# electrodynamometer

Discussion in 'Mathematics and Physics' started by PG1995, Oct 4, 2012.

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Hi

Regards
PG

2. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

The first looks like a typo. That is not the RMS current, and they even made the mistake of writing Erms instead of Irms.

When the coils are arranged as in (b), the current that flows through L1 and L2 is the same as that flowing through L3.
L3 is not a permanent magnet, but it's field polarity changes the same way that the other two coils change. Thus the net field orientations are always the same regardless of the polarity of the current. If L3 was a permanent magnet, then it would move back and forth, but because it is also changing polarity it keeps the same deflection direction.

For one direction of the current we have three magnets:
SN <-- SN <-- NS

for the other direction we have:
NS <-- NS <-- SN

So one coil pulls and the other coil pushes no matter what the polarity is.

Last edited: Oct 4, 2012
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3. ### PG1995Active Member

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Thank you, MrAl.

Could you please help me with the query related to the Q2 from my previous post above? Thanks.

Regards
PG

Last edited: Oct 5, 2012

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG,

There are two basic mechanisms here.
First is the sine wave drive sin(wt) turns into abs(sin(wt)) because of the coils being wound in series and we talked about this last time.
The second is that the mass of the moving coil integrates this abs(sin(wt)) as it tries to move the mass back and forth.

The abs signal is trying to move the coil back and forth rapidly, and if the coil had zero mass it would actually move from zero to maximum and back again as the abs(sin(wt)) kept changing. But the rotational inertia acts as an integration over time so that we end up with the average angle instead of constantly seeing the actual angle going from 0 to max and back.

So the equation would look something like:
(1/T)*Integral[0 to T](|sin(wt)|) dt=0.6366

We could look at this deeper where we would find that the rotational inertia of the mass acts to average out the changes, but since the changes are always positive we see a net average show up in the deflection of the needle.

Last edited: Oct 5, 2012
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6. ### steveBWell-Known MemberMost Helpful Member

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PG,

You may find this document helpful.

http://www.electro-tech-online.com/custompdfs/2012/10/14188_ch3.pdf

I have a question for MrAl also. You mentioned the torque being an averaging of abs(sin), but I wonder if it is actually an averaging of sin^2 since the meter seems to be a square law effect. I haven't taken the time to analyze this in detail yet, but I figured I'd run this thought by you first, before spending time on it.

If true, the squaring will double the frequency and make the averaging effect even better. There is also the damping effect which is discussed. Hence, not only is the mass unwilling to move at high frequency, but it seems there should be higher electromagnetic damping for any high frequency or fast movement.

Last edited: Oct 5, 2012
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7. ### PG1995Active Member

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Thank you, MrAl, Steve.

@Steve: Thanks for the PDF. I have skimmed through it and saved it to my hard disk.

I'm happy my question was able to make its way and was understandable this time.

Once again, I'm only considering positive half cycle of a sine wave. So, in a way we can say that there are two opposing forces at work - one is trying to move the pointer back and forth ( i.e. abs(sin(wt) ), and the other is rotational momentum which tends to offset this back and forth tendency, and the result of these two opposing effects is that the pointer stays at some 'average' position which lies somewhere between maximum deflection position which the pointer will show for peak value of the sine wave if it were massless and zero deflection position which the pointer will indicate for zero value of sine wave. Here, we can also include a third force of a tiny torsion spring which always tries to pull the pointer back to the zero position.

Now the problem is that how we can mathematize the above situation to reach the value for that 'average' deflection position of the pointer. I don't understand MrAl's equation, "(1/T)*Integral[0 to T](|sin(wt)|) dt=0.6366". The part of the reason for this is that I'm not good at math and secondly I don't see this how he reached this final equation. I have also attached another document about electrodynamometer taken from another book; it didn't help me much but perhaps you can find it useful to help me.

By the way, does persistence of vision have anything to with this? I hope I'm not saying something too silly!

And Steve mentioned square law effect and that book excerpt from my first post also mentions it. I know what inverse square law is but have never heard of this 'square law effect'.

I don't know if I understand correctly what you are saying but I thought I should mention this. In the "another document" it is said that no core is used for moving coil in an electrodynamometer so that there are no hysteresis and eddy current losses. And as is mentioned in this document from my previous post that damping effect is due to eddy currents in the aluminum core used for moving coil in a galvanometer.

Regards
PG

Last edited: Oct 6, 2012
8. ### steveBWell-Known MemberMost Helpful Member

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Before going too far, we should wait for MrAl's answer about whether there is a abs(sin) or sin^2 effect that causes the meter to move in one direction only.

However, I can make two comments here beforehand.

First, the square law effect is simply that the measurement response (in this case angle deflection) is proportional to the square of the measurement (in this case current). This is different from a D'Arsonval meter movement where deflection angle is proportional to current.

