Here we go again Newtons and Watts

[Leafhill]

Hi!

So I thought I knew the laws of physics and electronics, but I've recently stumbled on to a puzzling bit of "Math vs Pratical application". I'm messing up the forces and equations in my strained little head, so I'll let you guys have a go.

I'll go into practical applications of this and RC-helicopters if needed later. This should cut to the core of my problem:

I know a Newton is the force required to continuously accelerate a frictionless 1kg brick by 1 m/s^2 (horizontal motion). My question is how many watts would I need to do the same thing?

Say I have a motor with 85% efficiency mounted to a coil of wire tied to my frictionless 1 kg brick. How many watts do I give the motor to achieve that same continuous horizontal acceleration?

I've really tried to find solutions in this forum and the rest of the net, but all I come up with is watts needed lift stuff. I've found that Watt = Joule/s = Nm/s, but this doesn't really help me.

I would appreciate any help!
Thank you!

 

[berkeman]

Hi!

So I thought I knew the laws of physics and electronics, but I've recently stumbled on to a puzzling bit of "Math vs Pratical application". I'm messing up the forces and equations in my strained little head, so I'll let you guys have a go.

I'll go into practical applications of this and RC-helicopters if needed later. This should cut to the core of my problem:

I know a Newton is the force required to continuously accelerate a frictionless 1kg brick by 1 m/s^2 (horizontal motion). My question is how many watts would I need to do the same thing?

Say I have a motor with 85% efficiency mounted to a coil of wire tied to my frictionless 1 kg brick. How many watts do I give the motor to achieve that same continuous horizontal acceleration?

I've really tried to find solutions in this forum and the rest of the net, but all I come up with is watts needed lift stuff. I've found that Watt = Joule/s = Nm/s, but this doesn't really help me.

I would appreciate any help!
Thank you!

Work is force multiplied by distance: W = F * D

Power is work per unit time: P = W/delta_t

So the power required depends on how far and how fast you push the block or whatever.

Does that help?

 

[Leafhill]

Wow, fast reply! thanks!

Still have some issues to understand, though. My brick is not actually moving so distance is kind of irrelevant. Its got the 1m/s^2 acceleration force (1 N), but what if a spring is holding it in place? It's not moving anywhere, but the force is still there and my motor is working hard maintaining this force.

What am I missing?

 

[berkeman]

Wow, fast reply! thanks!

Still have some issues to understand, though. My brick is not actually moving so distance is kind of irrelevant. Its got the 1m/s^2 acceleration force (1 N), but what if a spring is holding it in place? It's not moving anywhere, but the force is still there and my motor is working hard maintaining this force.

What am I missing?

How can the motor be moving and the object not?

 

[Leafhill]

No, the motor is not moving, but it is connected to a power supply. This applies a force on its output shaft trying to pull on the object even if it is not able to actually move it.

 

[stewartcs]

No, the motor is not moving, but it is connected to a power supply. This applies a force on its output shaft trying to pull on the object even if it is not able to actually move it.

Energy is being consumed by the motor but no mechanical work is being done on the brick since it is not moving.

CS

 

[Leafhill]

Correct!
The object is not moving, but the force of one Newton is there and it is consuming power. How much?

I'm sure you allready know this, but to be sure we and anyone else reading this are on the same page I'll write it anyway. Inside the motor is a copper coil facing a magnet. These are unresponsive until an electric current is passed through the coil. Based on the direction of the current, it will push the coil (and the output shaft of the motor) clock wise or counter clock wise. This "push" or force will be there even if nothing is actually moving as long as the current is running through the coil.

DL

 

[saunderson]

My question is how many watts would I need to do the same thing?

i guess you have the average power in mind!? The power the motor must provide at any time is

$$P = \frac{\mathrm dW}{\mathrm dt} \qquad \mbox{with} \qquad W = \int \mathrm dr ~ F = \int \mathrm dt ~ v(t) \, F$$​

(i treat the problem as one dimensional) so

$$P = \frac{\mathrm dW}{\mathrm dt} = \int \limits_{t_0}^t \mathrm dt^\prime ~ a \, F = a F (t - t_0) \qquad \mbox{with} \qquad a = \mathrm{const}$$​

the average power must be

$$\left<P\right> = \frac{1}{T} \int \limits_{t_0}^{t_0+T} \mathrm dt^\prime ~ P(t^\prime)$$​

for the simple case of $$t_0=0$$ this yield

$$\left<P\right> = \frac{1}{2} a F T$$​

so you can prove it with the law of conservation of energy. Assumed that the mass is initially at rest and we stop our treatment at $$t=T$$, then the energy in your system must be

$$E=m a s \qquad \mbox{with} \qquad s = \frac{1}{2} a T^2$$​

so the average power is

$$\left<P\right> = \frac{\Delta W}{\Delta t} = \frac{E}{T} = mas \cdot \sqrt{\frac{a}{2s}}$$​

$$T$$ applied to the equation of the average power i derived above yields

$$\left<P\right> = \frac{1}{2} a F T = \frac{1}{2} a F \sqrt{\frac{2s}{a}} = \frac{1}{2} a^2 m \sqrt{\frac{2s}{a}}$$​

That's an indication that the power of the motor must change with time! To be exact

$$P(t) = aFt$$​

That are my thoughts!

 

[stewartcs]

Correct!
The object is not moving, but the force of one Newton is there and it is consuming power. How much?

I'm sure you allready know this, but to be sure we and anyone else reading this are on the same page I'll write it anyway. Inside the motor is a copper coil facing a magnet. These are unresponsive until an electric current is passed through the coil. Based on the direction of the current, it will push the coil (and the output shaft of the motor) clock wise or counter clock wise. This "push" or force will be there even if nothing is actually moving as long as the current is running through the coil.

DL

Since the motor is not rotating, only holding the brick stationary, the amount of power used to hold the brick stationary will be determined based on the locked rotor current of the motor.

The catch is you are presuming the motor can only put out enough torque that is equivalent to 1 N of force on the brick. The locked rotor current depends on the specifics of the motor so I don't know of any analytical way to get the "equivalent power" for 1 N of force on the brick. It would depend on the motor specification.

CS

 

[Leafhill]

Hi, stewartcs!

The motor output force does NOT depend on the motor specification. I can with high accuracy (using dedicated electric equipment) control how many amps and volts (or if you will watts) should be applied to the motor. The more I supply the motor, the more the motor will respond with a powerful force on the output shaft.

Saunderson: I'm not as versed in the math as you clearly are, but I don't see why the supplied power would need to be dependent on time. One amount of watts will give one amount of Newtons pulling the brick, no matter when or for how long it is done. Maybe you can explain in more detail what you mean?

 

[saunderson]

One amount of watts will give one amount of Newtons pulling the brick, no matter when or for how long it is done.

The connection between the force exerted on an object and the power which is dissipated is

$$P = \frac{\mathrm dW}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \Bigl( \int \mathrm dt ~ \frac{\mathrm d\vec r(t)}{\mathrm dt} ~ F(\vec r,t) \Bigr) = \vec v \cdot \vec F$$​

Maybe it is helpful for you to imagine the power as a kind of energy current. So, the faster the object is, the more energy must flow on the object.

You can treat the force $$F=ma$$ with $$a=\mbox{const}$$ as to be conservative. It's easy to see that

$$\Delta T = \frac{1}{2} m \Bigl( v_2^2-v_1^2 \Bigr)$$​

rises for higher velocities (presumed $$\Delta v = v_2 - v_1 = \mbox{const}$$), so the energy current (Power) must rise too.I hope i could help!?

 

[berkeman]

Hi, stewartcs!

The motor output force does NOT depend on the motor specification. I can with high accuracy (using dedicated electric equipment) control how many amps and volts (or if you will watts) should be applied to the motor. The more I supply the motor, the more the motor will respond with a powerful force on the output shaft.

Just to make sure that you understand something -- what is the efficiency of the motor when it has a locked shaft? How much useful work is it doing? (Remeber, W = F*D)

 

[Leafhill]

Hi, berkeman!

The motors I use is normally specified at 85% efficiency under normal load. This is not at locked shaft where it probably will be a bit lower. It will also heat up due to the higher current being pulled, but a proper motor controller can continuously monitor this and reduce this current to protect the motor.

