enter the stream. So it's not helping to hold the plane. But for base estimate, you can just assume that the entire area under the rotor is pushing the air to some constant velocity, and that gives you the lift. This will be correct to some efficiency factor. I'm trying to remember what the typical efficiencies for heli rotors are. I think they tend to be around 70%, varying quite a bit, of course.

 

[jack action]

I advice you to look up the equation for a rocket and think about it. You are completely missing the point here (and making a fool of yourself if you play cocky).
It's not about efficiency and perfectness. To propell a spaceship you need to eject mass. There is no other way. This mass carries away energy. To accelerate to a given speed you need to eject some fixed amount of mass (at a given velocity). This mass will have a well defined amount of energy that is "wasted".

I gave you the example of efficiency and perfectness to illustrate what I think the OP is asking (I might be wrong):

I have a really fundamental problem with understanding relationship between force & power.

So, from my point of view, the question is not about rockets and how they work. It's about the relationship between force, power and acceleration in general. Hence, the following comment:

Lots of words for something that appears very basic

Again, maybe I misunderstood the OP and if it is the case then, you may be right, I am missing the point and making a fool of myself.

But if I'm wrong and someone is interested in space traveling, I suggest reading http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html" .

With imagination, we can assume anything when posing a fictional problem.

 

[K^2]

If you read OP's post carefully, you'll notice that his confusion arises from the fact that constant power applied to propellant gives you constant thrust, while constant thrust applied to accelerating rocket requires increasing power.

You cannot resolve this paradox without considering what happens to propellant. That's the whole point.

You're "simplification" leads to nothing good, as it does not resolve the paradox. Or are you under the impression that rocket's thrust really is decreasing the faster it goes?

 

[Gerenuk]

@Gerenuk:
Power will depend on loading factor. That is weight/area of the rotor. I don't see area in your equation, so I can tell you right away that it's wrong.

Whatever the area is, I think it doesn't enter my power estimate?! By knowing the weight only, you can calculate both the power and area independently (of course apart from efficiencies).
If used
$$mg=\frac{m_\text{air}v_\text{air}}{t}$$
$$P=\frac{m_\text{air}v_\text{air}^2}{2t}$$
Is that OK?

So, from my point of view, the question is not about rockets and how they work. It's about the relationship between force, power and acceleration in general. Hence, the following comment:

Your equation is correct for total energy transfer to kinetic energy. People were just pointing out that you cannot do that for rockets for example, because that method requires you to waste energy by thrusting out fuel.
So it really important to look at the particular problem and not plug in for example the power of your vacuum cleaner and assume it relates anyhow to the amount of objects it can pick up.
I know too many students who look for the keyword "power" and "force" and as soon as they see it, they dig out the first equation they can find with P and F.

 

[K^2]

So where does the mass of the air come from? And your power goes as 1/t?

Let me use M for mass of the helicopter, A for area under the rotor, and v for velocity of the flow. I'll probably need differential dm for mass of the air, because there isn't really a total mass. The final result will contain ρ, the density of air.

Balance impulse on heli and air:

$$Mg dt = v dm$$

And energy required to produce flow:

$$dE = \frac{1}{2} v^2 dm$$

Expanding dm in terms of air that flows through rotor in time dt.

$$dm = A v \rho dt$$

And using it to solve for v:

$$Mg dt = A v^2 \rho dt$$

$$v = \sqrt{\frac{Mg}{A\rho}}$$

And finally finding power.

$$P = \frac{dE}{dt} = \frac{1}{2} v^2 \frac{dm}{dt}$$

$$P = \frac{1}{2} A \rho \left(\frac{Mg}{A\rho}\right)^{\frac{3}{2}}$$

$$P = \frac{1}{2} \frac{(Mg)^{\frac{3}{2}}}{\sqrt{A\rho}}$$

That looks about right. One more thing you can do is introduce loading factor:

$$L = \frac{Mg}{A}$$

Then the equation for power becomes significantly more self-explanatory.

$$P = \frac{1}{2} Mg \sqrt{\frac{L}{\rho}}$$

Power is directly proportional to weight and root of loading factor and is inversely proportional to root of density. Just what we want to see.

 

[Gerenuk]

So you basically have to same equation as I, just with the velocity plugged in.

