1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force & power

  1. Jun 18, 2010 #1
    I have a really fundamental problem with understanding relationship between force & power. Please help me with this example:

    An electrically powered space ship (e.g. 100% efficient ionic engine) is travelling 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 accelarating 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!
     
  2. jcsd
  3. Jun 18, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

     
  4. Jun 18, 2010 #3

    K^2

    User Avatar
    Science Advisor

    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.
     
  5. Jun 18, 2010 #4
    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.
     
  6. Jun 18, 2010 #5
    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!
     
  7. Jun 18, 2010 #6

    K^2

    User Avatar
    Science Advisor

    Rocket formula is needed any time you use propellant.
     
  8. Jun 18, 2010 #7
    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.
     
  9. Jun 18, 2010 #8

    rcgldr

    User Avatar
    Homework Helper

    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 (Vrocket - Vpropellent), where Vpropellent is the terminal velocity of the propellent, and it remains constant because as (+ Vrocket) increases, then (-Vpropellent) deceases.
     
  10. Jun 18, 2010 #9

    K^2

    User Avatar
    Science Advisor

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

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

    All of this is the reason why rocket formula exists and why it works.
     
  11. Jun 18, 2010 #10

    rcgldr

    User Avatar
    Homework Helper

    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 Vrocket - Vpropellent will be a constant (assuming a contant throttle application), with any inertial frame of reference.
     
  12. Jun 19, 2010 #11
    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.

    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?
     
  13. Jun 19, 2010 #12

    jack action

    User Avatar
    Science Advisor
    Gold Member

    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).
     
  14. Jun 19, 2010 #13
    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.
     
  15. Jun 19, 2010 #14

    K^2

    User Avatar
    Science Advisor

    Then you're dealing with accelerated reference frames. Though, in this case, it's not making a difference, so fair enough.
    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.
     
  16. Jun 19, 2010 #15

    jack action

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  17. Jun 19, 2010 #16

    K^2

    User Avatar
    Science Advisor

    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 space ship 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.
     
  18. Jun 19, 2010 #17

    jack action

    User Avatar
    Science Advisor
    Gold Member

    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 gonna 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 gonna 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.
     
  19. Jun 19, 2010 #18
    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
    [tex]P=\frac{mgv}{2}[/tex]
    where v is the velocity of the air. Hmm...
     
  20. Jun 19, 2010 #19
    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.
     
  21. Jun 19, 2010 #20
    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 space ship 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".
     
  22. Jun 19, 2010 #21

    K^2

    User Avatar
    Science Advisor

    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.
     
    Last edited: Jun 19, 2010
  23. Jun 19, 2010 #22

    jack action

    User Avatar
    Science Advisor
    Gold Member

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

    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:

    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" [Broken].

    With imagination, we can assume anything when posing a fictional problem.
     
    Last edited by a moderator: May 4, 2017
  24. Jun 19, 2010 #23

    K^2

    User Avatar
    Science Advisor

    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?
     
  25. Jun 20, 2010 #24
    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
    [tex]mg=\frac{m_\text{air}v_\text{air}}{t}[/tex]
    [tex]P=\frac{m_\text{air}v_\text{air}^2}{2t}[/tex]
    Is that OK?


    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.
     
  26. Jun 20, 2010 #25

    K^2

    User Avatar
    Science Advisor

    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:

    [tex]Mg dt = v dm[/tex]

    And energy required to produce flow:

    [tex]dE = \frac{1}{2} v^2 dm[/tex]

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

    [tex]dm = A v \rho dt[/tex]

    And using it to solve for v:

    [tex]Mg dt = A v^2 \rho dt[/tex]

    [tex]v = \sqrt{\frac{Mg}{A\rho}}[/tex]

    And finally finding power.

    [tex]P = \frac{dE}{dt} = \frac{1}{2} v^2 \frac{dm}{dt}[/tex]

    [tex]P = \frac{1}{2} A \rho \left(\frac{Mg}{A\rho}\right)^{\frac{3}{2}}[/tex]

    [tex]P = \frac{1}{2} \frac{(Mg)^{\frac{3}{2}}}{\sqrt{A\rho}}[/tex]

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

    [tex]L = \frac{Mg}{A}[/tex]

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

    [tex]P = \frac{1}{2} Mg \sqrt{\frac{L}{\rho}}[/tex]

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook