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Lifting Power

  1. Jul 8, 2015 #1
    I'm a little afraid to ask something that should be such a basic question, yet there seems to be an enormous discrepency in-between theory and reality, here.

    In many, many problems I've worked with "find the work to lift X", you simply take the change in potential energy. When lifting Mass [itex]m[/itex] by distance [itex]h[/itex], finding the work [itex]W[/itex] is a simple matter of
    [itex]W = mgh[/itex].

    But things get a lot more interesting when we ask not about the total energy, but the power required.

    If I take the above instance at face value, then I end up with the absurd result that a small, thumb-sized, say, [itex]P[/itex] = .5 Watt motor can be used in a crane to lift a 1,000-kilo elevator to the top of a skyscraper, given enough time, [itex]t = \frac{P}{W}[/itex]

    Experience, however, would insist that that's simply not the case. What should happen instead, is the motor will create a little bit of tension in the cable and stop there, being unable to overcome the weight of the load. No matter how much [itex]mgh[/itex] says that there's no power going on when something is stationary, I simply can't believe that a little 1-watt motor could hold a 1,000-ton weight up against gravity - and then slowly pull it up on top of that.

    So what would be a proper way to calculate the power needed to lift a certain mass? I can imagine calling on the Equivalence principle (saying that gravity is actually the result of an accelerated frame) and saying that we actually need to accelerate it at 1 g, so every second we need to impart the kinetic energy it'd take to bring it to 9.8 m/s, to counteract gravity pulling it at 9.8 m/s2, so [itex]P = \frac{1}{2}mg^{2}[/itex] to counteract its weight, and any amount of power in addition to accelerate it upwards.

    Is that a valid way to approach the problem in terms of power?
     
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  3. Jul 8, 2015 #2

    mfb

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    It can. You just need good gears with a huge ratio to reduce torque at the motor and friction enough.

    It doesn't even have correct units. You arbitrarily introduce a second and give it some special meaning here.
     
  4. Jul 8, 2015 #3
    You can (theoretically) create a machine with the necessary mechanical advantage for a small motor to
    raise any specified weight. Say, a small gear driving a very large gear with no friction involved.
     
  5. Jul 8, 2015 #4

    anorlunda

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    Actually, it's pretty easy to do with hydraulics. Small, motor, positive displacement pump, big hydraulic reservoir, and you can lift a lot of weight.

    That is why a man can lift a heavy vehicle using a hydraulic jack.
     
  6. Jul 8, 2015 #5

    sophiecentaur

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    There are two important ratios in studying machines. Velocity Ratio and Mechanical Advantage. Velocity Ratio is based just on the geometry of the set up and it boils down to distance moved by effort / distance moved by load. Mechanical Advantage, otoh, is Force applied to load / Effort and, because of Friction and having to lift the mass of the machinery in most cases, this will never be as 'good' as the VR suggests.
    MA/VR = Efficiency
    You can either talk in terms of Work (Force times distance) in and out or Power (Force times velocity) in and out. They are equivalent and you will get the right (same) answers as long as you use the MA and VR appropriately.
     
  7. Jul 8, 2015 #6
    In the back of my mind I knew the units didn't add up, which is why It troubled me so much to ask.

    I guess this is just a case where I just need to get over my gut instinct. It's just very difficult to take peoples' word for it that a 1-watt motor could hold up the weight of a 10,000 ton package and lift it up against gravity, however slowly. But, physics is as physics does, and cares little for human intuition.

    Let's look at something, though - what about a helicopter hovering in the air? It's constantly burning fuel, so it's consuming a certain amount of energy per second in the form of chemical (fuel) energy to maintain altitude. What's fundamentally different about that?

    Clip on extra weight to the helicopter and it's going to have to throttle its engine up to even greater power to maintain altitude. So there's got to be something different, here.

    EDIT: Would an analysis in an inertial frame help, here? In a free-fall frame, a motor sitting on a wench is co-accelerating with the weight it is lifting, so doesn't develop a difference in energy. The ground underneath it keeping it in place keeps the two objects from developing a difference in energy over time.