Second, as far as damping, if there is no significant metal surrounding the field from the moving piece, then you are correct that there will be no electromagnetic damping. However, not having damping is not good for a meter because the needle movement will "ping" and can get damaged by sudden changes in current. Hence, perhaps mechanical damping is used. Generally, "damping" is any force/torque that is proportional to speed, and damping is usually energy loss through heating. Hence, damping of any type would prevent any high amplitude and high frequency oscillations on the needle movement.

Note that, even though viscous damping (as I just described) is helpful, any type of static friction is bad because then the needle can get stuck when there are small changes in current.

Also, you mentioned you are trying to understand the math of this, but I recommend you first try to understand intuitively why a large mass can filter a high frequency force input. Imagine holding a spring in your hand at the top, and attached to the bottom of the spring is a heavy mass. Now imagine moving your hand up and down very very fast. The mass will not move any where near as much as your hand. However, if you move your hand up and down slowly, the mass will almost exactly follow your hand. If you can't visualize this, then go to the hardware store and buy a spring and try it experimentally.

The issue of damping is different and you can imagine trying to do the above experiment with the mass sitting in a container of molasses. Now even slow movement of the mass is difficult because the damping (viscosity of the molasses) prevents motion. Now, only very very slow oscillations from your hand will cause the mass to move.

Last edited: Oct 6, 2012
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9. ### MrAlWell-Known MemberMost Helpful Member

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Hi again Steve and PG,

Steve:
Squaring or taking the absolute value will result in higher frequency. But i think the increase in frequency is much less important than everything else we've been talking about because after all the lowest frequency component only doubles, and if it didnt we'd just increase the mass to make up for it.

Steve,PG:
Yes for the moving coil movement we have a squaring law. That come up when there are two magnetic forces acting together instead of just one. If we had a coil and magnet this would not be the case, because then the 'current' in the magnet would be like another coil with constant current, so we would not see the squaring effect. But with two coils acting together (one moving and the other 'two' considered to be one coil) we see the squaring effect I^2, or more specifically, the product of the ampere turns of coil 1 and the ampere turns of coil 2. So for two equal turns coils the force is squared.

The rotational spring (the spring coil) acts with a force in the opposite direction, thus opposing the I^2 force. The needle reaches a steady state when the two forces are equal:
Ipk^2/2=d*K
where
d is the distance moved and K is the spring constant.

Now since we can transform this rotational system to a linear one we can think of this as a mass being pushed forward by a squared force and held back by a spring. So what we have here is a basic spring mass system, and we'll get some damping by the meter movement itself.
So solving for the distance d we get:
d=Ipk^2/(2*K)

so we see the distance is proportional to the squared peak of the current. This means that the face of the instrument needs to be calibrated.

To calculate the value Ipk^2/2 i simply took the average of Ipk*sin(wt) which is the familiar form:
Iavg=(1/T)*Integral[0 to T](Ipk*sin(wt))^2 dt

This is the short way. The longer more involved way is to evaluate the spring mass damper system with a squared force excitation.

Now there seems to be some discrepancy in some of the references. Some say that the basic movement is measuring the RMS current. They make it sound like the movement of the needle (either angular in the rotational system or linear in the translational system) is proportional to RMS current. But others insist that the meter face has to be calibrated (ie make it so that whole numbers appear as a curve so as to have a non linear scale). What i have seen here is that the latter seems to be the true case, as i cant find a mechanism to convert from Ipk^2/2 to the required Ipk/sqrt(2) if it was going to be a linear scale (and linear meter face).
I think what is happening here is that once the meter face is calibrated, then it indicates RMS. After all if we calibrate the meter face as a square root function, we'll see the exact RMS value appear on the meter face.

Last edited: Oct 6, 2012
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10. ### PG1995Active Member

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Thank you very much, Steve, MrAl.

I'm sorry but I'm still trying to understand this. Please have a look here and kindly help me. Thanks.

Regards
PG

PS: I have realized there is a mistake in the attached diagram. I have numbered two different stages as "5". The end stage should be numbered "7", not "6".

Last edited: Oct 7, 2012
11. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

When you look at this you have to look at it as a system not as separate processes doing separate things.

The momentum of the mass will only push the needle beyond the rest position a tiny bit. The needle only moves a tiny amount for each half cycle anyway not anywhere near the whole distance.