 

[Lsos]

This post is nothing more than a curious piece, but you actually "can" have a "rocket" that doesn't expel mass, since you can also get thrust by ejecting photons out the back (a big f-flashlight).

Simple calculations will show that it's actually a doable concept, assuming we can overcome some technical hurdles (mainly the availability of a big f-flashlight and some advanced power source, although even fission would work).

I threw this out there since I suppose this can be compared to a "100% ionic engine"...since an ionic engine ejects less mass faster, where as this engine would eject 0 mass, the fastest.

Nevertheless, and perhaps this is my point, we find that even with this imaginary spaceship we cannot get away from the rocket equation, as the preposterous amount of power it requires would still result in a significant shedding of the ship's mass.

 

[Gerenuk]

This post is nothing more than a curious piece, but I figured out from my time on this site that you actually "can" have a "rocket" that doesn't expel mass, since you can also get thrust by ejecting photons out the back (a big f-flashlight).

We are talking about energy lost. Photons carry energy and this energy is just as well lost and cannot go into kinetic energy of the space ship.

 

[jack action]

If you read OP's post carefully, you'll notice that his confusion arises from the fact that constant power applied to propellant gives you constant thrust, while constant thrust applied to accelerating rocket requires increasing power.

You cannot resolve this paradox without considering what happens to propellant. That's the whole point.

You're "simplification" leads to nothing good, as it does not resolve the paradox.

Constant power applied to any object that is accelerating (not just rocket) doesn't give you a constant force. Why? Because power is not related to force, it is related to force AND velocity. If one goes up, the other must come down.

If there are no other forces to fight (drag, friction, etc), then force and acceleration are related by F = ma. and for an object that is under constant power (rocket or not) the relationship between power (of the rocket), acceleration (of the rocket) and velocity (of the rocket) will be P = Fv = mav.

The fact that the way a machine is designed to create the power of the rocket and that more than needed will have to be produced is for me irrelevant in the present problem. When I deal with a problem involving a car, if someone ask me how much power it must produce to go a certain speed against a certain drag, I don't say: "Wait, don't forget to take into account the friction losses of the engine and the heat losses of the cooling system to calculate the real power produced by the fuel that is burned". I go simply: power needed = drag force X vehicle speed. Even though I know that there is not, and that there will never be, such an engine that will be able to transfer 100% of its energy to propel a vehicle.

That is what I understood when the OP said:

I have a really fundamental problem with understanding relationship between force & power.

(...)

No external forces (like nearby stars) act on it. We turn on the engines which work with constant power: P

I didn't think I was dealing with a rocket scientist trying to built a spaceship, but with someone struggling with the concept of power, force and velocity. I thought he/she was only trying to illustrate his/her misunderstanding as best as he/she could. This particular problem about power arise very often among amateur physicist. In such a case, I don't see the need to give more information than needed as it often only mix up and discourage the curious student. Answering a question that wasn't asked rarely helps.

Or are you under the impression that rocket's thrust really is decreasing the faster it goes?

Yes. Thrust is proportional to the exhaust mass flow rate and the exhaust velocity with respect to the rocket. As the rocket goes faster, the exhaust velocity with respect to the rocket goes down, hence the thrust goes down. The same thing happens when the speed of my car increases, its traction force decreases; even though the engine produces the same torque and power. Why? Because this brings us back to the fundamental relationship: P = Fv. The power acting on the rocket will drop and be transferred to the exhaust gases, which will also comply to the very basic P = Fv. And yes, I know there will be some energy losses to heat up the exhaust gases.

 

[Lsos]

Yes. Thrust is proportional to the exhaust mass flow rate and the exhaust velocity with respect to the rocket. As the rocket goes faster, the exhaust velocity with respect to the rocket goes down, hence the thrust goes down. The same thing happens when the speed of my car increases, its traction force decreases; even though the engine produces the same torque and power. Why? Because this brings us back to the fundamental relationship: P = Fv. The power acting on the rocket will drop and be transferred to the exhaust gases, which will also comply to the very basic P = Fv. And yes, I know there will be some energy losses to heat up the exhaust gases.

Apparently you're not understanding it either, because as the rocket goes faster, the exhaust velocity with respect to the rocket does NOT go down. And that's the whole point of this discussion and the whole reason for all the confusion.