It seems that "useful work" is a relative term. I would say that keeping a spring tight is quit useful in some applications even if the spring is not going anywhere.

Maybe another way of approaching the problem is in order:

This is a slightly different scenario, but poses the same problem:
An small RC-helicopter (electric) is able to hover (float in the air without moving) consuming about 60 Watts of power from the buildt in battery. I have one of these in my living room :-)
Much of this power is lost as heat in the electric circuits and controllers and not to forget the low efficiency of the propeller. All losses can be known, but since I don't have exact specifications for this scenario we can assume that a total of 3W (5% of input power) end up as pure lift. How much does the helicopter weigh? Or rather, how many Newtons of lift is being produced with these 3 Watts?

Remember that the helicopter does not move (up/down or sideways), but is continuously fighting gravity of 9.81 m/s^2 midair.

Why do I feel this should be easy, but is still so difficult for me to calculate.

 

[saunderson]

Another possibility to convince you: how much is the acceleration of an object if the energy flow (Power) on it is constant? (as one dimensional treatment)

$$P = v(t) F(t) = m v(t) a(t) = m v(t) \dot v(t) = \frac{m}{2} \frac{\mathrm d}{\mathrm dt} \Bigl( v^2(t) \Bigr)$$​

integration of both sides yields

$$\frac{2P}{m} \cdot t = v^2(t) \qquad \Leftrightarrow \qquad v(t) = \sqrt{\frac{2P}{m} ~ t}$$​

so the acceleration of the object is

$$a(t) = \frac{P}{m} \Bigl(\frac{2P}{m} ~ t \Bigr)^{-\frac{1}{2}}$$​

Significant: acceleration is not constant! In addition the acceleration diminishes with time.

you can prove the validity of $$v(t), a(t)$$ by applying them to the first equation.(for simplicity i omit to set the initial conditions and the special theory of relativity)very crazy if you look at the singularity at the origin :)

 

[Leafhill]

Hi, Saunderson!

I'm sure your equations are correct, but I'm more practical than theoretical, and have difficulties accepting the results.

My view is something like this:
If you are a waiter and holding a tray of food in one hand, you are continuously working against gravity by applying a force to counteract it. This is burning watts and calories in your arm and although it may feel like the tray gets havier as time passes its still the same forces acting on the tray (gravity pulling down and your arm keeping it up).

This is a very interesting subject and I really appreciate all your answers. I'm really eager to get to the bottom of this:-)

 

[rcgldr]

The simple answer is power = force x speed or power = torque x angular_velocity at any moment in time. In order for your motor to accelerate the brick using a continuous 1 Newton of force and the corresponding constant acceleration, the power at any moment increases linearly with speed (until the motor reaches it's maximum power output). The previous posts work out the details.

 

[berkeman]

The motors I use is normally specified at 85% efficiency under normal load. This is not at locked shaft where it probably will be a bit lower.

Nope. Try a lot lower. Like zero % efficiency.

It seems that "useful work" is a relative term. I would say that keeping a spring tight is quit useful in some applications even if the spring is not going anywhere.

No useful work is done if the object isn't moving.

And on your helicopter example, air is most defintely being accelerated and moved. That is where the engine power is being put to use.

 

[saunderson]

If you are a waiter and holding a tray of food in one hand, you are continuously working against gravity by applying a force to counteract it.

Physically you do no work because you are physically a rigid body! At least at the macroscopic level.

 

[berkeman]

If you are a waiter and holding a tray of food in one hand, you are continuously working against gravity by applying a force to counteract it. This is burning watts and calories in your arm and although it may feel like the tray gets havier as time passes its still the same forces acting on the tray (gravity pulling down and your arm keeping it up).

There's a similar thread somewhere her on the PF that's been active for a couple of days. Something about pusing against a wall and the concept of work. I haven't read it, but from the title it sounds like it might help you out.


EDIT -- link: https://www.physicsforums.com/showthread.php?t=382155&highlight=wall

.

 

[Leafhill]

Hi, Jeff!