The simplest way to look at this is using a low pass filter to filter a rectified AC input voltage. The abs(Vac) turns into humps of sine waves only going positive. The capacitor in the low pass filter smooths out the humps so what we end up with is a fairly smooth DC value with some ripple. The ripple would be similar to the tiny back and forth movement of the needle, but depending on the mass (which the capacitor represents) it may not move enough to see with the naked eye. This doesnt mean the needle is moving back and forth so rapidly that we cant see it because of persistence of vision, it just means the needle doesnt move much at all. If the needle did move back and forth your eyes would see it as a wider needle that was a little blurry if the system could not damp it enough.
It is possible that you could see the needle move back and forth a tiny amount but you'd have to look close.
If you decrease the frequency enough you'll see the needle follow the sine humps exactly, up and down quite a bit. But if you increased the mass again you'd see it come to a rest point where it just moves a tiny tiny amount. That's because when you try to accelerate a mass you find that you can not do it in zero time, but it takes some finite amount of time to actually get it to change position. So when the humps come they try to accelerate the mass but have to work not only against the spring but also against the inertia of the mass, and that inertia introduces a sort of time delay.
If we apply a voltage all of a sudden to the meter we have to deal with the initial change in excitation first. That could cause an initial overshoot, but once it starts to settle it's going to move very little if the meter is designed right. If we didnt have any mass the needle would swing back and forth from zero to max and back again and we'd see the needle as a blur across the face.

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12. ### steveBWell-Known MemberMost Helpful Member

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PG,

In principle the needle will move, but the amplitude of the motion is very tiny, and it is very fast. You can't see the movement, and it is a small amount.

Let's do an example. If damping is B=2, and spring constant is K=1, with mass M=1, you get the following transfer function. Note that I chose the parameters to give a critically damped second order system.

Differential Equation:

Take Laplace transform and find transfer function leads to:

Plug this into Matlab using the command sys=tf(1,[1 2 1]) and use the bode command to get the frequency response. You can do this yourself, but I attached the plot.

Now look at the response at high frequency and see how the response rolls off. This is a low-pass filter. If the magnitude response is 80 dB down from the DC response, this is 4 orders of magnitude. Hence if the needle moves 1 cm due to the DC part of the force (average value) then the sinusoidal variation in the needle movement might be less than 1 μm.

Now, can you see 1 um motion even if it's slow? Or, can you see 60 Hz variations in general?

Also, consider that other nonlinear effects, such as static friction, might even make this small motion impossible in reality.

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13. ### PG1995Active Member

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Hi

Please have a look on the attachment. Thanks a lot.

Regards
PG

14. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG,

Your system that you drew (good drawing BTW, i always like to see that) is only one of three possible classes of systems of that basic nature.

Unfortunately, the one you drew is not the equivalent to the meter movement. That's mainly because you did not include any viscous damping.

Viscous damping is the thing that prevents oscillatory systems from maintaining an oscillation. It comes in several forms. For the mechanical system, it happens to be the sliding friction. For the rocket and cart system, this can be either friction in the wheels or even just a piece of felt that rubs up against the side of the cart (assuming a small rocket). That is what keeps the oscillations down. In the meter we are discussing, i think it is air vanes being used to damp the oscillations, but there's also going to be some sliding and sticking friction.

Also, you have to consider the amount of force being used to drive the system. A rocket is going to be a lot more powerful than the force from a small coil of wire with some light current running through it.

Surely you've seen an unregulated DC power supply using just rectifiers and a capacitor. The drive here is a full wave rectified sine, yet the output has only some change to it and is mostly DC.

http://en.wikipedia.org/wiki/Damping

Last edited: Oct 10, 2012
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15. ### JimBSuper ModeratorMost Helpful Member

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I dont know if you have access to test equipment, if you do try this:
Connect the output of a function generator to a moving coil meter.
Set the function generator to a low frequency (1hz) and adjust the amplitude so that the meter pointer is moving a few divisions about zero.

You will see that the meter follows the FG output.
Now increase the frequency, the pointer will move faster and the amplitude of the swing will reduce.
As the frequency rises to around 50/50hz the pointer will be hardly moving and jiggling quite quickly (if at all, depends on the mechanics of the meter).
Now increase the frequency to a few hundred hz and the pointer will just sit motionless at its zero position.

JimB

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16. ### PG1995Active Member

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Thank you, MrAl.

@Jim: I like the experiment you suggested. I will perform it if I'm able to get the apparatus. Thank you.

The system I showed in the diagram is just there to convey the point. For instance, you can replace the rocket with something else if in your view it will provide too much thrust. Let's include damping force that could be, as MrAl say, the friction between the wheels of the trolley and horizontal surface shown. I don't think this will help me to solve my confusion any better.

When there is no friction force, here when the force (or, rocket's thrust) changes from "O" to "A", it takes the trolley 10x distance toward right of the screen. And as the force changes from "A" to "B" the trolley will move back toward left of the screen because the on average spring's force is greater in magnitude than the thrust produced by the rocket.

Now when there is friction present. When the force changes from "O" to "A", the trolley will move less distance toward the right of screen, say 5x. But as the thrust changes from "A" to "B", the friction stays the same therefore the spring will pull the trolley back to the same position which it will occupy even when there were no friction.