It's not something that makes sense unless you put some time to think about it, and/ or start a thread to get some help figuring it out.

 

[jack action]

Apparently you're not understanding it either, because as the rocket goes faster, the exhaust velocity with respect to the rocket does NOT go down. And that's the whole point of this discussion and the whole reason for all the confusion.

It's not something that makes sense unless you put some time to think about it, and/ or start a thread to get some help figuring it out.

I shouldn't have brought back the rocket. I don't know how to tell this simpler than it is:

P = Fv
F = ma
a = dv/dt

3 possible cases:

1- Object at constant velocity:

dv = 0; a = 0; F = 0 ---> P = 0

2- Object at constant acceleration:

F = constant; P/v = constant ----> P is proportional to v

3- Object at constant power (which I think is the subject under discussion):

Fv = constant ----> F is inversely proportional to v. Which leads to a is inversely proportional to v

If you have a rocket under constant power - same amount of energy spent on the rocket's motion in a given time - the thrust will be smaller as the speed of the rocket goes up, whether you decrease the velocity of the exhaust or its mass flow rate: something has to give.

Why is that? Because for the rocket that goes faster, the displacement will be greater in the same time dt. And since we assume that the energy spent will be the same and that energy = force times displacement, therefore the force must go down for the faster rocket to respect the initial assumption that power is constant.

And that is true for any type of object going through any type of motion.

Right now, you all have a problem similar to this https://www.physicsforums.com/showthread.php?t=408584", except that the person couldn't accept the first case (a constant v means zero power). And just like I said in that thread, I will give up at this point, as you are not even trying.

 

[Lsos]

Yeah, something does have to give. What happens is that, to an observer on the earth, the rocket's power IS increasing.

However, the paradox, or source of confusion, is that despite the power increasing the amount of propeller burned per second is NOT increasing. If you ask the pilot of the rocket, he'll tell you that the throttle is at the same position, the pumps are pumping the same amount of fuel, and the rocket is experiencing the same thrust...and yet, to the guy on the earth, the rocket is going faster and faster.

To the guy on the earth, the power is increasing. To the guy on the rocket, the power is the same.

The reason for this is, of course, because we're talking about a rocket. By taking it's propellant with it, it's essentially taking it's frame of reference with it. As such, we don't care that it's going 100000mph away from the earth. The propellant doesn't know this. It doesn't know how fast it's moving in relation to the earth, the sun, the galaxy, light, or whatever. It just knows that it's sitting still on a rocket, and when it is ignited it will go out at a given velocity, regardless of how fast other objects in the universe are moving. It will be expelled out the back at the same relative speed to the rocket, and will continue to provide the same thrust.

It seems pretty obvious that this paradox is why the OP started this thread.

 

[rcgldr]

If you have a rocket under constant power - same amount of energy spent on the rocket's motion in a given time.

For constant power the energy spent and forces involved on the rocket and the spent fuel remain constant. The initial condition could be rocket in space with zero speed, and all of the energy goes into the spent fuel exhaust.

Yeah, something does have to give. What happens is that, to an observer on the earth, the rocket's power is increasing.

Not if that observer also takes into account the energy put into the spent fuel exhaust. The total energy gain of rocket and spent fuel per unit time will be constant for a constant power (throttle) setting. Fuel will be consumed and chemical potential energy converted into mechanical energy and heat at the same rate, regardless of the rockets speed relative to any inertial (non-accelerating) frame of reference.

Imagine a rocket and it's unspent fuel in outer space away from any gravitational effects. It's a closed system with no external forces, the center of mass of rocket and fuel will not move, and the momentum of the rocket and it's fuel will remain contant (zero if the frame of reference is the rockets initial velocity).

 

[Lsos]

Not if that observer also takes into account the energy put into the spent fuel exhaust.

Ah, true. So to an Earth observer the total power remains the same, but as the rocket goes faster, more of this power goes into the rocket rather than into the exhaust. Is this the right way to think of it?

 

[K^2]

No, the power that pushes the rocket traveling at considerable velocity may greatly exceed the power of the engine.

The point is that in the beginning, the engine works to accelerate both the rocket and the propellant it is carrying. Only the energy used to accelerate the rocket is useful at this point. The rest is stored in the propellant. Once the rocket builds up to considerable speed, that energy stored in the propellant helps you push the rocket.