    With the helicopter, on the other hand, there is no ground underneath it holding it in place, so it must use some form of stored energy (chemical energy of the fuel) to accelerate it upwards in the inertial frame, to keep it a constant height above the ground (which, in an inertial frame, is also accelerating upwards).

    So the question is; how much energy does it need to consume per second to keep itself aloft?
     
    Last edited: Jul 8, 2015
  8. Jul 8, 2015 #7

    SteamKing

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    You're trying to analyze too much at one time without fully understanding the basics.

    The number of watts a motor generates is not what keeps a small weight or a large weight suspended. That chore is dependent entirely on the statics of the situation, that is, do you have a strong enough cable such that it doesn't snap when it supports the load, is the brake of the hoist strong enough so that it doesn't break under the load, etc.

    By moving the weight upward against gravity, there is a change in the potential energy of the weight with respect to the ground, which is what mgh is, the change in potential energy of the weight, mg, due to the change in height, h, above the ground.

    Since the energy input is E = mgh, the power input on raising or lowering the load is going to change with the amount of time, Δt, it takes to raise or lower the load.
    Power P = mgh / Δt

    The amount of power a winch motor has determines how quickly or how slowly the change in potential energy of the weight occurs, assuming the machine is capable of working against friction, etc. A motor with a large amount of power, which is measured in watts or horsepower, will be able to raise the same weight more quickly than a less powerful motor.

    Careful here. Wenches (= girls or young women) don't like to support large loads. :rolleyes:

    Winches, however, do just fine, as long as you don't overload them. :wink:

    I don't know what you mean by "co-accelerating". A winch can be fixed to the ground, so that it is not moving, and still operate perfectly fine, as long as too much (static) load is not placed on it.

    Now, you're jumping from winches to helicopters, which work on different principles.

    Think of the blades of a helicopter as a giant propeller. The torque of the motor turning on the chopper spins the blades, which, depending on their setting, tend to move large quantities of air about. When the blades are turning at a certain speed, the movement of the air creates a force, known as lift, which acts in the opposite direction to gravity, which wants to pull the machine to earth.

    As long as the lift generated by the turning blades is equal to the weight of the helicopter, the net force acting on the helicopter is zero, and the chopper stays stationary with respect to the earth.

    If the speed of the motor is increased slightly, more lift is generated, and the net force acting on the craft is slightly positive, and the chopper will start to rise and change altitude.

    If you add more weight to the helicopter, spinning the blades at the same rate previously generates the same amount of lift, but the weight of the helicopter has changed, and the craft stays on the ground. The motor must be throttled up to increase the amount of lift which the blades generate, until the lift is greater than the total weight of the craft. Only then can the helicopter get off the ground.
     
  9. Jul 9, 2015 #8

    sophiecentaur

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    A hovering helicopter is doing NO work on the load. This means it has zero efficiency. This example shows how non-intuitive this stuff can be.
     
  10. Jul 9, 2015 #9

    sophiecentaur

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    I have to bring in that pesky Efficiency thing again. When you use gears, worms, screws, levers etc to produce a massive mechanical advantage, the extreme ratio needed (plus dead weight) will usually mean that the efficiency is very low. So the mechanism just won't work in reverse at all. For example, a simple car screw jack doesn't need a 'stop' to avoid the car settling down and the handle spinning round when you let go. There is so much stiction that it just won't work in reverse. Your 1W motor could eventually provide enough Joules of energy to the system to lift that heavy package a metre, say, but the total energy put in by the motor would need to be tens of times more than the 108 J of work actually done on the package.
    As you say, intuition often goes counter to what Physics can prove to you. But intuition is based on a very limited amount of actual bodily experience. That's why Magicians get away with it.
     
  11. Jul 9, 2015 #10
    And that's exactly what does happen - the motor cannot provide enough FORCE to overcome the weight of the load; there is no movement so no work is done.
     
  12. Jul 9, 2015 #11

    mfb

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    It is not - if we have appropriate gears or hydraulics, as explained before.

    The helicopter doesn't have that option available, it has to generate enough lift by accelerating air downwards, which requires constant power that heats the air eventually.
     