Regards
PG

Last edited: Oct 10, 2012
17. ### steveBWell-Known MemberMost Helpful Member

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One important point is that friction is not the same. Friction is proportional to speed. For sinusoidal variations, speed is proportional to frequency and it is also proportional to displacement amplitude. The actual friction force is time dependent and varies during the cycle. This means that if the motion did not reduce in amplitude as the frequency went up, the friction force would get very large at high frequency. This large friction can be seen as one important effect that reduces amplitude.

There is another experiment you can try in a swimming pool. Try to move your hand back and forth under water. Make large sweeps with your hand going slowly. Then make small motions going back and forth fast. Then try to move your hand at high frequency and large amplitude. You can probably do the first two things easily and the third will take a very great effort that leaves you out of breath very quickly.

Also, don't forget the effect of mass on acceleration for a given force. How far can a big heavy mass move when the force goes from 0 to A at a very high frequency. Before the mass has moved a tiny amount, the force is already cycling back to the negative side of the cycle.

Last edited: Oct 10, 2012
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18. ### PG1995Active Member

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Thank you, Steve.

Are you referring this formula, v=ω*r*cos(ωt), for simple harmonic motion?

Okay. When the rocket is turned on, the trolley moves toward right and stays there afterwards. Let's say the trolley 2 feet. Could you please explain how things will go on between "O" --> "A" --> "B"? Why would the trolley move a distance of 2 feet in the first place?

Regards
PG

Last edited: Oct 10, 2012
19. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

I'll see if i can get a schematic of this problem drawn up at some point. The thing is, it's not entirely simple because there's various feedback mechanisms. There's even a small "back emf" when the coil is moving, and that would cause a decrease in the apparent force too, but only while it is moving. There's also inductance, resistance, and of course friction and rotational inertia to consider. So the system would look like a sort of control system with feedback from the speed of the shaft and feedback from the position itself. So it would look like we had a control system with little sensors that sensed the speed of the shaft and it's position, and fed these signals back into the system and of course that greatly alters the response (the response we are looking for is position but we might also look at angular speed).
Within this system there will be several integrations, one for the inductance, one for the rotational inertia, and one for the translation from angular speed to position, so it's not as simple as a system that is completely feed forward like "push here and see the movement there".

Hopefully i'll be able to get this drawn up tomorrow some time so you can take a good look at what we are dealing with here. The final equation would allow use to analyze the movement of the needle for any kind of input, which could include inputs that actually do cause an unstable response, or even a very bad reading (too much error) over a short time period.

20. ### steveBWell-Known MemberMost Helpful Member

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Basically yes, but let's write it out, as follows.

In steady state, the position will vary sinusoidally as

Here
is the constant offset in position, analogous to a DC offset in voltage in an electrical system.

Hence, speed is a derivative of position which results in the following

Here
is the amplitude and
is the frequency. Hence, since friction force is proportional to speed, it is also proportional to positional amplitude and positional frequency.

The 2 feet distance is the DC part of the solution, as described above. There is an average force from the rocket that must be counterbalanced by the average spring force. The DC component has no friction force since friction is proportional to frequency and frequency is zero for the DC part of the solution.

The AC part of the solution operates around the DC value (in this case 2 feet). The combination of relatively high frequency with relatively high mass will create a condition where the sinusoidally varying force can not move the trolly very far as the force goes from 0 to A, and then from A back to B. Then, the negative part of the cycle creates force in the other direction before the trolly has had any time to move far. This large mass is not able to instantly stop and turn around, but it will decelerate due to the negative force. So, there is a lag between force and motion and the amplitude must be small. Also remember, that (with high frequency) the friction force would be come very large if the amplitude of motion was not small. Hence, only a very very high force could force large motion.

Now, is it possible to apply a small DC force and a very large AC sinusoidal force? Yes, of course, and the result could be large motion for the trolly. However, the meter example has a rectified function that is forced to give AC force less than the average DC force. So, there is no way to get large motion.

It's important to be able to visualize all this without resorting to math. Then the math can give you numerical answers as needed.

By the way, every engineer needs to master first order and second order type systems and know them very well. First order systems are quite simple, but second order systems can have characteristically different behaviors (underdamped, overdamped and critically damped). These must be mastered as you move forward in your studies.

Last edited: Oct 10, 2012
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21. ### PG1995Active Member

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Thanks a lot.

It makes some sense to me now. But there still exists a big loophole in my understanding and good thing is that you have already stated it in your equation so I don't have to explain it in detail. In short, where does the
come from? That's the component which makes the trolley move 2 feet toward the right and
is the component which is responsible for very small oscillations around the mean position which are hard to discern. Why would the trolley move 2 feet in the first place? Why doesn't it simply stay at zero feet position and oscillate there? Thank you for the help.

Best wishes
PG