Thanks! So Power = force * speed. OK, but my speed is zero m/s. Nothing is moving, but the spring is kept tight and my motor is consuming watts. Clearly there is power being consumed, but it doesn't look like these equations cover the cases I'm explaining. The power that is usually transferred into speed and acceleration is now only seen as a tight spring and thermal build up on the motor. I think that somehow this transfer need to be taken into account.

 

[berkeman]

Thanks! So Power = force * speed. OK, but my speed is zero m/s. Nothing is moving, but the spring is kept tight and my motor is consuming watts. Clearly there is power being consumed, but it doesn't look like these equations cover the cases I'm explaining. The power that is usually transferred into speed and acceleration is now only seen as a tight spring and thermal build up on the motor. I think that somehow this transfer need to be taken into account.

With the locked shaft, your motor efficiency is 0%. No useful output work is being done, and the motor is converting electrical energy into heat energy via I^2R losses.

 

[Leafhill]

Hi, Berkeman!

Thanks for the link. I'll check it out.

So it seams that I need to find ways of eliminating speed and motion from the equations. As I'm sure you are aware of by now I don't care much for motion. I simply don't need it, but I do need the power and I do need the force (May the force be with you:-). For now, I think I'll sleep on it and get back to you later!

Its 1:12 AM in Norway!
Sleep tight!

 

[collinsmark]

Hi, Saunderson!

I'm sure your equations are correct, but I'm more practical than theoretical, and have difficulties accepting the results.

My view is something like this:
If you are a waiter and holding a tray of food in one hand, you are continuously working against gravity by applying a force to counteract it. This is burning watts and calories in your arm and although it may feel like the tray gets havier as time passes its still the same forces acting on the tray (gravity pulling down and your arm keeping it up).

This is a very interesting subject and I really appreciate all your answers. I'm really eager to get to the bottom of this:-)

Like some others have implied (or even explicitly detailed), it is a matter of the definition of work, which is based on Newton's fundamental laws.

In your example, the waiter is only burning calories (or Watts) because humans are inefficient. If the human body was 100% efficient, standing around with a tray would burn no energy whatsoever.

Perhaps a better example is the weightlifter. Suppose a weightlifter lies on a benchpress bench and lifts a 100 kg barbell up 0.5 m, then down 0.5 m cyclically, for a total of 10 repetitions (reps). How much overall work does the weightlifter do on the barbell? The answer is 0. Each time the weightlifter lifted up the barbell, he did 490 Joules of work on it, increasing its potential energy by that much. But when the barbell went back down it did 490 Joules on him (and it lost 490 Joules of potential energy in the process)!

So why does the weightlifter feel tired afterwords? Again, it's because the human body is inefficient. If the human body was 100% efficient, the weightlifter might feel a bit tired after lifting the barbell up, but when the barbell went back down, the weightlifter would feel suddenly rejuvenated -- no less tired than when he started! By the way, this sort of concept is used in regenerative breaking by the way, which is becoming popular on various hybrid electric and electric automobiles. Many systems are capable not only of expending energy, but actually harnessing it if things are done in the opposite manner. Humans, not so much though.

Going to your motor example pushing the brick. If the motor was 100% efficient, and if the brick was not moving, it would require 0 Watts to keep the brick and spring in place, force or no force. You could substitute the motor with a chair or a clamp or something to keep the brick/spring compressed. (Force is still there, but obviously no work is being done.)

Moving on, let's just say you introduce a force with equal and opposite direction to your 1 N force. Suppose the block is placed on a rough surface. Further suppose the force of dynamic friction on that block 1 N. Now, if you push the block such that it reaches a constant velocity of 1 m/s, and you continue pushing it, you are pushing the block with 1 W of power!

$$\frac{d}{dt}E = \vec F \cdot \vec v$$

$$= (1.0 \ N)(1.0 \ m/s) = 1 \ W.$$

The number of Watts you need to push something, is the force multiplied by the object's velocity (assuming force and velocity are in the same direction). The faster it goes, the more Watts you need to maintain the same force.

 

[stewartcs]

Hi, stewartcs!

The motor output force does NOT depend on the motor specification. I can with high accuracy (using dedicated electric equipment) control how many amps and volts (or if you will watts) should be applied to the motor. The more I supply the motor, the more the motor will respond with a powerful force on the output shaft.

