electrodynamometer

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PG1995 said:
Why doesn't it simply stay at zero feet position and oscillate there? Thank you for the help.

PG
Click to expand...

If there is no DC component to the force, then it will stay at x=0, but your question relates to a rectified force which has a DC component (i.e. average value not zero). If you like, the force function looks like.

[latex] F(t)=F_{DC}+F_0\;\cos{\omega t}[/latex]
 

Thank you.

Is there really any DC component of force **broken link removed**? I was thinking that if there was a DC component then it would look like **broken link removed**. Please help me with this. Thanks.

Regards
PG
 
PG1995 said:
Thank you.

Is there really any DC component of force **broken link removed**? I was thinking that if there was a DC component then it would look like **broken link removed**. Please help me with this. Thanks.

Regards
PG
Click to expand...

There is a DC component in both cases. The DC component is simply the average value over a period. For any sinewave (or any symmetrical periodic waveform), the average value is the maximum value plus the minimum value divided by 2. The function sin(t) has 0 for the DC or average value. The max is +1 and the min is -1, and they add to zero.

The average value of sin^2(t) is 1/2 because the maximum is 1 and the minimum is 0. They add to zero and then divide by 2 to get 1/2. Or you can use the trig identity, as follows,

[latex] \sin^2t=\frac{1}{2}-\frac{\cos 2t}{2}[/latex]

The above makes the 1/2 average value more obvious.

Note that your rectified sine wave (abs(sin(t)) is not symmetrical, so calculation of average value requires doing out the integral in full. This result is well known and is about 63.7 % of the sine wave amplitude.
 
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Hello again,


It's very hard to look at some systems and tell what's going on without rolling your sleeves up and doing some down to earth analysis. Here is a block diagram and waveforms to show what is happening in the system.

Note how the shaft speed starts out fast, then settles down, then moves up and down very slightly.
Note how the angular position jumps up but then settles down to a nearly movement free motion.

We could look at other values of rotational inertia and other things. These waveforms were made with a particular value of inertia, and lowering the inertia means the position jumps up farther before it settles down. Increasing the inertia means it jumps up less before settling, and the speed does not increase as much before it drops to zero. This makes sense because it's like pushing on the back of your car...it hardly moves at first.

In the diagram, Ks is the rotational spring constant, J is the rotational inertia, F is the rotational friction, R is the wire resistance.
 
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Thank you, Steve, MrAl.


I think you misunderstood what I said in my previous post. I was asking about the DC component of force from different point of view. Let's see if I can put it any clearer. I think I also said this in once of my previous posts. In circuit analysis, RMS values of voltage or current are used and these values are equivalent of DC values. In other words, AC values are replaced with constant values. But in actuality in the circuit there is still AC, no RMS. Likewise, **broken link removed** you can see the force is constantly varying between "0 N" and the peak "+N". You can replace this constantly changing force with some average for your analysis but then you can't say in the system you have both DC force and non-constant force. **broken link removed** in my previous post, I will trying to emphasize this point. You have both components in that diagram: DC and alternating components. If this is confusing, then please let me know. I will try to rephrase it. But if you think you understand me now, then I would suggest you have a look on this post again. Thank you.

Regards
PG
 
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PG,

RMS is not the same as DC component. These are completely different concepts. DC component is an integral which is an average value. RMS is a square root of the average of the square of the signal.

If you are confusing these two different things, then it's no wonder you have been having trouble understanding this.
 
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Thank you for the reply, Steve.

No, I think I have an adequate understanding of the difference between RMS and DC values. But I think what I said previously is not in line because we have full wave rectified force (if there really exist such a thing! ).

For an AC waveform, the RMS or effective value of voltage or current is defined as steady, direct value of current or voltage which converts electrical energy to other forms at the same rate as the alternating current or voltage. For instance, electric supply to a household in the UK is quoted as 240 V and this is RMS value, not the peak value which is around 340 V. **broken link removed** you can see voltage values used around the world. Therefore, if a lamp is connected to the mains of 240 V or a battery with 240 V, you will have same power dissipation. For the difference between RMS and average values, you can look **broken link removed**.

So, probably I'm using wrong terminology or wording but my confusion is legitimate in this context. If you ask me for the DC value of force **broken link removed**, then I will multiply the peak value of force with 0.6366 to get average or DC value of force.

Best regards
PG
 
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PG1995 said:
If you ask me for the DC value of force **broken link removed**, then I will multiply the peak value of force with 0.6366 to get average or DC value of force.
Click to expand...

OK, so i think I've totally lost track of what you are trying to say to be honest, but let's just stop and take this one statement I've quoted from you above.

This is not correct. You can't just blindly take any waveform and multiply peak times 0.6366. The waveform you showed is a sine squared function not a rectified sine wave. A rectified sine wave does indeed have a 63.7 % value, and if you drive the system with abs(sin(wt)) then this is the correct DC component to use. However, sine squared has a 50 % DC average value.

However, this is just a detail. Whether we have 50% or 63.7%, there is a large average value of force that must be compensated for by a large average value from the spring force.

So, let's start fresh from here. Are you able to see why the needle does not oscillate around the zero point, but must oscillate around the DC average point?

Note that when we break a system into the sum of AC and DC components, we are using superposition which implies a linear system. A real system may or may not be well approximated as a linear system, however, assuming a linear system is a valid way for you to get the basic concepts down.
 