  13. Jul 9, 2015 #12
    I should probably have noted that I have covered classical mechanics in formal education, it's just this one particular problem has never really made sense to me, even though I know a simple [itex]P = E/Δt = mgh/Δt = FD/Δt = FV[/itex] is how everything says it should be solved. Which, while it does cover a motor mounted on the ground (probably by a mechanical device (winch) rather than a woman (wench), haha. My bad), doesn't seem to be a complete description for a more complex case of seeing how much chemical energy a helicopter must drain in the form of fuel to stay aloft.

    And so I am very familiar with forces and balancing those and such - those I find very intuitive.

    Hydraulics is really an excellent way to answer the question, actually. I've covered some basic fluid mechanics, and it's a lot more convincing in that case because it's easier to trace back and visualize how conservation laws require that hydraulics behave that way.

    I'm still rather curious about exactly how I'd go about trying to answer the helicopter problem, though.

    I'm going to take a crack at it, here. I understand I'm going to be making some massive simplifications, so this isn't much better than a ballpark estimate, but I'd like to know if this is a valid sort of approach to take.

    Let's say the helicopter has mass [itex]M_{h}[/itex], and a blade radius [itex]r_{b}[/itex]. The blades then cover an area of [itex]A_{b} = πr_{b}^{2}[/itex]

    Now let's isolate a single second of time to make the analysis a bit easier. In a single second of free-fall, the helicopter would develop a momentum of [itex]ρ_{h} = M_{h}v = M_{h}aΔt = M_{h}gΔt = M_{h}g(1 second)[/itex]

    By conservation of momentum, an equal amount of momentum of the air must be forced down to keep the helicopter stationary: [itex]p_{air} = M_{air}v_{air}[/itex]

    The mass of the air will be the density multiplied by the volume of air displaced. [itex]M_{air} = ρ_{airdensity}V_{air}[/itex]

    Now, this is why I chose to isolate a single second: finding the mass and velocity of the displaced air will be tricky, because the volume of the air displaced in a single second will be a cylinder, the area of the base of which will be [itex]A_{b} = πr_{b}^{2}[/itex] and the height of which will be the velocity of the air times the period of time, Δt, which we've chosen to be one second, [itex]h = v_{air}Δt[/itex].
    *note, do not confuse volume [itex]V_{air}[/itex] with velocity [itex]v_{air}[/itex]

    [itex]V_{air} = A_{b}h = πr_{b}^{2}v_{air}Δt[/itex]

    Back to the momentum of the air:
    [itex]p_{air} = M_{air}v_{air}[/itex]
    Substituting:
    [itex]M_{air} = ρ_{airdensity}V_{air}[/itex]
    [itex]V_{air} = A_{b}h = πr_{b}^{2}v_{air}Δt[/itex]

    [itex]p_{air} = ρ_{airdensity}V_{air} v_{air} = ρ_{airdensity}A_{b}h v_{air} = ρ_{airdensity} πr_{b}^{2}v_{air}^{2}Δt [/itex]

    Since we've isolated a single second under the assumption the momentums, thus forces, are balanced;
    [itex]p_{air} = p_{h} = M_{h}gΔt = ρ_{airdensity} πr_{b}^{2}v_{air}^{2}Δt [/itex]
    Now we can isolate the velocity of the displaced air using that last equality:
    [itex]M_{h}gΔt = ρ_{airdensity} πr_{b}^{2}v_{air}^{2}Δt [/itex]
    [itex]\frac{M_{h}g}{ρ_{airdensity} πr_{b}^{2}} = v_{air}^{2}[/itex]
    [itex]v_{air} = \sqrt{\frac{M_{h}g}{ρ_{airdensity} πr_{b}^{2}}}[/itex]

    Now, going back to our earlier expression we can find the mass of the air:
    [itex]M_{air} = ρ_{airdensity}V_{air} = ρ_{airdensity}πr_{b}^{2}v_{air}Δt[/itex]

    With mass and velocity, we can find the kinetic energy imparted into the air each second with the familiar
    [itex]K_{E} = \frac{1}{2}mv^{2}[/itex]

    If we try plugging in the density of air at 1.22 kg/m^3, and pull our other numbers off of wiki's description of the UH-60, we end up getting that there's about 1 MJ of kinetic energy in the air every second, meaning about 1 MW of power (using maximum takeoff mass). This lines up quite nicely with wiki's report that the helicopter uses two 1,400 kW engines, considering a realistic efficiency and that this was only a ballpark estimate with the "solid cylinder of sea-level air" assumption.
     