I wasn't trying to imply that it did. However since you brought that point up, in practical applications, you can't just arbitrarily apply any amount of power to the motor and get the same power out...you'll destroy the motor. All motors have ratings based on their design. That means that they can only output a certain amount of power. Trying to go beyond the design of it will cause it to undoubtedly fail.

Let's just look at the physics of your question:

Initially we have just a brick with 1 N of force applied to it and the brick is not moving (implies that there is another force that counter acts the 1 N). Since the brick is not moving, yet has 1 N force applied to it, there is no work being done on the brick. What you are trying to find out is how much power it will take to hold that brick still. By definition, no work and thus no power is required since it is not moving. The power required to move the brick is equal to the force times the velocity. However, since the brick isn't moving there is no work required to hold it steady and thus no power required.

Now add the motor to the system so that we have a brick connected to a wire which is connected to the shaft of a DC motor. The previous argument is still valid since the brick isn't moving, however, there is energy being converted (i.e. used) from the power supply to the motor to hold the brick constant now. The energy is now being dissipated as heat in the windings of the motor. How much heat you might ask? Well, it depends on the applied voltage which will induce a current in the armature. How much current depends on the armature resistance which means it depends on the motor specification. Hence my comment that the amount of "effective power" depends on the motor specification.

I think the part that has you confused is that there is no direct conversion between force and power. It requires another variable...velocity.

Hope this helps.

CS

 

[stewartcs]

Going to your motor example pushing the brick. If the motor was 100% efficient, and if the brick was not moving, it would require 0 Watts to keep the brick and spring in place, force or no force.

Regardless of the efficiency of the motor, it still uses energy to hold the spring constant. It's just dissipated as heat instead of being converted to shaft power.

CS

 

[rcgldr]

If the motor is not turning, such as a servo motor, then it's a very efficient heater, but otherwise no work is being done on an object being held in place.

An interesting thing about most electric motors is that they tend to have a range of rpm where maximum power output is almost constant and determined by the load involved, assuming a proper speed controller for the motor. If the load is larger, the motor will produce more torque at a lower rpm, consume more current at a lower voltage, and vice versa. This is commonly seen when choosing prop parameters on radio control models, which gets back to your original post about rc helicopters. Most electric rc helicopters will include recomendations for a specific motor, controller, and battery pack.

 

[Lsos]

Leafhill, you need to understand that if there is no motion, there in no need for energy, or power. Yeah you can use a motor to supply a force, you can use a propeller, you can use a rocket. Each of these will give you a different answer.

You can also use a spring. It will apply a force and use NO energy. Likewise, you can use a helicopter or a rocket to keep an object hovering. These can use megawatts of power. Or you can simply replace them with a table, or indeed, the ground itself. These apply the same force to counter gravity, they can apply it indefinitely, and they use no energy.

As for the motor against the spring example, all its energy is being wasted as heat. As was mentioned, it's essentially a heater. It produces no actual kinetic energy.

If the wires inside of it were superconducting, you could get the motor to supply the force against the spring without wasting any energy. You could unplug it from the outlet or from the battery and it would still apply its force, with the current flowing through it indefinitely. Only if some motion occurred would the current slow down, and eventually stop.

 

[collinsmark]

[From collinsmark: "Going to your motor example pushing the brick. If the motor was 100% efficient, and if the brick was not moving, it would require 0 Watts to keep the brick and spring in place, force or no force."]

Regardless of the efficiency of the motor, it still uses energy to hold the spring constant. It's just dissipated as heat instead of being converted to shaft power.
CS

I have do disagree. It may take energy to compress the spring initially, and that energy is stored as potential energy (not as heat). But once the spring is already compressed, it takes no work at all to maintain the force that keeps it compressed.

If the motor is 100% efficient (and the spring is an ideal spring), than nothing is converted to heat. Heat is not even involved.

Think of it this way. After the spring is initially compressed, replace the motor with a fixed clamp that maintains the constant force on the spring. If this final setup absorbed or released any energy per unit time, we would be able to use a fixed (unmoving) spring and clamp combination as air conditioner or a heater that didn't require any fuel or input power! We could use it as a power source for a perpetual motion machine. But unfortunately that is impossible. The truth is, assuming ideal efficiencies, that once the spring is compressed, it takes 0 Watts to keep it compressed.