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Thanks once again.

I need to think everything over before asking anything again. By the way, I have been using absolute value, |A*sin(t)| in all my posts. Here, when you said, "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.", then MrAl implied that any of the two values, squared or absolute, can be used. This made me decide to use rectified values because it seemed simple to me.

Regards
PG
 
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Hello again,


Just to make this clear, the part about using the abs(x) or x^2, the meter uses x^2 and i said you can use abs(x) but ONLY to see how the DC component comes about. abs(x) would not give the right scaling while x^2 will.

If you study the diagram i provided you'll understand this thing just about as good as you'll ever need too, because that includes all of the important aspects of the meter movements, including spring, mass, etc. To understand fully what is going on inside the meter you'll have to understand that diagram or one similar. I posted it because your questions are detailed but you seem to be not looking at all the details required to understand this correctly. The method is in the details, where we see how all of the parts work together and that makes it more clear how the 'average' needle position comes about.
In short both abs(x) and x^2 both have a DC component, and that DC component keeps the average force on the needle in one direction even though the needle might move back and forth slightly. The mass kills some of the force too because in order to move the mass the force has to overcome it's inertia, and the small force ABOVE and BELOW the average DC can not move the mass very quickly once the force nears equilibrium with the spring, so by the time the small force is over the mass hasnt moved much (either direction). Initially however the force is large compared to the reverse force of the spring, but as the needle moves out farther then the force becomes closer to the spring force and it stops moving.
Once it stops moving with only a small force above and below that of the spring's, the small force can not move the mass much.
 

Hi

I think in the experiment quoted above JimB wanted me to use AC signal and he was saying that if frequency of AC signal is too high then the needle of the meter won't move at all from its zero position.

I just wanted to know one thing that would this also happen for **broken link removed** when the frequency of 'rectified force' becomes too high? I mean to say that would the trolley too just like the needle in JimB's experiment just stay at the zero displacement position if the frequency of rectified force becomes so high? I think what JimB says only applies when you have an AC signal, and this isn't true when you have rectified signal or force. Please let me know so that I can proceed. Thank you for the help.

Regards
PG
 
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Once again, the needle does not move around zero when you have a rectified signal or force.
 
steveB said:
Once again, the needle does not move around zero when you have a rectified signal or force.
Click to expand...

Thank you. Yes, you are right. But I'm asking for something different.


If it's not clear, then please let me know. Perhaps, I can put it differently. Thank you.

Regards
PG
 
Hi,

If the frequency of a sine increases we start to see other effects start to dominate like the inductance, but with a rectified or squared sine there's always some DC present and that DC gets through anything other than a capacitance, and there's no series capacitance so it always gets through.
The average DC for a full wave rectified sine is 0.6366, and for a sine squared it is 0.5 (times the peak value).

Picture this:
You connect a battery to the device (pure DC) through a small value resistor. You necessarily get a constant deflection.
Now take a noise source who's average is zero and connect that so the noise rides on the DC value (with another resistor and a capacitor). What is going to happen. If the lowest noise frequency is 50Hz the needle wont move much, and if it's faster it will move even less.
You can even do this with a sine wave. Have the sine wave (with real average zero) riding on the DC level. If the sine frequency is high enough (50Hz or higher) the needle wont move much.
 
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PG1995 said:
If it's not clear, then please let me know. Perhaps, I can put it differently. Thank you.
Click to expand...

I guess you need to clarify.

It seems you are asking if the trolly moves around zero position just because the frequency gets very high. I've clearly said that no it does not move around zero at high frequency. The act of rectification creates a DC offset that does not go away as the frequency goes up. Higher frequency reduces the AC response only, so the trolly does not wiggle as much at higher frequency, but it does not spontaneously move back to the zero position.
 
Thank you, MrAl, Steve.


No, I did not intend to say that it will move around zero position. I think I phrased it poorly. Okay. Forget about JimB's experiment. Suppose, when the frequency of rectified force is 50 Hz, the **broken link removed** (please note that in the diagram it's written that the surface is frictionless but disregard it; it's not frictionless) moves 2 feet to the right of screen. It just oscillates a little bit there around a mean position but overall its displacement is 2 feet.

Now we reduce the frequency to 25 Hz while keeping the amplitude constant. What would happen? Would the displacement still be 2 feet? Here I can guess one thing that now amplitude of oscillations around some mean position has increased.

Now we increase the frequency to 100 Hz without changing the amplitude. What would happen now? Would the displacement still be 2 feet?

Thank you.

Regards
PG
 
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The average displacement will be 2 feet for every frequency. The amplitude of the variation around that average displacement point of 2 feet will be a large value at low frequency and a small value at high frequency.
 
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steveB said:
The average displacement will be 2 feet for every frequency. The amplitude of the variation around that average displacement point of 2 feet will be a large value at low frequency and a small value at high frequency.
Click to expand...

Thank you for letting me know this. Now I can proceed.

Best wishes
PG
 

goddamn u know everything

u live for this ****
 
lebevti said:
[Expletives overtyped]

god**** u know everything

u live for this ****
Click to expand...

Ha ha i had to laugh at that one I wish i did know everything
Not sure if that language is a good idea though, did you check the forum rules?
 
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