Help Identifying Diagrams?

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Need some help. I studied math a million years ago and my boss thinks I still remember anything about diff EQ besides my professor's haircut. I don't.

I have these diagrams (http://www.dubberly.com/stage/Maxwell2.jpg [Broken] and attached).
I need to label them with appropriate formulas.

I have a whole pile of indecipherable stuff from a book "On Governors" about steam engine governors written by J. Clerk Maxwell in 1868. I know something in this pile is an appropriate label for the diagrams, but it would take me weeks to even hazard a guess.

I think that these diagrams have some canonical differential equation that's applicable. (Like y=1/x is one canonical formula for the Negative Feedback diagram, but it's not a differential equation.)

Does this make any sense? Can anyone help me with equations that would be appropriate for these diagrams? (I'm going to be showing it to some Stanford students, so it needs to be, you know, pretty accurate.)

Thanks so much in advance...
 

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Bigger diagram...

...attached for your viewing pleasure.
 
Tom Mattson
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audcrane said:
Need some help. I studied math a million years ago and my boss thinks I still remember anything about diff EQ besides my professor's haircut. I don't.
:rofl:

Does this make any sense? Can anyone help me with equations that would be appropriate for these diagrams? (I'm going to be showing it to some Stanford students, so it needs to be, you know, pretty accurate.)
Well, I don't know the work by Maxwell, but it seems clear enough that you have a sinusoidal function that is modulated by an exponential function. If you want to try to fit the curves to analytical functions then it seems to me that you will need to know the coordinates of some of the points on the graphs. Can you put some kind of scale on the axes so that we may determine coordinates?
 
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Tom Mattson said:
:rofl:
Can you put some kind of scale on the axes so that we may determine coordinates?
The scale can be anything we want it to be. The diagrams were created by a graphic designer based on text descriptions in the document ("oscillation of continually increasing amplitude", etc.). Now that I look at it, the diagrams are somewhat suspect where T=0, that should probably be an inflection point, not just randomly stop.

At any rate, the diagrams are just a hint based on the copy, and can (and will) be changed if appropriate to match the equation. If that makes any sense?
 
Tom Mattson
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It's a bit difficult to see from your diagrams what's going on at T=0. The graph marked "Positive Feedback" asymptotic to the T-axis as T goes to negative infinity? And does the graph marked "Negative Feedback" cross the v-axis at T=0?
 
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Well, here is what Maxwell says:

It will be seen that the motion of a machine with its governor consists in general of a uniform motion, combined with a disturbance which may be expressed as the sum of several component motions. These components may be of four different kinds :—
1. The disturbance may continually increase.
2. It may continually diminish.
3. It may be an oscillation of continually increasing amplitude.
4. It may be an oscillation of continually decreasing amplitude.


The diagrams are a graphic designer's interpretation of that description, so they can, and should, be modified where they're impossible or inaccurate.

Based on Maxwell's description, I think that in the leftmost diagrams, the curve has to touch V (given that these are real applications and that they won't start at infinity when the steam engine starts). Similarly, I think that in the rightmost diagrams, the curve has to touch V. In the case of the lower right diagram, it seems that it should curve to touch V (to be a maximum, not an infleciton point, sorry).
 
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OK, that helps, but it certainly does not uniquely determine the functions that we need to use.

audcrane said:
1. The disturbance may continually increase.
There are an infinite number of monotonically increasing functions, so we have to pick one. An easy one to use would be:

[tex]f(x)=Ae^{k(t-t_0)}[/tex]

[itex]A[/itex] is a sort of "amplitude" for the exponential function and [itex]k[/itex] controls the rate of growth. The [itex]t_0[/itex] is there to shift the graph of the more "basic" function [itex]g(x)=Aexp(kt)[/itex] along the [itex]t[/itex]-axis. If [itex]t_0>0[/itex] then the graph is shifted to the right.

2. It may continually diminish.
You can use an exponential again here, but this time you want the exponent to be negative for [itex]t-t_0>0[/itex]. Hence:

[tex]f(x)=Ae^{-k(t-t_0)}[/tex].

3. It may be an oscillation of continually increasing amplitude.
Multiply an increasing exponential and a sinusoid.

[tex]f(x)=Ae^{k(t-t_0)}sin(\omega t+\phi)[/tex],

where [itex]\omega[/itex] is the angular frequency, which controls the rate of oscillation, and [itex]\phi[/itex] is the phase angle, which controls where on the [itex]t[/itex]-axis the sine wave "starts".

4. It may be an oscillation of continually decreasing amplitude.[/I]
Similarly:

[tex]f(x)=Ae^{-k(t-t_0)}sin(\omega t+\phi)[/tex]
 
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Ah-hah!

In this paper, I have a description of Jenkin's Governor (forgive me, I'm trying to figure out this formatting thing)...

A centrigual piece revolves on the principal axis, and is kept always at a constant angle by an appendage which slides on the edge of a loose wheel, B, which works on the same axis. [snip] F is a new constant (I assume force). [snip] V is velocity. [snip] Y is a coefficient depending on ... resistances varying with the velocity.

The solution is of the form

[tex]
x=A_{1}\e^{n_1 t} + A_{2}\e^{n_2 t} + A_{3}\e^{n_3 t} + Vt
[/tex]

Where [tex]n_1, n_2, n_3[/tex] are the roots of the cubic equation

[tex]
MBn^3 + (MY + FB)n^2 + FYn + FG = 0.
[/tex]

If n be a pair of roots of this equation of the form [tex]a \pm \sqrt{-1}b[/tex], then the part of x corresponding to these roots will be of the form

[tex]
e^{at}cos (bt + \beta).
[/tex]

If a is a negative quantity, this will indicate an oscillation the amplitude of which continually decreases. If a is zero, the amplitude will remain constant, and if a is positive, the aplitude will continually increase.
So perhaps that would be the label for the two diagrams on the right (although I would need to define a, b, and beta I'm thinking, yes)?

I'll look through the paper for something that looks similar to your suggestions for the leftmost diagrams.

Tom, I wish you lived on the left coast! I owe you a beer!
 
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Tom Mattson
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Regarding [itex]e^{at}cos (bt + \beta)[/itex].

audcrane said:
So perhaps that would be the label for the two diagrams on the right (although I would need to define a, b, and beta I'm thinking, yes)?
Yes, you would. In fact that function is exactly equivalent to the third function I posted.
 
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I have poured over this paper again and cannot find anything similar to your first equation that might work for the two leftmost diagrams. However, immediately following what I quote above, I find this:


One root of the [cubic equation above] is evidently a real negative quantity. The condition that the real part of the other roots should be negative is

[tex]\left(\frac{F}{M} + \frac{Y}{B}\right)\frac{Y}{B}-\frac{G}{B}>0.[/tex] (14)

This is the condition of stability of the motion.

...So if I add the condition that (14) is true, then the same equation applies to the leftmost diagrams as well? And probably also translates to your first equation, I bet!
 
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