 

[stewartcs]

I have do disagree. It may take energy to compress the spring initially, and that energy is stored as potential energy (not as heat). But once the spring is already compressed, it takes no work at all to maintain the force that keeps it compressed.

If the motor is 100% efficient (and the spring is an ideal spring), than nothing is converted to heat. Heat is not even involved.

Think of it this way. After the spring is initially compressed, replace the motor with a fixed clamp that maintains the constant force on the spring. If this final setup absorbed or released any energy per unit time, we would be able to use a fixed (unmoving) spring and clamp combination as air conditioner or a heater that didn't require any fuel or input power! We could use it as a power source for a perpetual motion machine. But unfortunately that is impossible. The truth is, assuming ideal efficiencies, that once the spring is compressed, it takes 0 Watts to keep it compressed.

Again, I'm not saying work is being done on the spring. I'm saying that the electric motor will consume energy with a locked shaft. The energy consumed is converted to heat and dissipated from the windings (which will eventual burn up with a locked shaft).

Take an electric motor, connect a spring to the shaft and then the anchor one end of the spring. Now apply a voltage to the terminals of the motor. What happens? Well, a current is induced in the armature due to the applied terminal voltage which results in electrical power being used (i.e. supplied from some source to the motor). I'm sure everyone here agrees that electrical motors use electrical power to work yes? Next, the output shaft will spin momentarily to stretch the spring and then stop. The shaft of the motor is now locked and is not moving, the spring is stretched (contains potential energy) and the current in the armature is still present (but a lot higher) as well as the voltage on the terminals...hence power is being supplied to the motor which means energy is being used (i.e. converted to heat in this case).

I've personally verified this with lab tests before and have actually seen the input power go up dramatically when the shaft became locked. It went up dramatically due to the armature current increasing dramatically which was a direct result of the locked shaft.

Again, as I've said all along, their is no work being done on the brick or spring or whatever object you want to put in its place. But the electric machine is still using energy (i.e. power) while holding the spring in place (i.e. locked shaft).

Now, if you turn off the power supply to the motor what happens? Well, the potential energy in the spring will cause the shaft to rotate in reverse until the spring reaches its natural length. What does this tell us? It tells us that the electric motor was applying a force to the spring to keep it stretch. How was it applying the force? Through the conversion of electrical power to torque on the shaft. Once the shaft became locked all of the energy is just converted to heat since motion is no longer involved. Since no motion is involved no mechanical work is being done on the spring.

If the resistance of the armature is zero or near zero (as with the case of a superconductor I imagine), the armature current will tend to infinity. The results in an indeterminate solution (i.e. infinite current). This is precisely why some shunt DC motors have starting resistors - to limit the starting current so as to avoid destroying the motor.

As far as replacing the motor with a fixed clamp goes...you're changing the entire problem and has nothing to do with what I was talking about. But yes, if you have a clamp instead of a motor I definitely agree that once the spring is compressed no electrical energy is being used...LOL!

CS

 

[rcgldr]

Again, I'm not saying work is being done on the spring. I'm saying that the electric motor will consume energy with a locked shaft. The energy consumed is converted to heat and dissipated from the windings (which will eventual burn up with a locked shaft).

Servo motors are desgined to be able to locked up without burning out. It's common for servos used with radio control models to have a much higher "holding" torque than moving torque. The rc helicopter models use four or five servo motors to move or hold various components in position, typcially three for collective + cyclic on the main rotor, and one for the collective on the tail rotor. If it's a fuel (versus electric) powered model, there's a fith servo for throttle control.

 

[collinsmark]

Again, I'm not saying work is being done on the spring. I'm saying that the electric motor will consume energy with a locked shaft. The energy consumed is converted to heat and dissipated from the windings (which will eventual burn up with a locked shaft).

Energy is only dissipated as heat in the motor, if the motor is less than 100% efficient.

If the motor is 100% efficient, then by definition, all energy input into the motor is converted to work (are you using a different definition of efficiency?). If such a motor does not do any work, it takes in 0 J of energy.