  14. Jul 9, 2015 #13

    jbriggs444

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    In a helicopter, one could attempt to reduce the inefficiency by increasing the rotor diameter -- pushing a greater quantity of air downward with a reduced speed. Lift goes roughly as mass flow rate times downwash velocity. Power goes roughly as mass flow rate times the square of downwash velocity. To a first approximation, a bigger rotor gives a better lift to power ratio. However, a rotor big enough to lift a 1000 ton weight using a 1 watt motor would be somewhere between ludicrous and impossible.
     
  15. Jul 11, 2015 #14

    mfb

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    Very far in the impossible range, at least with current technology and on Earth. Larger rotors also add mass. Human-powered helicopters are barely able to lift off (still with ground effect) - at less than 200 kg and a power of more than 100 W.
     
  16. Jul 11, 2015 #15

    sophiecentaur

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    The efficiency just gets worse and worse (lift / drag, in the case of flying machines) when your input force or power get very low.
    There are few worse examples for discussing basic Work problems than rockets and helicopters. Stick to gears, levers and inclined planes for the first few years of learning the subject.
     
  17. Nov 25, 2016 #16
    Hi MattRob,
    In your very first initial post you declared the equation [itex] P={1\over 2} m g^2 [/itex]

    That looks like an interesting equation. It looks like [itex] E={1\over 2} m v^2 [/itex] but I don't see how you simply replaced [itex] E [/itex] by [itex] P [/itex] and [itex] v^2 [/itex] by [itex] g^2 [/itex]

    Can you please show your derivation of that formula? Please show the math and describe it in many words being so explanatory that even those who are not familiar with the equations can still understand your derivation. I am familiar with the equations but I would enjoy seeing both.

    Thank you very much!
     
  18. Nov 25, 2016 #17

    russ_watters

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    For Thanksgiving, Mythbusters ran a marathon. In one episode, they lifted a car with an ordinary shop-vac!
     
  19. Nov 25, 2016 #18

    russ_watters

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    Ever change a tire on a car? However did you manage to lift one side of a car off the ground all by yourself?!
    A hovering helicopter is a bit like swimming against a river. But the power required is a[n inverse] function of the rotor size. You can have an arbitrarily large rotor and arbitrarily small power.
     
  20. Nov 26, 2016 #19

    mfb

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    The formula is wrong, as discussed in previous posts already.

    Note that this thread is from 2015.
     
  21. Dec 11, 2016 #20
    I think we need to break this down much more simply, because it is a very simple problem/question
    the 5 watt motor can lift the earth.... as Archimedes said, "the long lever....etc" you need deep gearing and that 5 watt motor will lift what ever object you want, the only thing that will vary will be how faster it will be done. power is torque x speed .... so you dont have any movement, there was no power used.. power is the rate of KE change. you can pick a second, but then, that will determine how much mass can be lifted by some distance. 5 watts over 1 second is 5 watt seconds.. or 5 Joules change in 1 second. Or in power terms, 550 lbs will be lifted 1 foot, in 1 second for 1 hp. that 5 watt motor is .07 HP..... thats 38lbs 1ft in 1 second. now, if you want to lift a huge weight like 38,000lbs, you would be able to move it 1/1000 of a foot in 1 second. (with the proper gearing for the motor reducing its max HP RPM / some gear ratio to give this linear speed)

    hope that helps

    the helicopter hovers and is doing no work, but is very inefficient because its still using energy to turn the rotors. that power is going up in moving and accelerating the air mass around the rotors . similar to a person leaning on a wall... you can get tired pushing, and nothign happens... no work is done, but you get tired, why? blood flow, cells are using food to contract muscles, heart uses energy to pump the blood and lungs to breath the air. in the end, no work is done mechanically, but energy is used to great the balance of fore on to the wall as the walls force acts back.
     
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