Take an electric motor, connect a spring to the shaft and then the anchor one end of the spring. Now apply a voltage to the terminals of the motor. What happens?

[Edit: What I mean here is what happens after the system reaches steady state. Or alternatively, if you just locked the shaft down directly.]

If the motor is 100% efficient, nothing. The terminals behave as an open circuit and no current flows.

This (hypothetical) 100% efficient motor will produce a torque however. Ignoring any momentary stress in the mechanics (which is reasonable to ignore since we're discussing things that are 100% ideal anyway) this torque is fully counteracted by the normal force of the locking mechanism on the shaft. Torque is applied, but no angular displacement -- meaning no work is done. Similarly, an electrical potential is applied, but no current is drawn, therefore no power is consumed.

[...]
The shaft of the motor is now locked and is not moving, the spring is stretched (contains potential energy) and the current in the armature is still present (but a lot higher) as well as the voltage on the terminals...hence power is being supplied to the motor which means energy is being used (i.e. converted to heat in this case).

Yes, this is quite true for motors operating at less than 100% efficiency. But if the motor's efficiency is at 100%, no electric power flows into the motor, even if a voltage is applied to the terminals.

I've personally verified this with lab tests before and have actually seen the input power go up dramatically when the shaft became locked. It went up dramatically due to the armature current increasing dramatically which was a direct result of the locked shaft.

Yes, so have I! Practical (real-world) DC motors especially have this characteristic (as opposed to other types of motors such as induction motors or synchronous motors)! practical DC motors produce produce their maximum torque at 0 RPMs. But if you look at a real-world DC motor's efficiency vs. RPM curve, you'll find that the efficiency drops to 0 at 0 RPM.

In other words, when you're running this (locked shaft) test you are testing with a motor that is operating at 0% efficiency.

Again, as I've said all along, their is no work being done on the brick or spring or whatever object you want to put in its place. But the electric machine is still using energy (i.e. power) while holding the spring in place (i.e. locked shaft).

Yes, but only true for motors having less than 100% efficiency. The same applies for human waiters and weightlifters.

 

[berkeman]

I think we should avoid trying to discuss 100% efficient electric motors. As you have just made clear, they are non-physical and not really of any help in tutoring newcomers in the physical concepts of work and power.

 

[collinsmark]

I think we should avoid trying to discuss 100% efficient electric motors. As you have just made clear, they are non-physical and not really of any help in tutoring newcomers in the physical concepts of work and power.

Yes, perhaps something in nature, that is 100% efficient, and everybody is accustomed to would make for a better example. Like gravity! :tongue2:

Imagine a block lying on the ground. Gravity exerts a force on it with a magnitude mg (where m is the mass of the block and g is the graviational acceleration (9.8 m/s₂). Now imagine that you attach the block to a very tall crane (using a rope of negligible weight). If you use the crane to raise the block higher, at a constant velocity v, the power required is

$$P = Fv = mgv,$$

ignoring the very brief intervals of acceleration.

So if you wanted to raise a 1 kg block up at 1 m/s, it takes

$$P = mgv = (1.0 \ kg)(9.8 \ m/s^2)(1.0 \ m/s) = 9.8 \ \mbox{Watts}$$

If you wanted to raise the block up at 2 m/s,

$$P = mgv = (1.0 \ kg)(9.8 \ m/s^2)(2.0 \ m/s) = 19.6 \ \mbox{Watts}$$

Now here is my key point: In both examples above, the force on the block is identical! It doesn't matter if you raise the block at a constant 1 m/s or 2 m/s or 50 m/s (ignoring the brief intervals of acceleration). The force is still 9.8 m/s². And if you raise the block up twice as fast, it requires twice the power (in Watts). But it get's there in 1/2 the time, so the total energy expenditure is identical.

At the risk of going back to the motors (:shy:), this is an important concept in automobiles. One of the specs used to described automobile engines is horsepower. Horsepower is a measure of power, just like Watts (watts and horsepower can be converted to one another by a simple constant, since they are both measures of the same thing). The maximum speed at which an automobile can climb a steep hill is a function of the automobile engine's horsepower. The more horsepower the car has, the faster it can get up very steep and very long hills (all else being the same such as the cars' mass).

Consider this, and it may illustrate the point. Consider a souped up Mazarati and a used 4-cylinder economy car, each climbing the same mountain hill, which is very long and very steep (and constant grade). The Mazarati climbs the hill at 60 mph while the economy car climbs the hill at 20 mph. Assuming both cars have equal mass, and ignoring air resistance and tire friction, both engines are producing the same force on the respective car. The mazarati's force is the same as the economy car's force. But the Mazarati uses more horsepower, so it gets up the hill faster.

 

[rcgldr]

My brick is not actually moving so distance is kind of irrelevant.

In your first post, it seemed the goal was to understand the motor used to drive the main rotor of a rc helicopter, so why are you concerned about the non moving case?

It would help here if you could better explain what the ulitmate concept that you're trying to understand in this thread. I think we've explained that without motion, there is no power output.

In the case of a radio control helicopter, if it's an aerobatic helicopter, most of the time it will be run in "idle up" mode were the rotor is maintained at a fairly fast and constant rate of rotation, with the primary change being the pitch of the rotor to create upforce or downforce (for inverted flight or areobatic maneuvers). The transmitter will have a throttle versus pitch curve mix in order to adjust the power input into the motor to keep the rotor moving at the same speed across the full range of pitch (usually -12 to +12 degrees).

 

[stewartcs]

Energy is only dissipated as heat in the motor, if the motor is less than 100% efficient.

If the motor is 100% efficient, then by definition, all energy input into the motor is converted to work (are you using a different definition of efficiency?). If such a motor does not do any work, it takes in 0 J of energy.

If we use a magical electric motor that uses magical conductors and magically has no other losses the motor would be about 100% efficient while operating. This means the armature resistance would be zero. If we take that very same magical electric motor and lock the shaft it will become 0% efficient.

Now assuming we are still applying the same terminal voltage to the motor when running or when it has a locked shaft what happens to the current in the armature? Where does the energy go that was being provided while running now that the shaft has been locked?

Torque is still being developed on the shaft which means the electric motor is being supplied with energy. Energy cannot be destroyed, so where does the energy go?

If the motor is 100% efficient, nothing. The terminals behave as an open circuit and no current flows.

The terminals would behave as short circuit. They are still connected but now have zero resistance. If it where open the resistance would be infinite.

This (hypothetical) 100% efficient motor will produce a torque however. Ignoring any momentary stress in the mechanics (which is reasonable to ignore since we're discussing things that are 100% ideal anyway) this torque is fully counteracted by the normal force of the locking mechanism on the shaft. Torque is applied, but no angular displacement -- meaning no work is done. Similarly, an electrical potential is applied, but no current is drawn, therefore no power is consumed.

Current is drawn in the armature as it always is drawn. The armature resistance is normally very low to begin with for practical electric motors (say 90% efficient). These 90% efficient machines still have very high locked rotor current flowing in them. To reduce the resistance even lower would result in even higher currents flowing. The circuit does not become open when lowering the resistance of the connect member.

Yes, this is quite true for motors operating at less than 100% efficiency. But if the motor's efficiency is at 100%, no electric power flows into the motor, even if a voltage is applied to the terminals.

What if the motor is 99.99999999999999999999999999999999999999999999999% efficient? Does it require power? You're argument involves an indeterminate form so your answer is nonsensical. Look at what happens mathematically as you approach zero resistance.

Yes, so have I! Practical (real-world) DC motors especially have this characteristic (as opposed to other types of motors such as induction motors or synchronous motors)! practical DC motors produce produce their maximum torque at 0 RPMs. But if you look at a real-world DC motor's efficiency vs. RPM curve, you'll find that the efficiency drops to 0 at 0 RPM.

In other words, when you're running this (locked shaft) test you are testing with a motor that is operating at 0% efficiency.

I agree that the efficiency drops to 0% when the shaft is locked as I have always agreed.

berkemen has a good point though, no machines in real life are 100% efficient so it seems a bit pointless to me to continue bringing up magical electric motors. All that I'm personally concerned with are the real motors that I deal with and how they respond. Discussions such as these have no value in the practical world for engineers. Interesting in academia though.

CS